YouTube Transcript:
Calculus 1 - Full College Course

This video is about working with rational expressions. A rational expression is a fraction,
usually with variables in it, something like x plus two over x squared minus three is a
rational expression. In this video, we'll practice adding, subtracting, multiplying
and dividing rational expressions, and simplifying them to lowest terms. We'll start with simplifying
to lowest terms. Recall, if you have a fraction with just numbers in it something like 21
over 45, we can reduce it to lowest terms by factoring the numerator and factoring the
denominator and then canceling common factors. So in this example, the three is cancel, and
our fraction reduces to seven over 15. If we want to reduce rational expression with
variables and add to lowest terms, we proceed the same way. First, we'll factor the numerator,
that's three times x plus two, and then factor the denominator. In this case of factors 2x
plus two times x plus two, we could also write that as x plus two squared. Now we cancel
the common factors. And we're left with three over x plus two. Definitely a simpler way
of writing that rational expression. Now next, let's practice multiplying and dividing. Recall
that if we multiply two fractions with just numbers in them, we simply multiply the numerators
and multiply the denominators. So in this case, we would get four times two over three
times five or 8/15. If we want to divide two fractions, like in the second example, then
we can rewrite it as multiplying by the reciprocal of the fraction on the denominator. So here,
we get four fifths times three halves, and that gives us 12 tenths. But actually, we
could reduce that fraction to six fifths, we use the same rules when we compute the
product or quotient of two rational expressions with variables on them. Here, we're trying
to divide to rational expressions. So instead, we can multiply by the reciprocal, I call
this flipping and multiplying. And now we just multiply the numerators. And multiply
the denominators. It might be tempting at this point to multiply out to distribute out
the numerator and the denominator. But actually, it's better to leave it in this factored form
and factored even more completely. That way, we'll be able to reduce the rational expression
to cancel the common factors. So let's factor even more the x squared plus x factors as
x times x plus one, and x squared minus 16. And that's a difference of two squares, that's
x plus four times x minus four, the denominator is already fully factored, so we'll just copy
it over. And now we can cancel common factors here and here, and we're left with x times
x minus four. This is our final answer. Adding and subtracting fractions is a little more
complicated because we first have to find a common denominator. A common denominator
is an expression that both denominators divided into, it's usually best of the long run to
use the least common denominator, which is the smallest expression that both denominators
divided into.
In this example, if we just want a common denominator, we could use six times 15, which
is 90 because both six and 15 divided evenly into 90. But if we want the least common denominator,
the best way to do that is to factor the two denominators. So six is two times 315 is three
times five, and then put together only the factors we need for both six and 15 to divide
our numbers, so if we just use two times three times five, which is 30. We know that two
times three will divide it and three times five will also divide it and we won't be able
to get a denominator any smaller because we need the factor of two three and five, in
order to ensure both these numbers divided, once we have our least common denominator,
we can rewrite each of our fractions in terms of that denominator. So seven, six, I need
to get a 30 in the denominator, so I'm going to multiply that by five over five, and multiply
by the factors that are missing from the current denominator in order to get my least common
denominator of 30. For the second fraction, for 1515 times two is 30. So I'm going to
multiply by two over two, I can rewrite this as 3530 s minus 830 s. And now that I have
a common denominator, I can just subtract my two numerators. And I get 27/30. If I factor,
I can reduce this to three squared over two times five, which is nine tenths. The process
for finding the sum of two rational expressions with variables in them follows the exact same
process. First, we have to find the least common denominator, I'll do that by factoring
the two denominators. So 2x plus two factors as two times x plus 1x squared minus one,
that's a difference of two squares. So that's x plus one times x minus one. Now for the
least common denominator, I'm going to take all the factors, I need to get an expression
that each of these divides into, so I need the factor two, I need the factor x plus one,
and I need the factor x minus one, I don't have to repeat the factor x plus one I just
need to have at one time. And so I will get my least common denominator two times x plus
one times x minus one, I'm not going to bother multiplying this out, it's actually better
to leave it in factored form to help me simplify later. Now I can rewrite each of my two rational
expressions by multiplying by whatever's missing from the denominator in terms of the least
common denominator. So what I mean is, I can rewrite three over 2x plus two, I'll write
the 2x plus two is two times x plus one, I'll write it in factored form. And then I noticed
that compared to the least common denominator, I'm missing the factor of x minus one. So
I multiply the numerator and the denominator by x minus one I just needed in the denominator,
but I can't get away with just multiplying by the denominator without changing my expression,
I have to multiply by it on the numerator and the denominator. So I'm just multiplying
by one and a fancy form and not changing the value here. So now I do the same thing for
the second rational expression. I'll write the denominator in factored form to make it
easier to see what's missing from the denominator. What's missing in this denominator, compared
to my least common denominator is just the factor two. So I multiply the numerator and
the denominator by two. Now I can rewrite everything. So the first rational expression
becomes three times x minus one over two times x plus 1x minus one, and the second one becomes
five times two over two times x plus 1x minus one, notice that I now have a common denominator.
So I can just add together my numerators. So I get three times x minus one plus 10 over
two times x plus 1x minus one. I'd like to simplify this. And the best way to do that
is to leave the denominator in factored form. But I do have to multiply out the numerator
so that I can add things together. So I get 3x minus three plus 10 over two times x plus
1x minus one, or 3x plus seven over two times x plus 1x minus one. Now 3x plus seven doesn't
factor. And there's therefore no factors that I can cancel out. So this is already reduced.
As much as it can be. This is my final answer.
In this video, we saw how to simplify rational expressions to lowest terms by factoring and
canceling common factors. We also saw how to multiply rational expressions by multiplying
the numerator and multiplying the denominator had divide rational expressions by flipping
and multiplying, and how to add and subtract rational expressions by writing them in terms
of the least common denominator. This video is about the difference quotient and the average
rate of change. These are topics that are related to the concept of derivative and calculus.
For function y equals f of x, like the function whose graph right here, a secant line is a
lie. that stretches between two points on the graph of the function. I'm going to label
this x value has a, and this x value as B. So this point here on the graph is going to
have an x value of a, and a y value given by f of a, the second point will have x value
B and Y value, f of b. Now the average rate of change for a function on the interval from
a to b can be defined as the slope have the secret line between the two points A, F of
A and B, F of B. In symbols, that slope m is the rise over the run, or the change in
y over the change in x, which is the difference in Y coordinates f of b minus F of A over
the difference in x coordinates b minus a. So this is the average rate of change. To
put this in context, if for example, f of x represents the height of a tree, and x represents
time in years, then f of b minus f of a represents a difference in height, or the amount the
tree grows. And B minus A represents a difference in in years, so a time period. So this average
rate of change is the amount the tree grows in a certain time period. For example, if
it grows 10 inches in two years, that would be 10 inches per two years, or five inches
per year would be its average rate of change its average rate of growth, let's compute
the average rate of change for the function f of x equals square root of x on the interval
from one to four.
So
the average rate of change is f of four minus f of one over four minus one, well, f of four
is the square root of four of one's the square root of one. So that's going to be two minus
one over three or 1/3. Instead of calling these two locations on the x axis, a and b,
this time, I'm going to call the first location, just x and the second location, x plus h.
The idea is that h represents the horizontal distance between these two locations on the
x axis. In this notation, if I want to label this point on the graph of y equals f of x,
it'll have an x coordinate of x and a y coordinate of f of x. The second point will have an x
coordinate of x plus h and a y coordinate of f of x plus h. a difference quotient is
simply the average rate of change using this x x plus h notation. So a different version
represents the average rate of change of a function, f of x on the interval from x to
x plus h. Equivalent way, the difference quotient represents the slope of the secant line for
the graph of y equals f of x between the points with coordinates x, f of x, and x plus h,
f of x plus h. Let's work out a formula for the difference quotient. Remember that the
formula for the average rate of change could be written as f of b minus F of A over B minus
A, where a and b are the two locations on the x axis. But now I'm calling instead of
a I'm using x instead of B, I'm using x plus h. So I can rewrite this average rate of change
as f of x plus h minus f of x over x plus h minus x. That simplifies a little bit on
the denominator. Because x plus h minus x, I can cancel the access and I get the difference
quotient formula f of x plus h minus f of x over h. The quantity h on denominator It
looks like a single entity, but it still represents a difference in x values. Let's find and simplify
the difference quotient for this function given first or write down the general formula
for the difference quotient. That's an F of x plus h minus f of x over h. I'm going to
compute f of x plus h first, I do this by shoving in x plus h, everywhere I see an x
in the formula for the function. So that's going to give me two times x plus h squared
minus x plus h plus three. Notice how I use parentheses here. That's important, because
I need to make sure I shove in the entire x plus h for x. So the entire x plus h needs
to be subtracted, not just the x part, so the parentheses are mandatory. Similarly,
the parentheses here signal that the entire x plus h is squared as it needs to be, I'm
going to go ahead and simplify a bit right now, I can multiply out the x plus h squared,
I can go ahead and distribute the negative sign. So if I multiply out, I'm gonna get
x squared plus XH plus h x plus h squared. Now I can distribute the two to get to x squared
plus 2x H plus two h x plus two h squared minus x minus h plus three, these two terms
are actually the same, I can add them up to get 4x H. And I think that's as simple as
I can get that part. Now, I'm going to write out F of x plus h minus f of x. So that's
going to be this thing right here, minus f of x. Again, I need to put the F of X formula
in parentheses to make sure I subtract the whole thing. I'll distribute the negative.
And now I noticed that a bunch of things cancel out. So the 2x squared and the minus 2x squared
add to zero, the minus x and the x add to zero,
and the three and the minus three, add to zero. So I'm left with 4x H plus two h squared
minus h. Finally, I'll write out the whole difference quotient by dividing everything
by H. I can simplify this further, because notice that there's an H in every single term
of the numerator. If I factor out this H, H times 4x, plus two h minus one divided by
h, these two H's cancel, and I'm left with a difference quotient of 4x plus two h minus
one. This difference quotient will become important in calculus, when we calculate the
difference quotient for smaller and smaller values of h, eventually letting h go to zero,
and ending up with an expression that has no H's in it, and represents the derivative
or slope of the function itself. In this video, we use the formula f of b minus F of A over
B minus A to calculate an average rate of change, and the related formula f of x plus
h minus f of x over h to calculate and simplify a difference quotient. This video will introduce
the idea of limits through some graphs and examples. When I worked in San Francisco,
I used to eat at Julia sushi and salad bar that had an interesting price structure, the
food there cost $10 a pound. But if you happen to load up your plate to exactly one pound,
you got a free lunch. So let's let y equals f of x represent the price of your lunch in
dollars as a function of its weight x in pounds. And I want to write an equation to describe
f of x as a piecewise defined function. It makes sense to use a piecewise defined function.
Because there are two situations, the weight x could be exactly one pound, or it could
be different from one pound. If the weight is exactly one pound, then f of x The price
is zero. But if x is different from a pound, then the price is 10 times x as a graph, my
function is going to follow the line y equals 10x. But when x is exactly one, my function
is going to have a value of zero and not 10. The open circle here represents a hole or
a place where a point is missing on the graph. Now when x is near one, but not equal to one,
then the F of X values that is the y values are very close to 10. In the language of limits,
we say that the limit as x approaches one of F have x is equal to Ted. More informally,
we can write, as x approaches one, f of x approaches 10. Notice that the value of f
at one is actually equal to zero, not 10. And so the limit of f of x as x goes to one,
and the value of f at one, are not equal. This illustrates the important fact that the
limit here as x goes to one doesn't care about the value of f, at one, it only cares about
the value of f, when x is near one, in general, for any function f of x, and for real numbers,
a and L, the limit as x goes to a of f of x equals L means that f of x gets arbitrarily
close to L, as x gets arbitrarily close to a, in other words, as x heads towards a, f
of x heads towards L. Let me draw this as a picture. In this picture, I can say that
the limit as x goes to a of f of x is L. Because by taking a sufficiently small interval around
a of x values, I can guarantee that my y values, my f of x values lie in an arbitrarily small
interval around L. Now the limit doesn't care about s value at exactly a. So if I change
the function, if I change the functions value at a, or even leave the function undefined
at a, the limit is still L. But the limit does care about what happens for x values
on both sides of a.
If, for example, the function didn't even exist for x values bigger than a, then we
could no longer say that the limit, as x approached a was out, the limit would not exist. In other
words, the limit of f of x is L, only if the y values are approaching now, as x approaches
a from both the left and the right. For the function g of x graph below, the limit as
x goes to two of g of x does not exist. Because the y values approach one as the x values
approach two from the left and the y values approach three, as the x values approach two
from the right. Although the limit doesn't exist, we can say that the left sided limit
exists. And we write this as limit as x goes to two from the left of g of x is equal to
one. The superscript minus sign here means that x is going to two from the left side.
In other words, the x values are less than two and approaching two. Similarly, we can
talk about right sided limits. In this example, the limit as x goes to two from the right
side of g of x is three. And here, the superscript plus side means that we're approaching two
from the right side. In other words, our x values are greater than to and approaching
to. In general, the limit as x goes to a minus of f of x equals L means that f of x approaches
L, as x approaches a from the left, and the limit as x goes to a plus of f of x equals
our means that f of x approaches our as x approaches a from the right limits from the
left or from the right are also called one sided limits. In this last example, let's
look at the behavior of y equals h of x graph below when x is near negative two. As x approaches
negative two from the right, our Y values are getting arbitrarily large, larger than
any real number we might choose. We can write this in terms of limits by saying the limit
as x goes to negative two from the right of h of x is equal to infinity. As x approaches
negative two from the left, the y values are getting smaller. In fact, they go below any
negative real number we might choose. In terms of limits, we can say that the limit as x
goes to negative two from the left of h of x is negative infinity. In this example, the
limit of h of x as x goes to negative two, not specifying from the left or the right
means we have to approach negative two from both sides. And so this limit does not exist,
because the limits from the left and the limit from the right are heading in opposite directions.
I want to mention that some people say that the limits from the right and the limits from
the left also do not exist. Because the functions don't approach any finite number. I prefer
to say that these limits do not exist as a finite number. But they do exist as infinity
and negative infinity. This video introduced the idea of limits, and one sided limits and
infinite limits. This video gives some examples of when limits fail to exist. For this function,
f of x graph below, let's look at the behavior of f of x in terms of limits, as x approaches
negative one, one, and two. Let's start with x approaching negative one. When x approaches
negative one from the last, the y value is going to be approaching about one half. When
x approaches negative one from the right, the y values sandvi, approaching one. So when
x approaches negative one, and we don't specify from either the left or the right, we can
only say that the limit does not exist, because these two limits from the left and right are
not equal.
Now let's look at the limit as x is approaching one. This time, we approach from the left,
we get a limiting y value of two, we approach from the right, the y values are going towards
two. So both of these left and right limits are equal to
and therefore the limit as x goes to one of f of x equals two. That's true, even though
f of one f of one itself does not exist. The limit doesn't care what happens at exactly
x equals one just what happens when x is near one.
Finally, let's look at the limit as x goes to two. So here on the left side, the limit
is going to negative infinity. And on the right side, it's negative infinity. So we
can say the limit as x goes to two is negative infinity. Or we can also say that the limit
as x goes to two does not exist. This is the correct answer. This is a better answer because
it carries more information. What values of x does a limit of f of x fail to exist? Well,
let's see. Negative one and two are the only two values. Let's talk about the ways that
limits can fail to exist, we've seen at least a couple different ways. So we've seen examples
where the limit from the left is not equal to the limit from the right. Here's our number
a where we're calculating the limit at. So that's one example we've seen. We've also
seen examples where they're vertical asymptotes. There's a vertical asymptote here at a, that
limit fails to exist because of the unbounded behavior because the y values are going off
to infinity. There's one other way that limits can fail to exist that comes up, sometimes
not quite as frequently. And that's wild behavior. Not a technical term,
just to descriptive term.
Let's look at an example that has this wild behavior, forcing a limit not to exist. And
the one of the most classic examples as the limit as x goes to zero of sine pi over x
or sometimes you'll see sine one over x. If you graph this on your graphing calculator
and zoom in near x equals zero, you're gonna see something that looks roughly like this.
It just keeps oscillating up and down and up and down, as x goes towards zero As x goes
towards zero, pi over x is getting bigger and bigger. And you're going to go through
these oscillations between one and negative one for the faster and faster.
From the other side, when x is negative, you'll see a similar kind of behavior just oscillating
faster and faster. As x goes to zero,
here, this top values up here at one, and the bottom value of these are all supposed
to hit it negative one. Now, when you try to decide what the limit is, as x goes to
zero, well, the y values are going through all possible real numbers and between negative
one and one infinitely often as x goes to zero, so there's no single number that the
limit can settle at. And so the limit as x goes to zero, of sine pi over x does not exist.
In this video, we saw three types of examples when limits fail to exist, they can fail to
exist because the one sided limits on the left and the right are not equal. Or they
can fail to exist because of vertical asymptotes. Also, limits can fail to exist when there's
wild behavior. And the function fails to settle down at any single value. This video is about
limit laws, rules for finding the limits of sums, differences, products, and quotients
and functions. Let's start with an example. Suppose that the limit as x goes to seven
of f of x is 30. And the limit as x goes to seven of g of x is two. What's the limit as
x goes to seven of f of x divided by negative three times g of
x.
Wilson's f of x is heading towards 30. And g of x is heading towards two, it makes sense
that the quotient should head towards 30 divided by negative three times two, or negative five.
in calculating this limit, by plugging in numbers for the components, we were actually
using the limit loss, which are now state. Suppose that c is a constant, just some number,
and that the limits as x goes to a of f of x and g of x exist as finite numbers that
is not as limits that are infinity or negative infinity. Then the limit of the sum f of x
plus g of x is equal to the limit of f of x plus the limit of g of x. In other words,
the limit of the sum is the sum of the limits. Similarly, the limit of the difference is
the difference of the limits. The limit of C times f of x is just c times the limit of
f of x. And the limit of the product is the product of the limits. The limit of the quotient
is the quotient of the limits, provided that the limit of g of x is not equal to zero.
Since we can't divide by zero. Let's use these limit laws in an example, to find the limit
as x goes to two of x squared plus 3x plus six divided by x plus nine. Well, the limit
rule about quotients allows us to rewrite this limit of a quotient as a quotient of
limits, provided that the limits of the numerator and denominator exist, and that the limit
of the denominator is not zero. But we'll see in a moment that these conditions do in
fact hold. Next, we can use the limit rule about sums to rewrite the numerator as a sum
of limits. And we can rewrite the limit in the denominator also as a sum of limits. Next,
we can use the limit rule about products to rewrite the limit of x squared as the square
of the limit. That's because x squared is really x times x. And so the limit of x squared
is really the limit of x times x, which by the product rule is a limit of x times the
limit of x, which is the limit of x quantity squared. Going back to the original problem,
we can now use the limit rule about multiplying by a constant to rewrite the limit of three
times x as three times the limit of x. And I'll just carry the rest forward. Now that
we've got things broken down into bite sized pieces, we can start evaluating some limits.
Notice that the limit as x goes to two of x is just two, because as x heads towards
2x, heads towards two, so we can replace all these limits of access by just the number
two. So we get two squared plus three times two. Now notice that the limit as x goes to
two of six, well, six doesn't have any x's in it. So as x two heads towards to six days
at six, and the limit here is just six. So in my original problem, I can replace the
set limit of six with six. And on the denominator here, I get to plus nine. And after a little
arithmetic, this simplifies to 16/11. Notice that we could have gotten this answer a lot
faster, just by substituting in the value of two into our original expression. And in
fact, that's the beauty of the lemon laws, they allow us to evaluate limits of rational
functions, just by plugging in the number that x is going towards, as long as plugging
in that number doesn't make the denominator zero. It's not nearly so simple. When plugging
in the value does make the done at zero. And in the future, we'll build up a bunch of algebraic
techniques for handling the situation. In this video, we've talked about the limit loss.
It's important to note, these limit laws only apply if the limits of the component functions
actually exist as finite numbers.
If the limit of one or both of the component functions don't exist, then a limit rules
just simply don't apply. And instead, we have to use other techniques to try to evaluate
the limit of a sum, difference product or quotient. This video is about the squeeze
theorem, which is another method for finding limits. Let's start with an example. Suppose
we have a function g of x. We don't know much about it. But we do know that for x value
is near one, g of x is greater than or equal to four times the square root of x drawn here
in red, and less than or equal to 3x squared minus 4x plus five, drawn here in blue. So
on the picture, G has to lie between the red and the blue curves for x values near one,
it could look something like this. What can we say about the limit of g of x as x goes
to one? Well, if you notice, the red curve and the blue curve have the exact same limit
of four as x goes to one. And since the green curve is squeezed in between the red and the
blue curve, its limit must also be for this example as a special case of the squeeze theorem.
Now, let's say the squeeze theorem in general, suppose that we have three functions, f of
x g of x and h of x. And let's suppose that f of x is less than or equal to g of x, which
is less than or equal to h of x, at least for x values near some number A. this inequality
doesn't necessarily have to hold for x equal to a, because we're going to be talking about
limits. And limits don't care what happens when x is exactly a just when x is near a.
Let's suppose also, that like in the previous example, f of x and h of x have the exact
same limit as x approaches a. So we're going to suppose that the limit as x goes to a of
f of x is equal to the limit as x goes to a of h of x. And we'll call this limit. Now.
The picture looks a lot like the previous example. Again, it doesn't matter exactly
what happens at x equals a, for example, g of x could have a hole there. And its value
could be for example, way up here. Since g of x is trapped here in between f of x and
h of x, which both have the same limit l at a, we can conclude that the limit as x goes
to a of g of x is equal to l also. And that's the squeeze theorem, also known as the pinching
theorem, and the sandwich theorem, three very descriptive names that capture the idea of
geovax being trapped here, in between lower and upper bounds. Now let's use the squeeze
there. To find the limit as x goes to zero of x squared sine one over x. Now, you might
remember that sine one over x by itself has this crazy oscillating behavior. In fact,
the limit as x goes to zero of sine one over x does not exist, because the function never
settles down to a single finite value. At first glance, you might think the limit of
x squared sine of one over x also wouldn't exist. In fact, it's tempting to try to use
the product rule and say that the limit of the product is the product of the limits.
But in fact, the product rule only applies when the component limits both exist. And
since the second limit doesn't exist, the product rule tells us absolutely nothing about
whether the limit that we're interested in exists or doesn't. So we can't use the product
rule. But it turns out, we can use the squeeze theorem. Now this example is a little trickier
than the first example, because in the first example, we were told what the upper and lower
bounding functions should be. And in this example, we have to come up with. But if we
look at a graph of x squared sine one over x, we can see that it does seem to be trapped
in an envelope here.
Let's use algebra to see what those two bounding functions might be. Now we know that sine
of one over x is always between one and negative one, just because sine of anything has between
one and negative one. And if we multiply this whole inequality by x squared, we get minus
x squared is less than or equal to x squared, sine one over x, which is less than or equal
to x squared. Notice that x squared is always positive, so we don't have to worry about
flipping any of the inequality signs when we multiply by this positive number. So x
squared and minus x squared are good bounding functions. And if we notice that the limit
as x goes to zero of x squared is zero, and the limit as x goes to zero of negative x
squared is also zero, we can conclude by the squeeze theorem, that the limit as x goes
to zero of x squared, sine of one over x is also zero, because it squeezed in between
these two functions with the same limit. the squeeze theorem is a great trick for evaluating
limits when you happen to have a function that you're interested in, trapped in between
two other functions with the same limit. The example in this video is a classic example,
where we have a crazy oscillating trig function, multiplied by a power of x, that x is one
of our bounding functions. In this video, we'll compute a bunch of limits using algebraic
tricks. All these limits are of the zero over zero indeterminate form kind. Recall that
this means that the limits are of the form the limit of f of x over g of x where the
limit of f of x equals zero and the limit of g of x also equals zero. For a first example,
let's look at the limit as x goes to one of x cubed minus one over x squared minus one.
Notice that the numerator and the denominator are both going to zero as x goes to one. To
calculate this limit, we want to simplify this expression. And one way to simplify it
is to factor it. So let's rewrite this as the limit as x goes to one of x minus one
times x squared plus x plus one. We're factoring the numerator here is a difference of cubes.
We can also factor the denominator as a difference of squares, x minus one times x plus one.
Now, as long as x is not equal to one, we can cancel out these two factors of x minus
one. And so the limit of this expression is just the same as the limit of this expression.
Now we can just plug in one, because plugging in one gives us a numerator of three and a
denominator of two. And that way, we've evaluated our limits. Next, let's look at the limit
of five minus z quantity squared minus 25. All divided by z. Once again, when we plug
in x equals zero in the numerator, we get zero and then nominator we also get zero.
This time instead of factoring, the trick is going to be to multiply out. So let's rewrite
this limit as the limit as t goes to zero
of
25 minus 10 z plus z squared minus 25. I just distributed to get this expression divided
by z. Since 25 minus 25 is zero, I just have the limit as t goes to zero of negative 10
z plus c squared over
z.
Now what? Well, I could factor out the Z here. One z is not zero, I can cancel here. So my
original limit is the same as this limit. Plugging in z equals zero, I just get negative
10 as my answer. This third example is also a situation where the numerator is going to
zero, and the denominator is also going to zero. In this case, I'm going to try to simplify
the expression by adding together the fractions and the numerator. So I'll need a common denominator,
which is r plus three, times three. So rewriting, I get the limit of one over r plus three,
I multiply that by three over three, in order to get the appropriate common denominator,
minus 1/3, which I have to multiply by R plus three over r plus three, all that over are
continuing to rewrite, I have in the numerator, adding together these fractions mature, have
the denominator of R plus three times three, I get three minus quantity r plus three. And
then this entire fraction is still divided by our distributing the negative sign, I have
three minus r minus three divided by R plus three times three, all divided by R. three
minus three is zero. So I can rewrite this as negative r over r plus three, times three.
And now instead of dividing by r, which is r of r one, I can multiply by the reciprocal,
one over R. R divided by R is one. So this expression simplifies to negative one, over
r plus three, times three. So finally, I'm in a good position, because now I can just
go ahead and let our go to zero. And by plugging in R equals zero, I have a limit of negative
one over zero plus three times three, or negative one, nine. This example is a little tricky,
because involve square roots can be hard to deal with. Well, there's one nice trick for
dealing with square roots that works here, which is the conjugate. So I'm going to take
the expression that we're given, and multiply by the conjugate of the numerator, in this
case, because the numerator is the place where the square root is by the conjugate of A minus
B, I just mean a plus b conjugate of A plus B is a minus b. Well, of course, if I multiply
on the numerator, by something, I also have to multiply the denominator by the same thing,
so that I won't alter the value of the expression. So this limit here is equal to the limit of
the expression down here that looks more complicated, than in a moment, if we're lucky, things will
clear up and become simpler. So Multiplying the numerators across, I get the square root
of x plus three squared.
I get
plus two times the square root of x plus three minus two times the square root of x plus
three minus four. On the denominator, I get x times the square root of x plus three
plus
2x minus the square root of x plus three minus two. Let's see what simplifies here. So the
square root of x plus three squared is just x plus three. And I see that this expression
and this expression are opposites, so they subtract to zero here. So on the numerator,
I just have x plus three, and then I still have the minus four on the denominator, nominator
looks a little messy. I'll just copy it over for now. Okay, so on my numerator, I'm getting
x minus one. Notice that when x goes to one, that numerator is still going to zero. And
in fact, as I let x go to one, that denominator, if I plug in here, everything is gonna also
cancel out to zero. So I still got a zero over zero indeterminate form. But maybe I
can use one of the previous tricks of factoring. Because look at here, if I factor an X out
of these two expressions as a square root of x plus three out of these two expressions,
and I could maybe factor a two out of these two, factoring by grouping, let's try that.
So we have x squared of x plus three times x minus one from these two expressions, and
then I have a plus two times x minus one. From these two expressions, this is looking
promising. So now I've got an x minus one on the top. And if I factor out the x minus
one from each of these two expressions, I'm going to have an x minus one on the bottom
times the square root of x plus three, plus two. Now, for x values near one, but not equal
to one, I can cancel those. And my limit simplifies to just one over square root of x plus three
plus two, plugging in one, I get a one on the numerator, and square root of four, which
is two plus two, four on the denominator, and we have calculated this limit. Last example
here, another zero over zero indeterminate form. Anytime I see an absolute value, I'm
going to want to take cases, because the absolute value of X plus five naturally follows in
the cases, if X plus five is greater than zero, in other words, x is greater than negative
five, then the absolute value of this positive number is just
itself.
On the other hand, if X plus five is less than zero, in other words, x is less than
negative five, than the absolute value of a negative number is its opposite. And we
make the expression x plus five, turn it into its opposite by putting a negative sign in
front. Now let's look at one sided limits. As x goes to negative five from the left,
we have
a situation where x is less than negative five, this situation right here. And so we
can rewrite the absolute value by taking the negative of the expression. We still have
a zero over zero and determinant form. But if we use the old factoring trick,
factor out a two cancel, we got the limit of two over negative one, which is just negative
two.
We do the same exercise on the right side. When we're on the right, then x is greater
than negative five. So we're in this situation here, where we can just replace the absolute
value with the stuff inside and again factoring the numerator and canceling the X plus five
We just get a limit
of two.
So we have a left limit, and a right limit that are different. And so in this example,
the limit does not exist. So we've seen a five different kinds of limits all of the
zero or zero indeterminate form. And we've used five different methods to evaluate them.
For the first example, we use factoring. For the second example, do the opposite of factoring,
we multiplied out. For the third example, we added together our rational expressions
to simplify things. The next example uses the old multiply by the conjugate trick. And
the last example, we used cases and looked at one sided limits.
The limit law about quotients tells us that the limit of the quotient is the question
of the limit, provided that the limits of the component functions actually exist, and
that the limit of the function on the denominator is not equal to zero. But what happens if
the limit of the function on the denominator is equal to zero? This video will begin to
answer that question. In fact, there are two different situations we'll want to consider.
It could be that even though the limit on the denominator is equal to zero,
the limit on the numerator exists and is not equal to zero. Or it could happen that both
limits are zero. We'll focus on the first situation, first starting with an example.
And we'll look at the second situation later on.
In this example, the limit of the numerator, negative 4x is just negative 12, which we
can see by plugging in three for x. But the limit as x goes to three of the denominator
is zero. So we're exactly in one of these situations where the numerator goes to a finite
nonzero number, but the denominator goes to zero. Let's see what happens as we approach
three from the left first. As we approach three from the left, x is going through numbers
that are slightly less than three numbers like 2.9 2.99 2.999, and so on. If we plug
in those numbers into the expression, here on our calculator, we didn't get answers of
116 1196, and 11,996. If even without a calculator, we could approximate these answers pretty
closely by just thinking about the fact that since x is very close to three, the numerator
is about negative four times three, so about negative 12. The denominator 2.9 minus three
is negative 0.1. That quotient of two negative numbers gives us a positive value of 120.
Similarly, we could approximate the value when x is 2.9 times as almost negative 12
divided by 0.01, which is 1200. And approximate the third value as 12,000. Either way, we
do it exact answers on our calculator or approximations in our head when noticing that these values
are positive numbers that are getting larger and larger as x goes towards three from the
left. This makes sense. Because if we look at our expression, as x goes towards three
from the left, the numerator is getting close to negative 12, which is a negative number.
And the denominator, since x is less than three will always be a small negative number,
negative over negative is a positive. And as x is getting really close to three, those
denominators are getting smaller and therefore the fractions are getting bigger and bigger
in magnitude. So we can conclude that our limit is positive infinity. We can make a
similar argument By looking at the limit, as x goes to three from the positive side,
it's supposed to be an X minus three here. So now x is going through a value is slightly
bigger than three 3.1 3.01 3.01. And again, we can plug directly into our calculator and
figure out the answers are negative 124. Negative 1204, negative 12,004. Or we can make a similar
approximating argument. This answer is approximately negative 12 over a positive point one, which
is negative 120, and so on. Like before, if we consider the signs of our numerator and
denominator, we can see that as x goes to three, our numerator is a negative number.
But our denominator is a positive number, since we're approaching three from the right
where x is bigger than three, and therefore, our quotient is a negative number, it's still
getting bigger and bigger in magnitude as x goes towards three, because the denominator
is still getting tinier and tinier, while the numerator stays pretty close to negative
12. So in this case, we're getting a negative number that's bigger and bigger magnitude,
so that makes a limit of negative infinity. Now, since our limit on the left is infinity,
and our limit on the right is negative infinity, the only thing we can say about the limit
as x goes to three
is that it does not exist. Now let's look at another example. The limit as x goes to
negative four of 5x, over the absolute value of x plus four, notice that the limit of the
numerator is just negative 20 by plugging in negative four for x, and the limit of the
denominator is zero. Because we've got an absolute value in our expression, here, it's
screaming out at us to look at cases. Remember that the absolute value of x plus four is
going to equal just x plus four, if x plus four is positive, in other words, if x is
greater than negative four, however, if x is less than negative four, then the expression
x plus four will be negative. So taking the absolute value has to switch at sign in order
to make a negative expression positive. Alright, so that's going to come in handy when we look
at the limit as x goes to negative four, from the left, and from the right. So when we approach
negative four from the left, x is going to be less than negative four. So we're going
to be in this situation here, where the absolute value gives us the opposite sign. That means
that the limit as x goes to negative four minus of this expression, the same as the
limit of 5x over negative x plus four. Now reasoning as before, as x is going to negative
four from the left, the numerator here is a negative number. Pretty close to negative
20. The denominator, since x is less than negative 4x plus four is negative, the negative
of it is positive. So our quotient is negative. And since denominator is getting really tiny,
while the numerators will pretty level at at negative 20. This limit is going to be
bigger and bigger magnitude a limit of negative infinity. Now let's look at the limit as x
goes to negative four from the right, in this case, x is just a little bit bigger than negative
four. So we're in this case where the absolute value doesn't change the expression. So we
can rewrite this As 5x
over x plus four,
now the numerator is still gonna be a negative number, the denominator, since x is slightly
bigger than negative four, slightly to the right, this expression is going to be positive
number. Negative or over a positive is a negative. And again, since the denominator is getting
tiny, the fractions getting huge in magnitude. And so this limit is negative infinity. Now,
in this case, look at what's going on, we've got a negative infinity limit on the left
and a negative infinity limit on the right. So we can conclude that the limit as x goes
to negative four of 5x, over the absolute value of x plus four is equal to negative
infinity. In fact, we can confirm that by looking at a graph. If we look, check out
the graph near x equals negative four, it's gonna look something like this, with a vertical
asymptote at x equals negative four, like expected. So we've seen that if the limit
of f of x is equal to something that's not zero, and the limit of g of x is equal to
zero, then the limit of the quotient could be negative infinity as it was in the past
example. It could also be infinity, or it could just not exist. supposed to say does
not exist, if the one sided limits are infinity on one side, and negative infinity on the
other. Now, what about this second situation that I mentioned the beginning when the limit
of f of x is zero, and the limit of f of g of x is zero? What can we say about the limit
of the quotient in this situation? Well, in fact, in this situation, the limit of the
quotient, it could exist and be any finite number, or infinity, or minus infinity, or
it could not exist at all. In fact, in this sort of situation, which is called a zero,
or zero indeterminate form, anything could happen. Which makes it in some ways the hardest,
but in some ways, the most fun situation of all. So in another video, we'll talk about
techniques for dealing with 00 indeterminate forms, and how to use algebra and other simplification
techniques to evaluate these these mysterious limits. So in this video, we've looked at
the limits of quotients, when the limit of the denominator is zero. We've done some examples
when the limit of the numerator was not zero, but the limit of the denominator was zero.
And we saw that these situations corresponded to vertical asymptotes, and gave us an answer
for the limit of the quotient of infinity, or negative infinity, or sometimes infinity
on one side and negative infinity on the other. We also hinted at fun things to come when
we look at the limits in this situation when the numerator and the denominator are both
heading towards zero. And when anything can happen. This video is about graphs and equations
of lines. Here we're given the graph of a line, we want to find the equation, one standard
format for the equation of a line is y equals mx plus b. here, m represents the slope, and
B represents the y intercept, the y value, where the line crosses the y axis, the slope
is equal to the rise over the run. Or sometimes this is written as the change in y values
over the change in x values. Or in other words, y two minus y one over x two minus x one,
where x one y one and x two y two are points on the line.
While we could use any two points on the line, to find the slope, it's convenient to use
points where the x and y coordinates are integers. That is points where the line passes through
grid points. So here would be one convenient point to use. And here's another convenient
point to use. The coordinates of the first point are one, two, and the next point this
is let's say, five negative one. Now I can find this slope by looking at the rise over
the run. So as I go through a run of this distance, I go through a rise of that distance,
especially gonna be a negative rise or a fall because my line is pointing down. So let's
see counting off squares, this is a run of 1234 squares and a rise of 123. So negative
three, so my slope is going to be negative three, over four. I got that answer by counting
squares. But I could have also gotten it by looking at the difference in my y values over
the difference of my x values. That is, I could have done negative one minus two, that's
from my difference in Y values, and divide that by my difference in x values, which is
five minus one, that gives me negative three over four, as before, so my M is negative
three fourths. Now I need to figure out the value of b, my y intercept, well, I could
just read it off the graph, it looks like approximately 2.75. But if I want to be more
accurate, I can again use a point that has integer coordinates that I know it's exact
coordinates. So either this point or that point, let's try this point. And I can start
off with my equation y equals mx plus b, that is y equals negative three fourths x plus
b. And I can plug in the point one, two, for my x and y. So that gives me two equals negative
three fourths times one plus b, solving for B. Let's see that two equals negative three
fourths plus b. So add three fourths to both sides, that's two plus three fourths equals
b. So b is eight fourths plus three fourths, which is 11 fourths, which is actually just
what I eyeballed it today. So now I can write out my final equation for my line y equals
negative three fourths x plus 11 fourths by plugging in for m and b. Next, let's find
the equation for this horizontal line. a horizontal line has slope zero. So if we think of it
as y equals mx plus b, m is going to be zero. In other words, it's just y equals b, y is
some constant. So if we can figure out what that that constant y value is, it looks like
it's to let's see, this three, three and a half, we can just write down the equation
directly, y equals 3.5. For a vertical line, like this one, it doesn't really have a slope.
I mean, if you tried to do the rise over the run, there's no run. So you'd I guess you'd
be divided by zero and get an infinite slope. But But instead, we just think of it as an
equation of the form x equals something. And in this case, x equals negative two, notice
that all of the points on our line have the same x coordinate of negative two and the
y coordinate can be anything. So this is how we write the equation for a vertical line.
In this example, we're not shown a graph of the line, we're just get told that it goes
through two points. But knowing that I go through two points is enough to find the equation
for the line. First, we can find the slope by taking the difference in Y values over
the difference in x values. So that's negative three minus two over four minus one, which
is negative five thirds. So we can use the standard equation for the line, this is called
the slope intercept form.
And we can plug in negative five thirds. And we can use one point, either one will do will
still get the same final answer. So let's use one two and plug that in to get two equals
negative five thirds times one plus b. And so B is two plus five thirds, which is six
thirds plus five thirds, which is 11 thirds. So our equation is y equals negative five
thirds x plus 11 thirds. This is method one. method two uses a slightly different form
of the equation. It's called the point slope form and it goes y minus y naught is equal
to m times x minus x naught where x naught Why not is a point on the line, and again
is the slope. So we calculate the slope the same way, by taking a difference in Y values
over a difference in x values. But then we can simply plug in any point. For example,
the point one, two will work, we can plug one in for x naught and two in for Y not in
this point slope form, that gives us y minus two is equal to minus five thirds x minus
one. Notice that these two equations, while they may look different, are actually equivalent.
Because if I distribute the negative five thirds, and then add the two to both sides,
I get the same equation
as above.
So we've seen two ways of finding the equation for the line is in the slope intercept form,
and using the point slope form. In this video, we saw that you can find the equation for
a line if you know the slope. And you know one point, you can also find the equation
for the line if you know two points, because you can use the two points to get the slope
and then plug in one of those points. To figure out the rest of the equation. We saw two standard
forms for the equation of a line the slope intercept form y equals mx plus b, where m
is the slope, and B is the y intercept. And the point slope form y minus y naught equals
m times x minus x naught, where m again is the slope and x naught y naught is a point
on the line. This video is about rational functions and their graphs. Recall that a
rational function is a function that can be written as a ratio or quotient of two polynomials.
Here's an example. The simpler function, f of x equals one over x is also considered
a rational function, you can think of one and x as very simple polynomials. The graph
of this rational function is shown here. This graph looks different from the graph of a
polynomial. For one thing, its end behavior is different. The end behavior of a function
is the way the graph looks, when x goes through really large positive, or really large negative
numbers, we've seen that the end behavior of a polynomial always looks like one of these
cases. That is why marches off to infinity or maybe negative infinity, as x gets really
big or really negative. But this rational function has a different type of end behavior.
Notice, as x gets really big, the y values are leveling off at about a y value of three.
And similarly, as x values get really negative, our graph is leveling off near the line y
equals three, I'll draw that line, y equals three on my graph, that line is called a horizontal
asymptote. A horizontal asymptote is a horizontal line that our graph gets closer and closer
to as x goes to infinity, or as X goes to negative infinity or both. There's something
else that's different about this graph from a polynomial graph, look at what happens as
x gets close to negative five. As we approach negative five with x values on the right,
our Y values are going down towards negative infinity. And as we approach the x value of
negative five from the left, our Y values are going up towards positive infinity. We
say that this graph has a vertical asymptote at x equals negative five. vertical asymptote
is a vertical line that the graph gets closer and closer to. Finally, there's something
really weird going on at x equals two, there's a little open circle there, like the value
at x equals two is dug out. That's called a hole. A hole is a place along the curve
of the graph where the function doesn't exist. Now that we've identified some of the features
of our rational functions graph, I want to look back at the equation and see how we could
have predicted those features just by looking at the equation. To find horizontal asymptotes.
We need to look at what our function is doing when x goes through really big positive or
really big negative numbers. Looking at our equation for our function, numerator is going
to be dominated by the 3x squared term when x is really big, right, because three times
x squared is going to be absolutely enormous compared to this negative 12. If x is a big
positive or negative number in the denominator, the denominator will be dominated by the x
squared term. Again, if x is a really big positive or negative number, like a million,
a million squared will be much, much bigger than three times a million or negative 10.
For that reason, to find the end behavior, or the horizontal asymptote, for our function,
we just need to look at the terms on the numerator and the term on the denominator that have
the highest exponent, those are the ones that dominate the expression in size. So as x gets
really big, our functions y values are going to be approximately 3x squared over x squared,
which is three. That's why we have a horizontal asymptote at y equals
three. Now our vertical asymptotes, those tend to occur where our denominator of our
function is zero. That's because the function doesn't exist when our denominator is zero.
And when we get close to that place where our denominator is zero, we're going to be
dividing by tiny, tiny numbers, which will make our Y values really big in magnitude.
So to check where our denominators zero, let's factor our function. In fact, I'm going to
go ahead and factor the numerator and the denominator. So the numerator factors, let's
see, pull out the three, I get x squared minus four, factor in the denominator, that factors
into X plus five times x minus two, I can factor a little the numerator a little further,
that's three times x minus two times x plus two over x plus 5x minus two. Now, when x
is equal to negative five, my denominator will be zero, but my numerator will not be
zero. That's what gives me the vertical asymptote at x equals negative five. Notice that when
x equals two, the denominators zero, but the numerator is also zero. In fact, if I cancelled
the x minus two factor from the numerator and denominator, I get a simplified form for
my function that agrees with my original function, as long as x is not equal to two. That's because
when x equals two, the simplified function exists, but the original function does not
it's zero over zero, it's undefined. But for every other x value, including x values near
x equals two, our original function is just the same as this function. And that's why
our function only has a vertical asymptote at x equals negative five, not one at x equals
two, because the x minus two factor is no longer in the function after simplifying,
it does have a hole at x equals two, because the original function is not defined there,
even though the simplified version is if we want to find the y value of our hole, we can
just plug in x equals two into our simplified version of our function, that gives a y value
of three times two plus two over two plus seven, or 12 ninths, which simplifies to four
thirds. So our whole is that to four thirds. Now that we've been through one example in
detail, let's summarize our findings. We find the vertical asymptotes and the holes by looking
where the denominator is zero. The holes happen where the denominator and numerator are both
zero and those factors cancel out. The vertical asymptotes are all other x values where the
denominator is zero, we find the horizontal asymptotes by considering the highest power
term on the numerator and the denominator, I'll explain this process in more detail in
three examples. In the first example, if we circle the highest power terms, that simplifies
to 5x over 3x squared, which is five over 3x. As x gets really big, the denominator
is going to be huge. So I'm going to be dividing five by a huge, huge number, that's going
to be going very close to zero, and therefore we have a horizontal asymptote at y equals
zero. In the second example, the highest power terms, 2x cubed over 3x cubed simplifies to
two thirds. So as x gets really big, we're going to be heading towards two thirds and
we have a horizontal asymptote at y equals two thirds. In the third example, the highest
power terms x squared over 2x simplifies to x over two. As x gets really big, x over two
is getting really big. And therefore, we don't have a horizontal asymptote at all. This is
going to infinity, when x gets through goes through big positive numbers, and is going
to negative infinity when x goes through a big negative numbers. So in this case, the
end behavior is kind of like that of a polynomial, and there's no horizontal asymptote.
In general, when the degree of the numerator is smaller than the degree of the denominator,
we're in this first case where the denominator gets really big compared to the numerator
and we go to zero. In the second case, where the degree of the numerator and the degree
of the denominator equal, things cancel out, and so we get a horizontal asymptote at the
y value, that's equal to the ratio of the leading coefficients. Finally, in the third
case, when the degree of the numerator is bigger than the degree of the denominator,
then the numerator is getting really big compared to the denominator, so we end up with no horizontal
asymptote. Finally, let's apply all these observations to one more example. Please pause
the video and take a moment to find the vertical asymptotes, horizontal asymptotes and holes
for this rational function. To find the vertical asymptotes and holes, we need to look at where
the denominator is zero. In fact, it's going to be handy to factor both the numerator and
the denominator. Since there if there are any common factors, we might have a hole instead
of a vertical asymptote. The numerator is pretty easy to factor. Let's see, that's 3x
times x plus one for the denominator, first factor out an x. And then I'll factor some
more using a guess and check method. I know that I'll need a 2x and an X to multiply together
to the to x squared, and I'll need a three and a minus one or also minus three and a
one. Let's see if that works. If I multiply out 2x minus one times x plus three, that
does give me back my 2x squared plus 5x minus three, so that checks out. Now, I noticed
that I have a common factor of x in both the numerator and the denominator. So that's telling
me I'm going to have a hole at x equals zero. In fact, I could rewrite my rational function
by cancelling out that common factor, and that's equivalent, as long as x is not equal
to zero. So the y value of my
whole
is what I get when I plug zero into my simplified version, that would be three times zero plus
one over two times zero minus one times zero plus three, which is three over negative three
or minus one. So my whole is at zero minus one. Now all the remaining places in my denominator
that make my denominator zero will get me vertical asymptotes. So I'll have a vertical
asymptote, when 2x minus one times x plus three equals zero, that is, when 2x minus
one is zero, or x plus three is zero. In other words, when x is one half, or x equals negative
three. Finally, to find my horizontal asymptotes, I just need to consider the highest power
term in the numerator and the denominator. That simplifies to three over 2x, which is
bottom heavy, right? When x gets really big, this expression is going to zero. And that
means that we have a horizontal asymptote at y equals zero. So we found the major features
of our graph, the whole, the vertical asymptotes and the horizontal asymptotes. Together, this
would give us a framework for what the graph of our function looks like. horizontal asymptote
at y equals zero, vertical asymptotes at x equals one half, and x equals minus three
at a hole at the point zero minus one. plotting a few more points, or using a graphing calculator
of graphing program, we can see that our actual function will look something like this. Notice
that the x intercept when x is negative one corresponds to where the numerator of our
rational function or reduced rational function is equal to zero. That's because a zero on
the numerator that doesn't make the denominator zero makes the whole function zero. And an
X intercept is where the y value of the whole function is zero. In this video, we learned
how to find horizontal asymptotes of rational functions. By looking at the highest power
terms, we learned to find the vertical asymptotes and holes. By looking at the factored version
of the functions, the holes correspond to the x values that make the numerator and denominator
zero, his corresponding factors cancel. The vertical asymptotes correspond to the x values
that make the denominator zero, even after factoring any any common any common factors
in the numerator denominator. This video focuses on the behavior of functions and graphs, as
x goes through arbitrarily large positive and negative values. You may have touched
on these ideas before in the past when you studied horizontal asymptotes. But in this
video, we'll talk about the same ideas in the language of limits. In this first example,
what happens to the function f of x drawn below as x go through larger and larger positive
numbers? Well, the arrow here on the end is supposed to mean the trend continues. So as
x gets bigger and bigger, the values of y, that is f of x, get closer and closer to one,
we can write this in the language of limits by saying the limit as x goes to infinity
of f of x is equal to one. Now what happens to this function as x goes through larger
and larger negative numbers, by larger and larger negative numbers, I mean numbers that
are negative but are larger and larger in magnitude. So for example, negative five,
negative 10, negative 100, and negative a million and so on. Well, assuming this trend
continues, it looks like f of x, even though it's oscillating, it's settling down at a
value of two. So we say that the limit as x goes to negative infinity, of f of x equals
to limits in which x goes to infinity, or negative infinity are called limits at infinity.
The phrase limits at infinity should be contrasted with the phrase infinite limits. An infinite
limit means that the y values, or the F of X values go to infinity, or negative infinity.
limits and infinity correspond to horizontal asymptotes, as drawn above, while infinite
limits correspond to vertical asymptotes. The only exception to this when we don't have
a horizontal or vertical asymptote is when x and f of x are both going infinite at the
same time, we'll see an example of this on the next page. Let's figure out the limits
of infinity for these two functions, g of x and h of x. The function g of x is actually
the function e to the minus x, and it has a horizontal asymptote, heading right here
at y equals zero. So the limit as x goes to infinity of g of x equals zero. But as we
head to the left, and x goes through larger and larger negative values, our Y values don't
settle down to a particular finite value, they get arbitrarily large. And so we say
that the limit as x goes to minus infinity of g of x is equal to infinity. Now let's
look at the graph of y equals h of x. Please pause the video for a moment and try to figure
out the limits of infinity for this function. The limit as x goes to infinity of h of x
is negative infinity. Because as x goes to infinity, the y values get below or more negative
than any finite number. Now as X goes to negative infinity, the y values oscillate, and never
settle down at a particular number. So we say that the limit as x goes to minus infinity
of h of x does not exist. Finally, let's look at some limits of infinity of functions without
looking at their graphs first, to find the limit as x goes to infinity of one over x.
Let's think about what happens to one over x. As x gets bigger and bigger through positive
numbers. As x gets bigger and bigger, one over x gets smaller and smaller. So the limit
as x goes to infinity of one over x equals zero. To find the limit of one over x as x
goes to negative infinity, let's look at what happens as X goes to negative numbers that
are larger and larger in magnitude. Now one of our x goes through numbers that are negative,
but they're still getting smaller and smaller. magnitude. So the limit as x goes to negative
infinity of one over x is also zero. We can use similar reasoning to find the limit as
x goes to infinity of one over x cubed. As x goes to infinity, x cubed also goes to infinity.
So one of our x cubed has to go to zero. To find the limit as x goes to infinity of one
over the square root of x, notice that as x goes to infinity, the square root of x still
goes to infinity. So one over the square root of x also goes to zero. In other words, both
of these limits are equal to zero. Both of these examples are actually closely related,
because both have the form of the limit as x goes to infinity of one over x to the R,
where R is a number greater than zero. In the second example, the square root of x is
really x to the R, where R is one half. In fact, the limit as x goes to infinity of one
over x to the R is always equal to zero. Whenever R is bigger than zero, we can even say the
same thing about the limit as x goes to negative infinity of one over x to the R. As long as
we avoid exponents, like one half that don't make sense for negative numbers. But for other
values of r, as x goes to negative infinity, x to the R is getting bigger and bigger and
magnitude. And so one over x to the R is getting smaller and smaller and magnitude and heading
towards zero. Notice that this is no longer true. if r is less than zero, for example,
something like r equals negative two, because one over x to the minus two is really x squared.
And the limit as x goes to infinity of x squared is going to be infinity, not zero.
In this video, we looked at examples of limits as x goes to infinity, and x goes to minus
infinity. And we saw that those limits could be zero. Any other number, infinity, negative
infinity, or not exist. This video gives some algebraic techniques for computing the limits
at infinity of rational functions. Let's find the limit as x goes to infinity of this rational
function. The numerator and the denominator of this rational function are each getting
arbitrarily large as x goes to infinity. One way to see this is by estimating the graphs,
the graph of the numerator looks like a parabola pointing upwards. And the graph of the denominator
looks like some kind of cubic. So something like this, for both of these graphs, as x
goes to infinity, y also goes to infinity. So this is an infinity over infinity indeterminate
form. And just like the zero have over zero indeterminate forms we saw earlier, and infinity
over infinity and determinant form could turn out to be absolutely anything. So we're going
to use algebra to rewrite this expression in a different form that makes it easier to
evaluate. Specifically, we're going to factor out the highest power of x that we can find
from the numerator, and then from the denominator. In the numerator, I'm going to factor out
the highest power I see in the numerator, which is x squared. When I factor x squared
out of 5x squared, I get five, when I factor x squared out of negative 4x, that's like
dividing negative 4x by x squared, so I get negative four divided by x. You can check
this works by distributing the x squared and making sure we get back to the original expression.
Now the highest power of x SC, and the denominator is x cubed. So I'll factor out an x cubed
for each from each of those terms, I get a two minus 11 over x plus 12 over x squared.
Because factoring out an x cubed is the same as dividing each term by x cubed and then
writing the x cubed on the side. Now, we can rewrite again by canceling an x squared from
the top and the bottom to get the limit of one over x times five minus four over x over
two minus 11 over x plus 12 over x squared. Now as x goes to infinity, four over x goes
to zero, because I'm dividing for by larger and larger numbers. Similarly 11 over x goes
to zero, and 12 over x squared goes to zero. So I end up with one over x, which is itself
going to zero times something that's going to five halves. So my limit is going to be
zero times five halves, which is just zero. I've actually been applying limit laws to
do these last steps, which is fine, because my component limits exist as finite numbers.
Something that wasn't true from my original expression when I had infinity over infinity.
In this example, we're asked to find the limit as x goes to negative infinity of a different
rational expression. I encourage you to stop the video and try it for yourself first. In
this example, the highest power of x in the numerator is x cubed, and the highest power
in the denominator is also x cubed. Factoring out the x cubed from the numerator, we get
x cubed times three plus six over x plus 10 over x squared, plus two over x cubed. And
factoring out the x cubed from the denominator, we get x cubed times two, plus one over X
plus five over x cubed. Now the x cubes cancel, and all these parts go to zero. So when the
dust clears here, our limit is just three halves. In this next example, the highest
power in the numerator is x to the fourth, and the highest power in the denominator is
x squared. So we factor out the x to the fourth from the numerator and the x squared from
the denominator and cancel as much as we can. A lot of these pieces are going to zero.
So our limit is the same as the limit of x squared times one over negative five. As x
goes to negative infinity, x squared is positive and goes towards positive infinity. multiplying
by negative fifth turns it negative, but doesn't change the fact that the magnitudes are getting
arbitrarily large. Therefore, our final limit is negative infinity. Now let's look at the
same three examples again, more informally, using a heuristic to get the same conclusions.
In the first example, the term 5x squared dominates the numerator, because x squared
is much larger than x when x is large. In the denominator, the highest power of x to
x cubed dominates, because x cubed is much larger than x squared, or x when x is large.
If we ignore all the other terms in the numerator and denominator, and just focus on the important
terms, which have the highest powers, then we can rewrite our limit as the limit of 5x
squared over 2x cubed, which is the same as the limit as x goes to infinity of five over
2x, just by canceling Xs, which is zero as x goes to infinity. Similarly, if we just
focus on the highest power terms in the numerator and denominator in the second example, we
get the limit of 3x cubed over 2x cubed, which simplifies to the limit of three halves, which
is just three halves. In the third example, the highest power terms are x to the fourth
and negative 5x squared. And we rewrite the limit using only these highest power terms
and simplify, and we get the limit as x goes to negative infinity of x squared over negative
five, which is negative infinity as before, for rational functions, in general, looking
at the highest power terms, lets you reliably predict the limits that infinity and negative
infinity when the degree of the numerator is less than the degree of the denominator,
then the limit as x goes to infinity, or negative infinity, is zero. As an example, one above,
when the degree of the numerator is equal to the degree of the denominator, then the
limit is just the quotient of the highest power terms, which is how we got three halves
as the limit in the second example. And finally, if the degree of the numerator is greater
than the degree of the denominator, then the limit is going to be plus or minus infinity.
Like it was in the third example. These shortcut rules are really handy. But it's important
to also understand the technique of factoring out highest power terms. Since this technique
can be used more generally. This video gave two methods for computing limits and infinity
of rational functions. First, there's the formal method of factoring out highest power
terms and simplifying. Second, there's the informal method of looking at the degree of
the numerator and the degree of the denominator, and focusing on the highest power terms. In
the past, you may have heard an informal definition of continuity. Something like a function is
continuous, if you can draw it without ever picking up your pencil. In this video, we'll
develop a more precise definition of continuity based on limits. It can be helpful to look
at some examples of functions that are discontinuous, that is functions that fail to be continuous.
In order to better understand what it means to be continuous. Please pause this video
and try to draw graphs of at least two different functions that fail to be continuous in different
ways. One common kind of discontinuity is called a jump discontinuity. A function has
a jump discontinuity, if its graph separates the two pieces with a jump in between them.
This particular function can be described as a piecewise defined function with two linear
equations, f of x equals to x when x is less than or equal to one, and f of x equals negative
x plus two, when x is greater than one. Another common kind of discontinuity is called a removable
discontinuity.
You may have encountered these before, when you learned about rational functions with
holes in them, for example, the function f of x equals x minus three squared times x
minus four divided by x minus four, which has a hole when x equals four, and otherwise
looks like the graph of x minus three squared. This kind of discontinuity is called removable,
because you could get rid of it by plugging the hole just by defining F to have an appropriate
value when x equals four. So in this case, you'd want f of x to be the same when x is
not equal to four, but you'd want it to have the value of one when x equals four. And that
would amount to plugging the hole and making it continuous. In this original example, our
function had a removable discontinuity because it wasn't defined when x equals four. But
a function could also have a removable discontinuity, because it's defined in the wrong place at
x equals four, for example, too high or too low, to fit the trend of f. a discontinuity
can also occur at a vertical asymptote, where it's called an infinite discontinuity. For
example, the rational function g of x is one over x minus two has an infinite discontinuity
at x equals two. Occasionally, you may encounter a discontinuity is that don't look like any
of these. For example, the graph of the function y equals cosine of one over X has a wild discontinuity
at x equals zero, because of the wild oscillating behavior there. So for a function to be continuous
at x equals a, we need it to avoid all of these problems. To avoid a jump discontinuity,
we can insist that the functions limit has to exist at x equals a way to avoid a hole
or removable discontinuity, we can insist that F has to be defined at x equals A to
avoid the other kind of removable discontinuity in which f of a is defined, but it's in the
wrong place. We can insist that the limit of f of x as x goes to a has to equal f of
a. Sometimes the definition of continuity is written with just the third condition,
and the first two conditions are implied. Notice that these three conditions not only
exclude jump this continuity and removal this continuity is they also exclude infinite discontinuities
and wild discontinuities. For example, in our third example, the function can't be continuous
at x equals two, because it fails to have a limit at x equals two and it fails to have
a value at x equals two. In our wild, this continuity example, the limit also fails to
exist at x equals zero, so the function can't be continuous there. So what are the places
where this function f is not in us and why, please take a moment to think about it for
yourself. The functions not continuous at negative three, because the function is simply
not defined there, the functions not continuous at x equals one, because of that jump discontinuity.
In the language of limits, we say that the limit of f of x does not exist, when x equals
two, the limit of the function exists and equals three, but the value of the function
is down here at negative one. So the function is not continuous, because the limit doesn't
equal the value at x equals three, the function is not continuous, because once again, the
limit doesn't exist. Notice that x equals negative two. Even though the function turns
a corner, the function still continuous. Because the limit exists and equals two, and the value
of the function is also two. The function drawn here is not continuous at x equals negative
two, because the limit doesn't exist at x equals negative two, the limit from the left
is one, while the limit from the right is zero. But it is true that the value of the
function at x equals negative two is equal to the limit of the function from the left
side. That is, f of negative two is equal to the limit as x goes to negative two from
the left of f of x.
Notice that we can't say the same thing about the right limit. The limit from the right
is zero, while the value of the function is one, and those are equal. In this situation,
we say that f is continuous from the left, but not from the right. By the same reasoning
at x equals one, the function is not continuous, but it is continuous from the right, because
the limit from the right is equal to the value of the function. Notice that f is not continuous
from the left here, because the limit from the left is about one and a half, while the
value of the function is one. In general, we say that a function f is continuous from
the left at x equals j. If the limit as x goes to a from the left of f of x is equal
to f of a, and a function is continuous from the right at x equals a. If the limit as x
goes to a from the right of f of x equals f of a, in practical terms of function is
continuous from the left if the endpoint is included on the left piece, and a function
is continuous from the right, if the endpoint is included on the right piece. This video
gave a precise definition of continuity at a point in terms of limits. Namely, a function
is continuous at the point x equals a. If the limit as x goes to a of the function is
equal to the functions value
at A.
In a previous video, we gave a definition for continuity at a point. In this video,
we'll discuss continuity on an interval and continuous functions. We say that a function
f of x is continuous on the open interval BC, if f of x is continuous at every point
in that interval. For x to be continuous on a closed interval BC, we require it to be
continuous on every point in the interior of BC, but we just require it to be continuous
from the right at B. And from the left at sea. We can also talk about f being continuous
on half open intervals. For example, on half open interval, BC, which is open at B and
closed at sea, or the other way around, or the happy to open it all from B to infinity
and so on. In all of these cases, we require F to be continuous on the interior of the
interval, and left or right continuous on the closed endpoints of the interval
as appropriate.
So on what intervals is this function geovax continuous? Well, it's continuous. On this
part, the arrows indicate it keeps on going. So I'd say it's continuous from negative infinity
to negative one, not including the endpoint negative one. It's also continuous here, and
we can include the endpoint this time. So this is from negative one to one and Then
again, on this last section, we can't include one. It's not continuous there. It's not even
defined there. So what kinds of functions are continuous? What kinds of functions are
continuous everywhere? And by everywhere, I mean, on the entire real line negative infinity
infinity? Well, polynomials are a great example. Also, sine x and cosine x, the absolute value
of x is another common example. There are certainly many other functions that are continuous
on the whole real line. I'll let you see if you can come up with some more examples. Now,
if we ask the second question, what kinds of functions are continuous on their domains,
we get a lot more answers, not only polynomials, but also all rational functions, things like
f of x equals 5x minus two over x minus three squared times x plus four is a good example
of a rational function, even though it's not continuous everywhere. Because it's not continuous
when x equals three or negative four, it is continuous on its whole domain, because three
and four are not in the domain of this rational function. In addition, all trig functions,
inverse trig functions, log and either the x functions. And pretty much all the functions
we normally encounter are continuous on their domains, although their domains are not necessarily
the whole real line. For example, for natural log of x, the domain is just zero, infinity,
and that's where the function is continuous. In addition, the sums, differences, products
and quotients of continuous functions are continuous on their domains, so for example,
y equals sine of x plus the natural log of x is continuous where it's defined, the compositions
of continuous functions are continuous on their domains. So for example, the function
y equals ln of sine of x is continuous, where defined turns out to be a bunch of disjoint
intervals where sine is positive. Since continuity is defined in terms of limits, it's sometimes
possible to use our knowledge of which functions are continuous to calculate limits. For example,
if we want to find the limit as x goes to zero of cosine of x, because cosine is continuous,
we can evaluate this limit just by plugging in zero for x. And cosine of zero is one.
We're using the definition of continuity here to say that the limit of the function is equal
to the value of the function. The second example is a little trickier, because the function
inside is not continuous at x equals two. In fact, it's not defined at x equals two.
But as X approaches 2x squared minus four over 2x minus four times pi, can be rewritten
as x plus two times x minus two over two times x minus two times pi, which is the same thing
as x plus two over two pi. For x not equal to two. So as X approaches two, this expression
here, approaches two plus two over two times pi, which is just two pi. In other words,
the limit as x goes to two of x squared minus four over 2x minus four pi is just two pi.
And therefore, the limit as x goes to two of cosine of this expression is just cosine
of two pi, which is again equal to one. We're using here the fact that cosine is continuous,
and a property of continuous functions, which says that the limit as x goes to a of f of
g of x is equal to f of the limit as x goes to a of g of x. If f is a continuous function.
In other words, for continuous functions, you can pass the limit inside the function.
That's all for continuity on intervals and continuous functions. The intermediate value
theorem says that if f is a continuous function, on the closed interval a b and if n is any
number, in between F of A and F of Bay, n f has to achieve that value and somewhere.
In other words, if n is a number between F of A and F of b, then there has to be a number
c in the interval a, b, such that f of c equals n. In our example, there are three such possible
values for C, it could be right here, since f of that number equals n, or it could be
here, or here, I'll just mark the middle one. The intermediate value theorem can only be
applied to continuous functions. If the function is not continuous, then it might jump over
and, and never achieve that value. When application of the intermediate value theorem is to prove
the existence of roots or zeros of equations, we call that a real root of p of x is a real
number c,
such that P of C is zero, we're going to want to apply the intermediate value theorem
with n equal to zero. Our polynomial is defined on the whole real line, not just an interval.
But the trick here is to pick an interval a b, so that P of A is negative, and P of
B is positive, or vice versa. So that the intermediate value theorem will tell us that
P has to pass through zero in between. I'm just going to use trial and error here and
calculate a few values of p. So P of zero is easy to calculate, P of zero is just seven.
P of one is going to be five minus three minus 12 plus seven, which equals negative three.
So in this very lucky example, the first two numbers that we pick will work for our A and
B, so we can just let A be equal 01. Because P of zero is a positive number, and P of one
is a negative number. So actually, the graph should look a little different. The graph
looks more like this. But in any case, by the intermediate value theorem, there has
to be a number c, in between, in this case, zero and one, where our polynomial p achieves
this intermediate value of zero. And that number See, we don't know what it is, but
we know it somewhere in the interval zero to one, that value c gives us a real root
for our polynomial. There may be other real roots, but we've proved there exists at least
one. The intermediate value theorem has lots of other applications besides finding roots.
For example, suppose you have a wall that runs in a circle around the castle, and the
height of the wall varies continuously as a function of the angle. Surprisingly, the
intermediate value theorem can be used to show there must be somewhere two diametrically
opposite places on the wall with exactly the same height. So if you can figure out a way
to show this in this video, we stated the intermediate value theorem, which holds for
continuous functions, and talked about a couple of applications. This video introduces the
trig functions, sine, cosine, tangent, secant, cosecant, and cotangent. For right triangles.
For a right triangle with sides of length A, B and C angle theta as drawn, we define
sine of theta as the length of the opposite side over the hypothesis. The side that's
opposite to our angle theta has measure a and I have partners is this side here with
measure C. So that would be a oversee for this triangle. Cosine of theta is defined
as the length of the adjacent side over the length of high partners. This side here is
the side adjacent to theta. Of course, the high partners is also adjacent to theta, but
it's special as the high partners so we don't think of it as the adjacent side. So that
would be B oversea. tangent of theta is the opposite side length over the adjacent side
length. So that would be a over b. The pneumonic to remember this is so A toa. That's sine
is opposite over hypotenuse. Cosine is adjacent over hypotenuse. And tangent is opposite over
adjacent. In fact, there's a relationship between tangent and sine and cosine. Namely,
tangent of theta is equal to sine of theta over cosine of theta. If you want to see why
that's, that's because sine of theta over cosine of theta is given by sine, which is
opposite over hypotenuse, divided by cosine, which is adjacent over hypotenuse. If we compute
these fractions by flipping and multiplying,
the high partners, length cancels, and we just get opposite over adjacent, which is
by definition, tangent of theta. There are three more trig functions that are defined
in terms of sine, cosine and tangent. First of all, their sequence of theta, by definition,
that's one over cosine of theta. So it's going to be one over the adjacent over hypotenuse,
which is the high partners over the adjacent, which in this triangle is C over B. cosecant
of theta is defined as one over sine theta. So that's one over the opposite over the hypotenuse,
which is the high partners over the opposite. And for this triangle, that's going to be
C over a. Finally, cotangent of theta is defined as one over tan theta, that's going to be
one over opposite over adjacent, flip and multiply, I get adjacent over opposite birth,
which in this case is b over a. So notice that the values for cotangent cosecant and
secant are the reciprocals of the values for tangent, sine and cosine, respectively. Let's
use these definitions to find the exact values of all six trig functions for the angle theta
in this triangle. I'll start with sine of theta. That's the opposite of our hypothesis.
Well, for this angle theta, the opposite side is down here, and as measured to the high
partners has measured five. So sine theta is two over five. Cosine theta is adjacent
over hypotenuse, but I don't know the value of this side length. But fortunately, I can
find it using the Fagor in theorem says I have a right triangle here, I know that a
squared, I'll call this side length A plus B squared, where b is this other leg of the
triangle is equal to c squared, where c is the hypothesis. So here I have a squared plus
two squared equals five squared, which means that a squared plus four equals 25. A squared
is 21. So A is plus or minus the square root of 21. But since I'm talking about the length
of a side of a triangle, I can just use the positive answer. Returning to my computation
of cosine theta, I can write it as adjacent, which is the square root of 21. Over hypotony
News, which is five, tangent theta is the opposite over the adjacent, so that's going
to be two over the square root of 21. To compute secant of theta, that's one over cosine theta,
so that's going to be one over square root of 21 over five, which is five over the square
root of 21. The reciprocal of my cosine value. cosecant theta is one over sine theta, that's
going to be the reciprocal of my sine, so five halves, and cotangent theta is one over
a tan theta, so it's going to be the reciprocal of my 10 value square root of 21 over two.
Finally, we'll do an application. So if we have a kite that's flying at an angle of elevation
as the angle from the horizontal bar of 75 degrees with a kite length string of 100 meters,
we want to find out how high it is. I'll call the height Why? Well, we want to relate the
known quantities this angle and this hypothesis to The unknown quantity, the unknown quantity
is the opposite side of our triangle. So if we use sine of theta equals opposite of our
hypothesis, then we can relate these known amounts sine of 75 degrees to our unknown
amount y, which is the opposite and are known amount of 100 meters. solving for y, this
gives that y is 100 meters times sine of 75 degrees, we can use a calculator to compute
sine of 75 degrees, be sure you use degree mode and not radian mode, when you type in
the 75. When I do the computation, I get a final answer of 96.59 meters up to two decimal
places. Notice that we're ignoring the height of the person in this problem.
To remember the definitions of the trig functions, you can use the pneumonic, so tau, and the
fact that secant is the reciprocal of cosine cosecant the reciprocal of sine and cotangent
the reciprocal of tangent. In this video, I'll use geometry to compute the sine and
cosine of a 30 degree angle, a 45 degree angle and a 60 degree angle. One way to compute
the sine of a 45 degree angle is to use a right triangle with a 45 degree angle. This
particular right triangle has I have partners of length one. Since all the angles of the
triangle have to add up to 180 degrees, and we already have 90 degrees and 45 degrees,
the remaining angle must also be 45 degrees. So we have an isosceles triangle with two
sides the same length, I'll call that side length a. If we want sine of 45 degrees, let's
use this 45 degree angle here, then sine is opposite over hypotenuse. So if I can figure
out how long this side length is, I'll be able to compute sine of 45 degrees. Now the
Pythagorean theorem says this side length squared plus that side land squared equals
I have hotness squared. So we have that a squared plus a squared equals one squared.
All right, that is two A squared equals one. So a squared is one half, and a is plus or
minus the square root of one half. Since we're talking about the length of sides of triangles,
I can just use the positive square root it's customary to rewrite this as the square root
of one over the square root of two, which is one over the square root of two, and then
rationalize the denominator by multiplying the top and the bottom by the square root
of two. That gives me a square root of two in the numerator, and the square root of two
squared in the denominator, which is the square root of two over two. So the side lengths
are the square root of two over two. Now I can figure out the sine of 45 degrees, by
computing the opposite over the hypotenuse. That's, I'm looking at this angle. So opposite
is square root of two over two hypothesis one. So the sine of 45 degrees is the square
root of two over two. Cosine of 45 degrees is adjacent over hypotenuse. That's this side
length over this hypothesis. So that's the square root of two over two over one again,
what would happen if instead of using this triangle with hypotony is one, we use this
triangle, also a 4545 90 triangle with hypotony is five, not quite drawn to scale. Please
pause the video for a moment and repeat the computation with this triangle. This time,
I'll call the side length B. Pythagoras theorem tells me b squared plus b squared equals five
squared. So two, b squared equals 25. And b squared equals 25 over two, B is going to
be the plus or minus the square root of 25 over two again, I can just use the positive
version. And so B is the square root of 25 over the square root of two, which is five
over the square root of two, rationalizing the denominator, I get five root two over
two. Now, sine of my 45 degree angle is opposite overhype hotness,
which is
five square root of two over two divided by five. That simplifies to the square root of
two over two as before, and a similar computation shows that cosine of 45 degrees is also square
root of two over two as before. This makes sense because sine and cosine are based on
ratios of sides. And since these two triangles are similar triangles, they'll have the same
ratios of sides. To find the sine and cosine of 30 degrees, let's use this 30 6090 right
triangle with hypotony is one. If we double the triangle, we get an angle of 30 here,
so a total angle of 60 degrees here, and this angle is also 60 degrees. So we have a 60
6060 triangle, that's an equilateral triangle, all side lengths are the same. Since this
side length has length one, this side length is also one. This entire side length is one,
which means this short side of our original triangle has length one half. Going back to
my original triangle, let's use the Pythagorean Theorem to find the length of its longer side
x. But agrin theorem says x squared plus one half squared equals one squared. So x squared
plus a fourth equals 1x squared is three fourths. And so x is plus or minus the square root
of three fourths is the positive version, and get the square root of three over the
square root of four, which is the square root of three over two. Now using our original
triangle, again, let's compute the sine of this 30 degree angle here. We know that sine
of 30 degrees is opposite of our hypothesis, the opposite of this angle is one half and
the hypothesis is one. So we get a sine of one half over one, which is one half cosine
of 30 degrees is adjacent over hypothesis. So that's the square root of three over two
divided by one. To find sine of 60 degrees and cosine of 60 degrees, we can actually
use this same green triangle and just focus on this upper corner angle of 60 degrees instead.
So sine of 60 degrees. Opposite overhype hotness, but this time the opposite to this angle is
the square root of three over two. Cosine of 60 degrees adjacent over hypotenuse gives
us one half. I'll summarize the results in this table below. Notice that a 30 degree
angle corresponds to pi over six radians since 30 degrees times pi over 180 is pi over six.
Similarly, 45 degrees corresponds to pi over four radians and 60 degrees corresponds to
pi over three radians. I recommend that you memorize the three numbers one half, root
two over two, and root three over two. And the fact that one half and root three over
two go together. And root two over two goes with itself. From that information, it's not
hard to reconstruct the triangles, you know that a 4545 90 triangle is my sauciest triangle.
So it must have the side lengths were the same number goes with itself. And a 30 6090
triangle has one side length smaller than the other. So the smaller side must be one
half and the larger side must be root three over two since root three is bigger than one
and so root three over two is bigger than one half. Doing a visual check, you can easily
fill in the angles, the smaller angle must be the 30 degree angle, and the larger one
must be the 60 degree one. In this video, we computed the sine and cosine of three special
angles 30 degrees, 45 degrees, and 60 degrees. This video defined sine and cosine in terms
of points on the unit circle. a unit circle is a circle with radius one.
Up to now we defined sine and cosine and tangent in terms of right triangles. For example,
to find sine of 14 degrees in theory, you could draw a right triangle with an angle
of 14 degrees and then calculate the sign as the length of the opposite side over the
length of pi partners. But if we use this method to try to compute sine of 120 degrees,
things go horribly wrong. When we draw this 120 degree angle, and this right angle, there's
no way to complete this picture to get a right triangle. So instead, we're going to use a
unit circle, that is a circle of radius one. The figure below illustrates how right triangles
and a unit circle are related. If you draw right triangles, with larger of hypotenuse
one with larger and larger angles, then the top vertex sweeps out part of a unit circle.
Let's look at this relationship in more detail. In this figure, I've drawn a right triangle
inside a unit circle, the high partners of the triangle is the radius of the circle,
which is one, one vertex of the right triangle is at the origin, another vertex of the right
triangle is at the edge of the circle, I'm going to call the coordinates of that vertex
A, B. Now the base of this right triangle has length a, the x coordinate, and the height
of the right triangle is B, the y coordinate. If I use the right triangle definition of
sine and cosine of theta, this right here is the angle theta, then cosine of theta is
adjacent over hypotenuse, so that's a over one, or a. Notice that a also represents the
x coordinate of this point on the unit circle. At angle theta from the x axis. I'll write
that down. For sine of theta, if I use the right triangle definition, that's opposite
overhead partners, so B over one, which is just B. But B also represents the y coordinate
of this point in the on the unit circle at angle theta. For tangent theta, if we use
the right side triangle definition, its opposite over adjacent. So that's B over A, I can think
of that as the y coordinate of the point over the x coordinate of the point. Now for angles
theta, that can't be part of a right triangle, because they're too big, they're bigger than
90 degrees, like now I'll call this angle here theta, I can still use this idea of x
and y coordinates to calculate the sine and cosine of theta. So if I just mark this point,
on the end of this line at angle theta, if I mark that to have coordinates x and y, then
cosine theta, I'm still going to define as the x coordinate of this point, sine theta
as the y coordinate, and tangent theta as the ratio of the y coordinate over the x coordinate.
When we use this unit circle definition, we always draw theta starting from the positive
x axis and going counterclockwise. Let's use this unit circle definition to calculate sine,
cosine and tangent of this angle fee. In our figure, we have a unit circle. And these numbers
are supposed to represent the x&y coordinates of this point on the unit circle at the endpoint
of this line segment, that lies at angle fee for the positive x axis. Sign fee is equal
to the y coordinate cosine fee is equal to the x coordinate. And tangent of fee is given
by the ratio of the two, which works out to negative 0.3639 up to four decimal places.
This video gives a method for calculating sine cosine and tangent in terms of the unit
circle. Starting from the positive x axis, you draw the angle theta going counterclockwise.
You look at the coordinates of the point on the circle where that angle ends. And the
cosine of that angle theta is the x coordinate, sine of theta is the y coordinate, and tangent
of theta is the ratio. This video gives three properties of the trig function sine and cosine
that can be deduced from the unit circle definition. Recall that the unit circle definition of
sine and cosine for angle theta is that cosine theta is the x coordinate. And sine of theta
is the y coordinate for the point on the unit circle at angle theta. The first property
is what I call the periodic property. This says that the values of cosine and sine are
periodic with period two pi. And what that means is that if you take cosine of an angle
plus two pi, you get the same thing as just if you took cosine of the angle. So when we
write this down, we're assuming that theta is measured in radians. If we want to measure
theta in degrees, the similar statement is that cosine of theta plus 360 degrees is equal
to cosine of theta, we can make the same statements for sine, sine of an angle plus two pi is
equal to sine of the original angle, here, the angle being measured in radians. If we
want to measure the angle in degrees, the statement is that sine of theta plus 360 is
equal to sine of theta, we can see why this is true from the unit circle definition of
sine and cosine. This is our angle theta, then theta plus two pi, the plus two pi adds
a full turn around the unit circle to our angle, so we end up at the same place, theta
and theta plus two pi are just two different names for the same location on the unit circle.
And since sine and cosine give you the y and x coordinates of that point on the unit circle,
they have to have the same value. Similarly, if we consider an angle theta, and an angle
theta minus two pi, the minus two pi means we go the other direction around the unit
circle clockwise, we still end up in the same place. And therefore, cosine of theta minus
two pi, the x coordinate of that position is the same thing as cosine of theta, sine
of theta minus two pi is the same thing as sine of theta, the same statements hold if
we add or subtract multiples of two pi. For example, cosine of theta plus four pi is still
the same thing as cosine of theta. This time, we've just gone to turns around the unit circle
and still gotten back to the same place. So if we want to find cosine of five Pi, that's
the same thing as cosine of pi plus four pi, which is the same thing as cosine of pi. Thinking
about the unit circle, pi is halfway around the unit circle. So cosine of pi means the
x coordinate of this point right here. Well, that point has coordinates negative one, zero,
so cosine of pi must be negative one. If I want to take sine of negative 420 degrees,
well, that's sine of negative 360 degrees, minus 60 degrees, which is the same thing
as sine of minus 60 degrees. Thinking about the unit circle, minus 60 degrees, means I
start at the positive x axis and go clockwise by 60 degrees that lands me about right here.
And so that's one of the special angles that has an x coordinate of one half a y coordinate
of negative root three over two. And therefore sine of negative 60 is negative root three
over two the y coordinate. The next property I call the even odd property, it says that
cosine is an even function, which means that cosine of negative theta is the same thing
as cosine of theta, while sine is an odd function, which means that sine of negative theta is
the negative of sine of theta. To see why this is true, let's look at an angle theta.
And the angle negative theta. A negative angle means you go in the clockwise instead of counterclockwise
direction from the positive x axis.
The coordinates of this point by definition, r cosine theta sine theta, whereas the coordinates
of this point are cosine negative theta, sine of negative theta. But by symmetry, these
two points have the exact same x coordinate, and therefore cosine of theta must equal cosine
of negative theta, while their y coordinates have the same magnitude, but opposite signs.
This one's positive and this one's negative. Have, therefore, sine of negative theta is
the negative of sine of theta. Let's figure out if tan of theta isn't even or odd function.
Well, we know that tan of negative theta, tangent by definition, is sine over cosine.
Well, we know that sine of negative theta is the negative of sine of theta, whereas
cosine of negative theta is cosine of theta. Therefore, we're getting negative sign theta
over cosine theta, which is negative tan of theta. Since tan of negative theta is the
negative of tan of theta, tan theta is an odd function. The last property on this video
is the Pythagorean property, which says that cosine of theta squared plus sine of theta
squared is equal to one. A lot of times this property is written with this shorthand notation,
cosine squared theta plus sine squared theta equals one. But this notation, cosine squared
theta just means you take cosine of theta and square it. This property is called the
Pythagorean property, because it comes from the Pythagorean Theorem. Let me draw a right
triangle on the unit circle. I'll call this angle theta. So the coordinates of this endpoint
here are cosine theta sine theta. Since this is supposed to be a unit circle, the hypotenuse
of my right triangle has length one, the base of my right triangle is just cosine theta,
same thing as the x coordinate of this point. And the height of my triangle is the y coordinate
of the point sine theta. Now the Pythagorean theorem says that this side length squared
plus that so that squared equals one squared, since one squared is the same thing as one,
that gives me the Pythagorean property. But tigrayan property is handy for computing values
of cosine given values of sine and vice versa. And this problem, we're told that sine of
t is negative two sevenths. And T is an angle that lies in quadrant three. When we say the
angle lies in quadrant three, that means the terminal side of the angle lies here in quadrant
three. One way to find cosine of t is to use the fact that cosine squared t plus sine squared
t is equal to one. That is cosine of t squared plus negative 2/7 squared is equal to one,
I can write this as cosine of t squared plus 4/49 is equal to one and so cosine of t squared
is equal to one minus 4/49, which is 4540 nights. taking the square root of both sides,
that goes for the cosine t is plus or minus the square root of 45 over 49, that's plus
or minus the square root of 45 over seven. Now since we're in the third quadrant, we
know that cosine of t, which represents the x coordinate of this point, must also be negative.
Therefore, cosine of t is going to be negative square to 45 over seven. It's also possible
to solve this problem using the Pythagorean theorem for right triangles directly. If we
look at the fact that sine of t is negative two sevenths and ignore the negative sign
for now, we can think of this information as telling us that we have a right triangle
angle theta, whose opposite side is two, and whose high partners is seven.
We call this side here a them but Tiger in theorem says us a squared plus two squared
is seven squared. So a squared plus four is 49. So a squared is 45. And a is plus or minus
the square root of 45. Since I'm worrying about a triangle, I'm going to use the positive
value. Now, cosine of t is going to be adjacent over hypotenuse. So that's going to be the
square root of 45 over seven. Now I go back to thinking about positive and negative signs.
And I noticed that since I'm in the third quadrant, my co sign is be negative, so I
just stick a negative sign in front. This alternative solution, this is many of the
same ideas as the previous solution, and ultimately gets us the same answer. This video gives
three properties of trig functions, the periodic property, the even odd property, and the Pythagorean
property. This video is about the graphs of sine and cosine. I want to graph the functions
y equals cosine t and y equals sine t, where t is in radians, I'm gonna think of this being
the t axis, and this being the y axis. One way to do this is to plot points. So I'll
fill in this chart, using my knowledge of special angles on the unit circle. These points
will be easier to graph, if I convert them all to decimals. Now plot the points for cosine
and connect the dots to get a graph of y equals cosine t from t equals zero to t equals two
pi. To continue the graph for t values less than zero or bigger than two pi, I could plot
more points. Or I could just use the fact that the cosine values repeat. If I add or
subtract two pi to the my angle T, I'll be at the same place on the unit circle. So my
cosine will be exactly the same. Therefore, my values of cosine, which are represented
by my y values on this graph, repeat themselves. For example, when my T value is two pi plus
pi over six about like here, it's cosine is the same as the cosine of just pi over six.
So I'll take this dot here and repeat it over here. Similarly, the when t is like two pi
plus pi over four, I get the same value of cosine is when it's just pi over four. So
this.is going to repeat. And I can continue repeating all my dots. This one repeats over
here at two pi plus, say pi over three. And so my whole graph will repeat something like
this. It also repeats on this side, something like this. Since subtracting two pi from my
t values will also give me the same value of cosine. We can also plot points to get
a graph for sine and extend it by repetition. Going forward, I'll usually write the function
sine and cosine as y equals cosine of x and y equals sine of x. When I read it this way,
notice that x now refers to an angle, while y refers to a value of cosine, or sine. That's
a different meaning of x and y, compared to when we're talking about the unit circle,
where x refers to the cosine value, and y refers to the sine value. Now let's look at
some properties of the graphs of sine and cosine. The first thing you might notice is
that the graph of cosine and the graph of sine are super similar to each other. In fact,
you can think of the graph of cosine as just being the graph of sine shifted to the left
by pi over two. So we can write cosine of x as the sine function of x plus pi over two,
since adding pi over two on the inside, move the graph horizontally to the left by pi over
two. Or we can think of the graph of sine as being constructed from the graph of cosine
by shifting the cosine graph right by pi over two, that means we can write sine of x as
equal to cosine of x minus pi over two, since subtracting pi over two on the inside, shifts
the cosine graph to the right by pi over two.
Next, let's look at domain and range. The domain of sine and cosine is all real numbers.
All right, that is negative infinity to infinity, but the range is just from negative one to
one. That makes sense, because sine and cosine come from the unit circle. The input values
for the domain come from angles. And you can use any numbers and angle positive or negative
as big as you want, just by wrapping a lot of times around the circle. The output values
for the range, that is the actual values of sine and cosine come from the coordinates
on the unit circle. And those coordinates can't be any bigger than one or any smaller
than negative one. So that gives us a range. As far as even an odd behavior, you can tell
from the graph. Here's cosine, that it's symmetric with respect to the y axis and so it must
be even. Whereas the graph of sine is symmetric with respect to the origin and must be odd.
The absolute maximum value have these two functions is one and the absolute minimum
value is negative one. We can also use the words midline amplitude and period to describe
these two functions. The midline is the horizontal line, halfway in between the maximum and minimum
points. Here, the midline is y equals zero, the amplitude is the vertical distance between
a maximum point and the midline. You can also think of the amplitude as the vertical distance
between a minimum point and the midline, or as half the vertical distance between a midpoint
and a max point. For the cosine function and the sine function, the amplitude is one. A
periodic function is a function that repeats at regular horizontal intervals. The horizontal
length of the smallest repeating unit is called the period for Y equals cosine of x, the period
is two pi. Notice that the period is the horizontal distance between successive peaks, or maximum
points, or between successive troughs, or minimum points. algebraically, we can write
cosine of x plus two pi equals cosine of x and sine of x plus two pi equals sine of x
to indicate that the functions repeat themselves over an interval of two pi and have a period
of two pi. In this video, we graphed y equals cosine of x and y equals sine of x. and observe
that they both have a midline at y equals zero, an amplitude of one and a period of
two pi. sine u sort of functions are functions that are related to sine and cosine by transformations
like stretching and shrinking and shifting. This video is about graphing these functions.
Let's start by graphing the function, y equals three sine of 2x. This function is related
to the function y equals sine x. So I'll graph that first. Now, the three on the outside
stretches this graph vertically by a factor of three, while the two on the inside compresses
that horizontally by a factor of one half. If instead I want to graph y equals three
sine 2x plus one, this plus one on the outside shifts everything up by one unit. Let's compare
the midline amplitude and period of our original y equals sine x are transformed y equals three
sine 2x. And our further transformed y equals three sine 2x plus one, the original sine
has a midline at y equals zero, an amplitude of one and a period of two pi. For the transformed
function, y equals three times sine of 2x. The two on the inside shrinks everything horizontally
by a factor of one half. So it changes the period of two pi into a period of one half
times two pi,
which is pi.
Since the two on the inside only affects x values and horizontal distances, it doesn't
affect the midline, which is a y value, or the amplitude, which is a vertical distance.
But the three on the outside does affect these things. Well, in particular, it affects the
amplitude, since everything is stretched out vertically by a factor of three, the amplitude
of one get stretch to an amplitude of three. In this case, the midline doesn't actually
change, because multiplying a y value of zero by three is still a y value of zero. Now on
the third function, we've taken the second function and added one on the outside, so
that shifts everything up by one. Therefore, instead of having a midline at y equals zero,
we now have a midline at y equals one. The amplitude doesn't change though it's still
three because shifting everything up by one doesn't affect the distance between the mid
mind and the end the maximum point. Also, the period is still pi since the period is
a horizontal measure, and adding one on the outside only affects vertical things. Now
next, let's graph the function y equals three times sine of two times quantity x minus pi
over four. This function is very closely related to the First function we graphed on the previous
page, that was y equals three sine of 2x. In fact, if we give the name f of x to that
function, and maybe we can call g of x, this other function, then we can get g of x by
taking f of x and plugging in x minus pi over four in for x. In other words, g of x is f
of x minus pi over four. This relationship gives me an idea for graphing g of x, the
function we want to graph, we can first graph f of x, we already did that on the previous
page. And then we can shift its graph to the right by pi over four, because that's what
you do when you subtract a number on the inside of a function. So here's the graph of y equals
three sine 2x. Recall that it's just the graph of sine stretched vertically by a factor of
three, and shrunk horizontally by a factor of one half. Now, to graph the function that
I want, I'm going to shift this graph over by pi over four to the right. Notice that
since I had my function written in factored form, I could just read off the horizontal
shift. But if I had written it instead, as y equals three sine 2x minus pi over two,
which is algebraically equivalent, it would be easy to get confused and think that I needed
to shift over by pi over two. So it's best to factor first, before figuring out what
the shift is, we're factoring out the coefficient of x. If instead, we wanted to graph this
function, same as the one we just graphed, it's just with a minus one on the outside,
that minus one would just bring everything down by one. Let's take a moment to look at
midline amplitude, and period for the original parent function, y equals sine of x, and our
final transformed function, y equals three sine of two times quantity x minus pi over
four minus one, our original sine function has midline at y equals zero amplitude of
one and period of two pi. For our transform function, the three on the outside stretches
vertically, so it makes the amplitude three. The minus one on the outside shifts everything
down by one. So it brings the midline, y equals zero, down to Y equals negative one, the two
on the inside, shrinks everything horizontally by a factor of one half. So the period becomes
one half times two pi, which is pi. Finally, there's a horizontal shift going on our transformed
function shifts to the right, by pi over four, this horizontal shift is sometimes called
the phase shift. The function we just analyzed was y equals three sine 2x minus pi over four
minus one, which could also be written as y equals three sine 2x minus pi over two minus
one. This is a function of the form y equals a sine B x minus c plus d, where b is positive.
If we have a function of this form, or the similar function with cosine in it, then we
know that the midline is going to be at y equals D. That's because the original midline
of sine or cosine at y equals zero gets shifted up by D, we know that the amplitude is going
to be a because this A multiplied on the outside stretches everything vertically by a factor
of A.
to be a little more accurate, we should say the amplitude is the absolute value of A in
case a is negative. If a is negative, then that amounts to a vertical reflection or a
reflection over the x axis. We know that the period of the original sine or cosine is two
pi. And we know that this factor of B amounts to a horizontal shrink by a factor of one
over B or I guess it could be a horizontal stretch by a factor of one over b If b is
less than one, so because we're starting with a period of two pi, and we're multiplying
by one over B, our new period is going to be two Pi over B. The trickiest thing is the
horizontal shift. And to get that right, I like to factor out this B for my equation.
So instead of writing y equals a cosine bx minus c plus d, I'm going to write y equals
A cosine B times quantity x minus c over b plus d. Similarly, if it's a sine function,
I write y equals a sine B times x minus quantity c over b plus d, then I can read off the horizontal
shift as C over B. And that'll be a shift to the right, if C over B is positive and
a shift to the left, if C over b is negative, this might seem backwards from what you're
used to, but it's because we have that minus sign there. So if C over B is positive, we're
actually subtracting on the inside. So that shifts right, if C of b over b is negative
minus a negative is actually adding something, and that's why it shifts it to the left. So
as one final example, say I wanted to graph y equals 1/3, cosine of one half x plus three
minus five, that would have a midline at y equals minus five, an amplitude of 1/3, a
period of two pi divided by one half, which is four pi, and a horizontal shift. Better
rewrite this horizontal shift of six units to the left, the horizontal shift is sometimes
called the phase shift. And that's all for graphs of sinusoidal functions. This video
is about graphing the trig functions, tangent, secant, cotangent and cosecant. To gain an
intuition for the graph of y equals tangent of x, I think it's handy to look at the slope
of a line at angle theta on the unit circle. The slope of this line is the rise over the
run. But the rise is given by sine of theta, and the run is given by cosine of theta. So
the slope is given by sine theta over cosine theta, which is simply tan of theta. So if
I want to graph y equals tan of x, I can think of x as being the angle and y as being the
slope of the line at that angle. Notice if the angle is zero, the slope is zero. But
as the angle increases towards pi over two, the slope gets bigger and bigger heading towards
infinity. As the angle goes from zero towards negative pi over two, the slope is getting
negative and heading towards negative infinity at exactly pi over two and negative pi over
two, we have a vertical line. And so the slope is undefined.
Using this information, let's graph a rough sketch of y equals tan x. Remember, we're
thinking of x as the angle and y as the slope, we're going to go between an angle of negative
pi over two and pi over two. So we said that the slope was zero when the angle is zero,
and then it heads up towards positive infinity as we go towards the angle goes towards pi
over two with an undefined value at pi over two, it goes negative heading towards negative
infinity as the angle heads towards negative pi over two, also with an undefined value
at negative pi over two. You can also verify that for angles slightly bigger than pi over
two, we have the same line as for angles that are approaching negative pi over two, and
therefore this picture repeats and it turns out that tangent is periodic With period not
to pi, like sine and cosine, but just pi, the period of pi makes sense because if you
take a line and rotate it by 180 degrees, it's the same line with the same slope, and
therefore has the same value of tangents. In this graph of y equals 10x, notice that
the x intercepts, all right values of x have the form negative two pi, negative pi, zero,
pi, two pi, etc, you can write that as pi times k, where k is an integer, that is a
positive or negative whole number or zero. This makes a lot of sense because tangent
of x is sine of x over cosine of x. And so you're going to get x intercepts, that's where
y is zero, which is where the numerator is zero, and sine x is zero, at values of the
form pi, two pi, and so on. From the graph, you can see the vertical asymptotes are at
values like negative three pi over two, negative pi over two, pi over two, and three pi over
two, these values can be written as pi over two times k, where k is an odd integer. Again,
this makes sense from the definition of tangents since vertical asymptotes, will occur where
the denominator is zero, and cosine x is zero, at numbers, like negative pi over two pi over
two, three pi over two, and so on, the domain of tangent is the x axis for which it's defined.
So that's going to be everything except for the vertical asymptotes, we can write that
as x such that x is not equal to pi over two times k, for K, an odd integer. The range
or the y values go all the way from negative infinity to infinity. And the period, as we
mentioned previously, is pi. Since the smallest repeating unit has a horizontal width of pi,
to graph y equals secant x, I'm going to remember that secant is one over cosine. So if I start
with a graph of cosine, I can take the reciprocal of the y values to get the graph of secant,
the reciprocal of one is one, the reciprocal of zero is undefined, so I'm not going to
have a value at pi over two, negative pi over two, three pi over two, or negative three
pi over two. When I take the reciprocal of numbers, just less than one, I'm going to
get numbers just greater than one, but I would take the reciprocal of positive numbers getting
close to zero, I'm going to get really big positive numbers going up towards infinity.
Similarly, on the other side, over here, I have numbers close to zero, but negative,
so their reciprocals will be negative numbers heading towards negative infinity. The reciprocal
of negative one is negative one. And similarly here, so I'm getting kind of positive and
negative buckets and upside down buckets as the graph of my sequence.
Notice that secant has a period of two pi, which makes sense, since cosine has a period
of two pi, it has a range that goes from negative infinity to negative one inclusive, and from
one to infinity. That makes sense because the range of cosine is between one and negative
one, and we're taking the reciprocal of those values. The domain is everything except for
the vertical asymptotes. Now the vertical asymptotes are where cosine is zero. So that
is at values of the form pi over two, three pi over two, etc. That's values of the form
pi over 2k, where k is an odd integer. So the domain is going to be x values such that
x is not equal to pi over 2k. For K and odd integer. The x intercepts of secant Well,
it doesn't have any, because you can't take one over something and get the numbers zero
for your y value. We've seen the graph of y equals Tana x and y equals secant x. This
is the graph of y equals cotangent x. It looks similar to the graph of tangent x, it's just
a decreasing function instead of an increasing one, and it has its vertical asymptotes. And
it's x intercepts in different places. Finally, this green graph is the graph of y equals
cosecant x. It's related to the graph of sine x, since cosecant is one over sine x, and
in fact, if I draw the graph of sine x in between, you can see how it kind of bounces
off. Because it's the reciprocal. I encourage you to memorize the general shape of these
graphs, you can always figure out the details by thinking how about how they're related
to the graphs of cosine of x, and sine x. This video is an introduction to solving trig
equations. Let's start with the equation two cosine x plus one equals zero, I want to find
all the solutions in the interval from zero to two pi, and then get a general formula
for all solutions, not just those in that interval. Let me start by rewriting this equation
to isolate the tricky part, which is cosine of x. So I'm going to write to cosine x equals
negative one, and then divide both sides by two. Now I'm looking for the angles x between
zero and two pi, whose cosine is negative one half. Since negative one half is one of
the special values on the unit circle, I can use my knowledge of the unit circle, to see
that the angle between zero and two pi must be either two pi over three, or four pi over
three, my answer needs to include both of these values. There are no other spots on
the unit circle whose cosine is negative one half. But there are more angles, because we
can always take one of these angles and add multiples of two pi to it. So if I want to
find all solutions, I can take these two principles solutions, two pi over three, and four pi
over three, and simply add multiples of two pi to them. For example, two pi over three
plus two pi, or two pi over three minus two pi, two pi over three plus four pi, and so
on. A much more efficient way to write this is to write two pi over three plus two pi
times k, any integer that is any positive or negative whole number or zero. Similarly,
I can write four pi over three plus two pi k, to capture all solutions, based on the
principal solution for pi over three by adding and subtracting multiples of two pi.
This is my final solution. Next, let's look at a tricky equation involving tangent. As
usual, I'm going to start out by cleaning things up and isolating the tricky part, which
in this case is tangent. So let me add tangent to both sides. That'll give me three, tan
x equals the square root of three. And so tan x is the square root of three over three.
The square root of three over three looks suspiciously similar to value the value of
square to three over two, which is a special value on my unit circle. So my suspicion is
that my unit circle will again help me find this value of x without a calculator. Recall
that tan x is sine x over cosine x. So I'm looking for angles on the unit circle between
zero and two pi with a ratio of sine over cosine will give me square root of three over
three, I actually only need to look in the first quadrant and the third quadrant, because
those are the quadrants where a tangent is positive. And I really only need to look at
angles whose either sine or cosine has a squared of three in it. So by trial and error, I can
see that tan pi over six, which is sine pi over six over cosine pi over six will give
me one half over root three over two That's the same thing as one half times two over
three, which is one over root three. If I rationalize that, I get root three over three,
so that value works. If I try tan of pi over three, instead, I get root three, which is
not equal to root three over three. So pi over three doesn't work. Similarly, I can
work out some the values in the third quadrant, and see that seven pi over six works. But
four pi over three does not. So my answer to part A includes just the two values, pi
over six, and seven pi over six. Now if I want to find all solutions, not just those
in the interval from zero to two pi, I noticed that I can take one of these principal solutions,
and add multiples of two pi to it, because that'll give me the same angle. So I get pi
over six plus two pi k, and pi over six, sorry, seven pi over six plus two pi, K, any integer.
This is a correct answer. But it's not as simple as it could be. Notice that seven pi
over six over here on the unit circle is exactly pi more than pi over six. So instead of taking
both of these and adding multiples of two pi to them, I could get all the same answers
by just taking one of them and adding multiples of pi to it. So a more efficient answer is
to say that x equals pi over six, plus pi times k, for K any integer. This will still
capture all the same solutions. Because when k is even, I'll get this family solutions.
And when k is odd, I'll get this family. For example, when k is one, pi over six plus one
times pi is just the original seven pi over six. If you think about the fact that tangent
has a period of pi, instead of two pi, it makes a lot of sense that you should be able
to write the solutions in this form. In this video, we solved basic trig equations by first
isolating sine, or tangent, or the same thing would work with cosine. And then using the
unit circle, to find principal solutions. Principal solutions are just solutions between
zero and two pi. And then adding multiples of two pi to these principal solutions to
get all solutions. For tangent, we noticed that it was equivalent to just use one principal
solution and add multiples of pi instead of two pi. In this video, we'll introduce the
idea of the derivative using graphs, secant lines and tangent lines.
So I have a function here, drawn in black, the function is y equals f of x. But actually
here f of x equals x squared. I also have a tangent line to my function drawn in red.
This tangent line is the tangent line at the point 1.5 2.25. By a tangent line, I mean
a line that touches the graph of my function at this one point, and heads off in the same
direction as the function. Well, normally to compute the slope, we need two points.
But for the tangent line, we really only have the exact coordinates of this one point, we
could approximate the slope by guessing the coordinates of some other points on the red
line. But in the long run, we'll end up with a more accurate estimate if we do something
else. So what we're going to do instead is we're going to calculate the slope of a secant
line. A secant line is a line that goes through two points on my graph. So in this case, my
secant line is going through my original point, and this other point at x equals three. So
that's the point three, three squared or three nine. Okay, so here's my secant line. To calculate
the slope of my secant line, I use the fact that slope is rise over run. Or in other words,
it's the change in y over the change in x. And that's going to be written in function
notation. F of three minus f of 1.5 is giving me the change in y and three minus one point
gives me the change in x. Wilson's f of x is x squared, this is the same thing as three
squared minus 1.5 squared over three minus 1.5. So that's just nine minus 2.25, over
1.5, which ends up as 4.5. So 4.5 is the slope of this second line. Well, the idea here is
that the slope of my secant line is an approximation of the slope of my tangent line. But in this
example, really the secant line that I've used, it's, its slope is only a very rough
approximation of the tangent line, not very accurate, all. So how could I get a better
approximation of the slope of my tangent line? Well, one thing I could do is I could use
as my second point, instead of using this point way out here, I could use a point closer
to my first point. So for example, I could use the point two, f of two. In other words,
the point two, four, as my second point for my second line, so let me draw that. And let
me calculate its slope, which after some arithmetic is going to give me an answer of 3.5. Well,
I could continue to pick second points for my secant line, closer and closer to my first
point. And I should end up with more and more accurate approximations of my tangent line,
let me make a little chart for this. For my next secant line, I could take my second point,
something pretty close to 1.5, say 1.6. And after some arithmetic, I get a slope of 3.1.
To take my second point to be 1.51, that's going to give me a slope of 3.01, and so on.
To write this more generally, if I take my second point as x, then the slope of my secant
line is going to be given by again the rise over the run. So that's going to be f of x
minus f of 1.5 divided by x minus 1.5. Change in y over the change in x, that's my slope.
Now, there's no reason I necessarily have to take my second point to be on the right
side of my first point, I could be using stead points on the left side here. Continue with
my chart, letting my second point be one, I can do the same computations to get a slope
of a secant line of 2.5. Here, I could get even closer on the left, say something like
1.4 and get a slope of 2.9, and so on. In general, if I have a point x f of x, and I
use a secant line through that point in our original point, I calculate the slope as change
in y over change in x, which is going to be f of 1.5 minus f of x divided by 1.5 minus
x.
Actually, I can rewrite this a little bit to make it look more like the expression up
here. If I multiply the numerator and denominator by negative one, then I can rewrite this as
f of x minus f of 1.5 divided by x minus 1.5. So that these two expressions look exactly
the same. So the only difference here on my mind is that over here, I was thinking of
x as being a little bit bigger than 1.5. And here I'm thinking of x as being a little smaller
than 1.5. But I get the exact same expression for the slope of the secant line either way.
Now, this process of picking points closer and closer to our original point from the
left, and from the right, should remind you of limits. And indeed, the slope of the tangent
line is the limit as x goes to 1.5 of the slope of my secant lines, which are given
by this expression. This quantity is so important that's given its own name, it's called the
derivative of f of x at x equals 1.5. So in other words, the derivative which is written
as f prime at 1.5, is the limit as x goes to 1.5 of f of x minus f of 1.5 over x minus
1.5. Now based on our numerical tables, for example, here, we can see that that limit
seems to be heading towards three, whether X approaches 1.5, from the right or from the
left. So I'll write down the answer of three. You know, it's pretty strong. If you wanted
to have a really precise argument. We'd actually need to use algebra to compute this limit.
Exactly using the formula for the function itself, f of x equals x squared. And we'll
do examples like that in a future video. But for now, the main point is just that the slope
of the tangent line is the limit of the slope of the secant lines, which is given by this
formula. For now, let's look at an animation that shows how the slope of our secant lines
approach the slope of our tangent line. So this black curve here is the function y equals
x squared. The red line is the tangent line through the point where x equals 1.5. And
the blue line is a secant line that goes through the point with x coordinate 1.5. And a second
point with x coordinate 2.5. The points are shown here on the right. So I'm going to use
this slider here and drag my second point closer to my first point. So notice how as
the x coordinate, my second point gets closer and closer to 1.5. My secant line is getting
closer and closer to my tangent line. So the slope of my tangent line really is the limit
of the slope of my secant lines, as my x coordinate of my second point goes to 1.5. So this is
true, even if I start with my second point on the left instead of the right, as I drag
that second point closer and closer to the first point, the slope of my secant lines
gets closer and closer to the slope of that red tangent line. We saw in our example, that
the slope of our tangent line, or the derivative at 1.5, was given by the limit as x goes to
1.5 of f of x minus f of 1.5 divided by x minus 1.5. Well, in general, the derivative
of a function y equals f of x at an x value a is given by f prime of A equals the limit
as x goes to a of f of x minus F of A over x minus a, the function is said to be differentiable
at A. If this limit exists. In particular, both the limit from the left and a limit from
the right have to exist and be equal for the function to be differentiable
at x equals a.
There's another equivalent version of the definition of derivative that's very common
and very useful. If we're looking at the graph of a function, I'm trying to calculate the
slope of the secant line between the points A, F of A and x f of x, then let's introduce
the letter H to be the quantity x minus a. So h represents the run, when I'm calculating
the slope of this secant line, I can write H equals x minus a, or equivalently x equals
a plus H. And so I can rewrite the definition of derivative in terms of H as f prime of
A equals the limit as x goes to a of f of a plus h minus F of A divided by h, just by
substituting in this expression for x. And h for x minus a. Well as x goes to a, x minus
a is going to zero. In other words, h is going to zero. So this is equivalent to the limit
as h goes to zero of f of a plus h minus f of a over h. One way to think of this is that
we're just relabeling this point right here, as the point A plus h, f of a plus H. And
the slope of the tangent line is still the limit of the rise over the run as the run
goes to zero, this is the definition of derivative that we'll use most frequently going forward
when we actually calculate derivatives based on the definition in future videos. But for
now, let's look at some examples to practice recognizing the derivative rather than computing
it. So each of these following two expressions are supposed to represent the derivative of
some function at some value a. So for each example, we're supposed to find the function
and figure out the value of A. Now remember, we've got two definitions of derivative going
on. They're both equivalent, but they look different. One of them looks like derivative
of f at a is the limit as x goes to a of f of x minus F of A over x minus a. And the
other version is the limit as h goes to zero of f of a plus h minus f of a over h. Now,
you might notice That our first expression looks more like this first definition, because
x is going to some number that's not zero. And we have both x and a number in the denominator
here. Whereas our second expression looks more like the second definition, we've got
the age going to zero, and we've just got the age on the denominator here. Okay, so
let's look at this first one here. First, let's figure out what A is here. It seems
like a has got to be negative one. Since x is approaching negative one. That kind of
makes sense, because now here on the denominator, x plus one could be thought of as x minus
negative one. So that's our x minus a with a is negative one. Okay, great. So we've got
a, now we need to find an F. And we will need the numerator here to look like f of x minus
f of a, well, let's try the simplest thing, we can, let's try f of x equals x plus five
squared, then X plus five squared is our f of x. and f of a is our f of negative one
is going to be negative one plus five squared, which is 16. So that matches up perfectly,
we've got our f of x here, our f of a here, and our x minus a at the bottom. That's exactly
the definition of derivative done the first one. All right, now the second one. Now, again,
we need to figure out what a is. And we need to find figure out what f is. This part of
the expression here is supposed to be f of a plus h. So I'm going to guess, make a guess
here, that f of x should be three to some power, let's just try three to the X and see
how that works. Now we need this nine to be f of a. So nine has to be three to the A.
And the only way that can work is if a is two. So continuing on the top, we need f of
a plus h, that is f of two plus h,
two B three to the two plus H and that actually works perfectly. If our f of x is three to
the x, it's all falling into place. So we've got our f of two plus h, r f of two, and r
h. And it all works where f of x being three to the x and a being two. In this video, we
introduced the idea of derivative as a slope of a tangent line. And we gave two equivalent
definitions of the derivative in terms of limits. We'll continue with some interpretations
of derivatives in the next video. This video is algebraically intensive, and is concerned
with calculating derivatives using the limit definition of derivative. Well, there are
actually two versions of the limit definition of derivative. And I'll mostly use this one,
the limit as h goes to zero of f of a plus h minus f of a over h. If you're interested,
you can try reworking the problems using the alternative definition of derivative. The
limit as x goes to a of f of x minus F of A over x minus a first example, find the derivative
of f of x, which is one over the square root of three minus x at x equals negative one.
Well, in other words, we want to find f prime of negative one. So that's the limit as h
goes to zero of f of negative one plus h minus f of negative one over h. Using our definition
of f, that's the limit of one over the square root of three minus negative one plus h minus
one over the square root of three minus negative one, all over h. Let me clean this up a bit.
So this is one over the square root of, let's see, it's three minus negative one, so that's
three plus one, or four minus h minus one over the square root of four
over h. And I guess I can replace the square root of four with two.
Now, unfortunately, I can't just evaluate this by plugging in H equals zero, because
if I try that, I get one of these zero over zero indeterminate forms. You'll run across
these a lot when calculating derivatives by definition, it kind of makes sense because
remember the context where computing slopes of secant lines as these points get closer
and closer together. So our rise and our runs are both going to zero. So it makes sense,
we'll get these zero or zero indeterminate forms, we have to use our algebraic tricks
that we learned before, for rewriting our expression in a way that we can calculate
the limit. And I see two things going on here, there's square roots lurking, and there's
also fractions. So it's anybody's guess which trick I might want to apply First, the trick
for square roots, which would be multiplying the top and the bottom by the conjugate. And
the trick for fractions, which would be adding together my fraction for the common denominator.
I guess I'll try my fraction trick first. So my common denominator for my two fractions
here, is just the product of these two denominators. So that's the square root of four minus h
times two. Let me rewrite my fractions with this common denominator. Continuing here,
I get the limit of two minus the square root of four minus h over squared, a four minus
h times two, all over h. instead of dividing by H, let me multiply by one over h. And let's
see here, let's see if we can evaluate by plugging in H equals zero at this stage. Unfortunately,
when I try to plug in, I'm still getting the zero of zero indeterminate form. But I'm not
out of tricks, I haven't used the conjugate trick yet. So let's try multiplying the top
and the bottom by the conjugate of the top. Once I multiply out here, I'll get four plus
two square to four minus h minus two square to four minus h, minus the square root of
four minus h squared, that's going to cancel out nicely. And on the bottom, I have two
h squared of four minus h times two plus square to four minus h, I'll leave that factored
for now. That's good thing, I have unlimited space here, I'm kind of need it. Now on the
numerator, I'm going to get four minus four minus h, is carrying the denominator along
for the ride now, oh, I see something good. I see in the numerator, we're getting a four
minus four plus H. Subtracting out those fours to zero, and then cancelling out, my H is
that divide by each other, I think I finally got something that I can evaluate without
getting a zero over zero indeterminate form. All right, so here we've got. So as h goes
to zero, I'm just going to get one over two times the square root of four times two plus
a squared of four, which equals 1/16. So by now, you may have forgotten what the original
problem was, I think I have it, let's go back up here, we were looking for the derivative
of f of x, which was one over a squared of three minus x at x equals negative one, we
set up the limit definition, did a bunch of algebra first adding together fractions, then
using the conjugate trick, and eventually found that that derivative equaled 1/16. The
algebra doesn't get much harder than this problem here. In this next example, we're
asked to find the equation of the tangent line to y equals x cubed minus 3x at x equals
two. So the slope of the tangent line is given by the derivative, f prime of two. So let's
calculate the derivative first. f prime of two is the limit as h goes to zero of f of
two plus h minus f of two, all over h. So that's the limit of two plus h cubed minus
three times two plus h minus
two cubed minus three times two. I'm just plugging in first two plus h for x in the
definition of my function, and then I'm plugging in two for x. Now all of that needs to be
over h. Once again, if I try to plug in zero for H in these expressions, I'm just gonna
get something that all cancels out to zero at the top, and also have zero at the bottom,
one of my classic zero over zero indeterminate forms. So instead, I need to use algebra to
simplify things. And hope I can calculate the limit after that. So a good trick for
simplifying here is to multiply out two plus h cubed multiplies out to two cubed, plus
three times two squared times h, plus three times two times h squared plus h cubed. I'm
getting this from the formula for multiplying out of cubic which I've memorized. But you
can also get it more slowly just by writing out two plus h times itself three times and,
and distributing. Now I need to subtract three times two, and three times h. And finally,
I needed to subtract my two cubed and then add my three times to all this over h. Now
if you have my terms cancel out to zero here, two cubed minus two cubed. And let's see I've
got a minus three times two and a plus three times two. And I notice all the terms that
are left have H's and so I'm going to factor out an H from the top from the remaining terms
here. And that gives me let's say, three times two squared, so that's 12, plus six H, plus
h squared minus three over h. Now, h divided by h is one. So I'm just left with the limit
as h goes to zero of 12 plus six h plus h squared minus three, as h goes to zero, I
can just plug in H zero, and I get 12 minus three, which is nine. So my slope of my tangent
line, my derivative is nine. I'm not quite done, I still need to find the equation of
the tangent line, I just know that its slope is nine. So the equation of the tangent line
equation of any line is something like y equals mx plus b. and here m is nine. So I have y
equals 9x plus b, I just need to find the intercept B. Now, usually, to find the intercept,
I need a plug in a point. what point do I have here to plug in? Well, remember, we're
talking about a tangent line here. So we've got the point of tangency, the point where
x equals two, and the corresponding y value is y equals two cubed minus three times two,
or two. So my tangent line has to go through the point to two, which means if I plug in
this point for x and y, I get two equals nine times two plus b, which means that B has to
equal negative 16. So the equation of my tangent line then becomes y equals 9x minus 16. I
found that by first calculating the derivative to get my slope, and then using the point
of tangency, plugging in the x value to get the y value, and plugging Matt in to get b
to finish off the equation. So in this video, we used our tricks for evaluating limits algebraically
to compute some derivatives, using the definition of derivative. This is pretty labor intensive.
So fortunately, pretty soon, we'll learn some shortcut methods for calculating derivatives
without resorting to the definition. We've seen that the derivative of a function y equals
f of x at a point, x equals A represents the slope of a tangent line through the point
A f of a. But if the function f of x represents some practical quantity, like distance as
a function of time, or fuel efficiency as a function of speed, then the derivative will
also have a practical interpretation. This video is about interpreting the derivative
in different contexts.
One of the most famous contexts for interpreting derivatives is problems involving motion.
So let's say I'm on a bike ride, heading straight north from campus. And let's suppose that
y equals f of x represents my distance from campus. So x is the time and hours and y or
f of x is my distance and miles, the distance away from campus. Distance north of campus.
It's kind of fun to see what this graph here means in terms of my bike ride. In particular,
what's going on up here, where my y values reach their maximum. And what about here,
where my function is constant. Please pause the video for a moment and see if you can
make up a story that fits the graph. So here at the top, my distance is no longer increasing,
it's actually starting to decrease. So I must have turned around and be heading back towards
campus again, over here, where my f of x is constant, I probably stopped at a coffee shop
to take a rest or maybe I'm fixing a flat tire. Now let's get to the questions at hand.
Consider these two points, three, f of three, and four F of four, we want to interpret the
slope of the secant line through those two points. Slope is change in y over change in
x. And y here is distance and x here is time. So change in distance over change in time.
Sounds a lot like speed, or more accurately, velocity. Velocity just means speed in a certain
direction, and is positive. If distance is increasing, and negative if it's decreasing.
Speed is the absolute value of velocity, and there's always positive or zero. So in our
case, the velocity here must be negative, because our distance is decreasing. And we
could estimate it very roughly as about, say four minus 12 over four minus three, so about
negative eight miles per hour. But what is this negative eight miles per hour refer
to.
Since we're looking at the change in distance, over this entire hour long interval, the slope
of my secant line gives my average velocity over this interval. It doesn't give my exact
velocity, and exactly three hours or exactly four hours, only my average velocity. If I
want to know my exact velocity at exactly three hours, I need to look instead at the
slope of the tangent line at x equals three. The velocity at an exact instant of time is
sometimes called the instantaneous velocity to distinguish it from the average velocity
over a time interval. Let's think for a minute about why the velocity at exactly three hours
is given by the slope of the tangent line. We saw in a previous video that the slope
of the tangent line is the limit of the slope of the secant lines More precisely, the limit
as x goes to three of f of x minus f of three over x minus three. Wow, well, each of these
ratios represents an average velocity on the interval from three to x. And so the limit
is the limit of average velocities on tinier and tinier intervals of time, one minute,
one second 100th of a second, in the limit, as the length of the time interval goes to
zero, we're going to get the exact velocity at exactly three hours. So to repeat, in this
example, the slope of the secant line represents the average velocity over time interval, and
the derivative at x equals three, written f prime of three, which is also the slope
of the tangent line, that derivative represents the instantaneous velocity at x equals three.
More generally, if f of x represents any quantity that's changing, then the slope of the secant
line represents an average rate of change. While the slope of the tangent line, f prime
of a represents an instantaneous rate of change. Let's see how that works in a couple of other
examples. Let's suppose that f of x represents the temperature of a cup of coffee and degrees
Fahrenheit as a function of time and minutes since he said it on the counter. So let's
interpret the first equation. f of zero is 140. Well, that just means that at time zero,
the temperature is 140 degrees. What about the equation f of 10 minus f of zero is negative
20. That's saying that the temperature goes down by 20 degrees as x the time goes from
zero to 10 minutes. Now, what about this quotient here being equal to negative two? Well, this
quotient looks a lot like the slope of a secant Why, right, so it must be an average rate
of change. And in this context, we have the temperature is decreasing by an average of
two degrees per minute, as x changes from zero to 10 minutes. Finally, the derivative
of f at 15 is negative point five means that at exactly 15 minutes, the temperature is
decreasing at a rate of point five degrees per minute. negative numbers here always mean
decreasing, and f prime is an instantaneous rate of change. Let's look at another example.
Please pause the video and try this one for yourself. Here g of x represents the fuel
efficiency of a Toyota Prius and mpg as a function of x, the speed in miles per hour
that is traveling g of 45 is 52 means that at 45 miles per hour, the fuel efficiency
is 52 miles per gallon. The second statement is saying that as speed increases from 35
to 45 miles per hour, fuel efficiency goes up by 10. That's 10 miles per gallon. The
third statement says that the average rate of change
of fuel efficiency is two miles per gallon per mile per hour, as speed increases from
35 to 40 miles per hour. So going up from 35 to 40, gives you better fuel efficiency
here. On the other hand, when you're going 60 miles per hour, your fuel efficiency is
decreasing at a rate of two miles per gallon per mile per hour. So I bet the optimal fuel
efficiency here occurs somewhere between 40 and 60 miles per hour. In this video, we've
interpreted the slope of the secant line as the average rate of change, and the slope
of the tangent line with the derivative as an instantaneous rate of change. I hope that
these general principles will help you interpret the derivative in a variety of contexts that
you might encounter throughout your life. In this video, we'll think of the derivative
of a function as being a function in itself, we relate the graph of a function to the graph
of its derivative, and we'll talk about where the derivative does not exist. We've seen
that for a function f of x and a number A, the derivative of f of x at x equals A is
given by this formula. But what if we let a very, if we compute f prime of a lots of
different values of a, we can think of the derivative of f prime as itself being a function,
I'm going to rewrite this definition of derivative with x in the place of a just to make it look
a little more like standard function notation. So f prime as a function of x is the limit
as h goes to zero of f of x plus h minus f of x over h. This isn't anything substantially
different from what we've been doing before, it's just a difference in perspective. So
let's do one more example of computing the derivative by hand using the definition, but
a general number x instead of a specific value, the function we're going to use is f of x
equals one over x. And first, let's just write down the definition of derivative in general.
So f prime of x is given by this formula. And using the fact that f of x is one over
x, I can rewrite this as one over x plus h minus one over x all over h. So as usual,
this is a zero over zero indeterminate form. If I plug in zero for H, I'm just going to
get one over x minus one over x on the numerator, which is zero. and plugging in zero for H
gives me zero on the denominator two. So I'll need to use some algebra to rewrite things
to get an A form that I can evaluate it. Let's add together our fractions in the numerator
here. The common denominator I need to use is x plus h times x. So I multiply this fraction
by x over x and the next fraction by x plus h over x plus h. All that's over h. And now
continuing, I get x minus x plus h over x plus h times X and instead of dividing the
whole thing by h here, multiply by one over h to get the limit have x minus x minus h
over x plus h times x times h. Now I can subtract off my x's here. And after I do that, I can
divide my minus h by my H, to get just a minus one on the numerator here. So that's just
the limit of negative one over x plus h times x. And now I'm in a good position because
I can plug in H equals zero and get something that makes sense. Namely, I'm getting a limit
of negative one over x plus zero times x or negative one over x squared as my derivative.
In this example, we're given the graph of a function that's supposed to represent the
height of an alien spaceship above the Earth's surface, we want to graph the rate of change.
The rate of change means the derivative of our function, but we're not given an equation
to work with. So we'll just have to estimate the derivative based on the shape of the graph
by thinking about slopes of tangent lines. I'll start by drawing a new set of axes where
I can graph my derivative. And I'll consider my original function, which I'll call g of
x, piece by piece.
For x values between zero and two, my original function g of x looks like a line, it has
slope negative one, since the rise is negative two, while the run is two, for any point on
the straight line segment, the tangent line will also be a straight line with slope negative
one, and therefore, the derivative will be negative one. For x values between zero and
two, I'm going to ignore the time being what happens when x is exactly zero or exactly
two, and just look at the interval of X values between two and three. Here, g of x is completely
flat. So tangent lines at any of these points will have slope zero. And I'll draw a derivative
of zero. When x is between two and three, I'll postpone worrying about the derivative
when x is exactly three. And just think about the derivative when x is between three and
five, where g of x is flat again, so it's tangent lines will have slopes of zero. And
I'll draw again, a derivative is zero when x is between three and five. Now things get
a little more interesting. As x increases from five to about seven, g of x is an increasing
function. The slope of tangent lines here are positive, starting at about, say, a slope
of three, and decreasing to a slope of zero, when x is seven, I can draw that down here.
As x increases from seven, the tangent lines now have negative slopes, going to a maximum
negative slope of about negative one here, and then heading towards a slope of zero,
when x is just shy of 10. My estimates of three and negative one for the slopes of my
tangent lines are just rough estimates based on approximating the rise over the run. As
x increases from 10, the tangent line slope is positive, and getting steeper and steeper,
so my derivative is going to be positive and increasing. That's the basic shape of the
derivative. Now let's see what happens at these special points like 235 and zero. To
figure out the derivative at x equals two, let's go back to the definition of derivative
as the limit of the slopes of the secant lines. If I draw a secant line, using a point on
the left, I'll just get this line that lines up with this line and has a slope of negative
one. But if I compute the slope of a secant line, using a point on the right, I'll get
a slope of zero. So the limit from the left and the limit from the right of the slopes
of my secret lines will be different. And so my limit does not exist and my derivative
does not exist. And so I'll just draw this as an open circle at x equals two. Next, let's
look at the derivative when x equals three. Remember that the derivative at three is the
limit as h goes to zero, of g of three plus h minus g of three over eight. Well, if H
is bigger than zero, then g of three plus H is going to be about a half, because three
plus H is to the right of three. On the other hand, if H is less than zero, g of three,
plus H is two, because three plus H is actually a number less than three, g of three itself
is equal to one half, based on the filled in bubble here. so if we calculate the limit,
as h goes to zero from the positive side, we get the limit of one half minus one half
over age, which is just the limit of zeros, so that's zero. On the other hand, if we compute
the limit from the left, we get the limit of two minus a half over eight, which is the
limit of three halves over h. And as h goes to zero, that limit is negative infinity.
So once again, the left limit and write limit are not equal. And so the limit of the slopes
of the secant lines does not exist, and there's no derivative at x equals three. And I'll
draw an open circle there to
add x equals five, again, we have a corner. And by the same sort of argument, we can conclude
that the derivative does not exist. And finally, when x equals zero, we can only have a limit
from the right not the left. And so by that sort of technical reason, we don't have a
derivative at that left endpoint either. So we've drawn a rough graph of the rate of change
of the height of our alien spaceship, as it comes closer to Earth beams down to pick up
Earthlings and then makes us escape up to the mothership. It's interesting to observe
that the domain of the original function g of x is from zero to infinity, but the domain
of g prime is somewhat smaller, and just goes from zero to two, then from two to three,
then from three to five, and finally, from five to infinity, missing some places where
the function originally existed. We saw in the previous example, that the derivative
doesn't necessarily exist at all the x values where the original function exists.
Please pause the video for a moment and try to come up with as many different ways as
you can, that a derivative can fail to exist
at an x value x equals a. Well, one kind of boring way that a function can fail to
have a derivative at x equals A is if f of x itself fails to exist. at x equals a, for
example, if it has a hole, like in this picture, we saw in the previous example, with the alien
spacecraft, that a derivative can fail to exist when the function turns a corner. When
we tried to evaluate the derivative in that example, by taking the limit of the slope
of the secant lines, the limit from the left, and a limit from the right did not agree.
A famous example of a function that turns a corner is the absolute value function. For
the absolute value function, f prime of x is negative one, since the slope here is negative
one, when x is less than zero, and it's positive one when x is greater than zero, but f prime
of zero itself does not exist. A function with a casp also fails to have a derivative
at the cusp. In the alien spaceship example, we also saw that F can fail to have a derivative
at a discontinuity. But there's another way that a derivative can fail to exist, even
when f has no cost per corner discontinuity. Let's look at the function f of x equals x
to the 1/3 graphed here, what's going on at x equals zero. At that instant, the tangent
line is a vertical with a slope that's infinite or undefined. So the limit of the slopes of
the secant lines will fail to exist because it'll be infinite. A function is called differentiable
at x equals a, if the derivative exists at a function is differentiable on an open interval,
if f is differentiable at every point in that interval. So all of the examples on the
previous slides are examples of places where a function is not differentiable. All of these
examples are important. But I'm going to focus on the example involving discontinuity. In
general, if f of x is not continuous at x equals a, then f is not differentiable at
x equals a. This is what we saw in the example involving the jump discontinuity. an equivalent
way of saying the same thing is that f is differentiable at x equals a, then f has to
be continuous at x
previous slides are examples of places where a function is not differentiable. All of these
examples are important. But I'm going to focus on the example involving discontinuity. In
general, if f of x is not continuous at x equals a, then f is not differentiable at
x equals a. This is what we saw in the example involving the jump discontinuity. an equivalent
way of saying the same thing is that f is differentiable at x equals a, then f has to
be continuous at x
equals a.
equals a.
However, if all we know is that f is continuous at x equals a, then we can't conclude anything
about whether or not is differentiable there, f may or may not be differentiable at x equals
a. Remember the square root example, the square root of x is continuous at x equals zero,
but it's not differentiable there because of the corner. In this video, we related the
graph of a function to the graph of its derivative. By thinking about the slopes of tangent lines,
we also looked at several ways that a derivative can fail to exist at a point and noted that
if a function is differentiable, it has to be continuous. This video gives a proof that
differentiable functions are continuous. What we want to show here is that if a function
is differentiable at a number x equals a, then it's continuous at x equals a. Let me
call the function f of x. And I'm going to start out by writing down what it means for
f of x to be differentiable at x equals a. That means that the limit as x goes to a of
f of x minus F of A over x minus a exists and equals this finite number that we call
f prime
However, if all we know is that f is continuous at x equals a, then we can't conclude anything
about whether or not is differentiable there, f may or may not be differentiable at x equals
a. Remember the square root example, the square root of x is continuous at x equals zero,
but it's not differentiable there because of the corner. In this video, we related the
graph of a function to the graph of its derivative. By thinking about the slopes of tangent lines,
we also looked at several ways that a derivative can fail to exist at a point and noted that
if a function is differentiable, it has to be continuous. This video gives a proof that
differentiable functions are continuous. What we want to show here is that if a function
is differentiable at a number x equals a, then it's continuous at x equals a. Let me
call the function f of x. And I'm going to start out by writing down what it means for
f of x to be differentiable at x equals a. That means that the limit as x goes to a of
f of x minus F of A over x minus a exists and equals this finite number that we call
f prime
of a. Now I'm going to multiply both sides of this equation
of a. Now I'm going to multiply both sides of this equation
by the limit as x goes to a of x minus a. Now wait a second, before I go any further
on to make sure this is legit. Does this limit actually exist? Well, yeah, because the limit
as x goes to a of x exists, that's just a, and the limit as x goes to a of a exists,
that's a also. So the limit of the difference has to exist. And actually, that limit of
the difference is the difference of the limits, which is just going to be a minus a or a zero.
So I've actually just multiplied both sides by zero and a fancy form.
by the limit as x goes to a of x minus a. Now wait a second, before I go any further
on to make sure this is legit. Does this limit actually exist? Well, yeah, because the limit
as x goes to a of x exists, that's just a, and the limit as x goes to a of a exists,
that's a also. So the limit of the difference has to exist. And actually, that limit of
the difference is the difference of the limits, which is just going to be a minus a or a zero.
So I've actually just multiplied both sides by zero and a fancy form.
This is actually a surprisingly useful thing to do. Because I have a product of a limit
of two limits here, both of which exist. So by the product rule for limits, I can rewrite
this as the limit as x goes to a
This is actually a surprisingly useful thing to do. Because I have a product of a limit
of two limits here, both of which exist. So by the product rule for limits, I can rewrite
this as the limit as x goes to a
of x minus a times f of x minus F of A over x minus a. And canceling these two copies
of x minus a, which is fine to do when x is near a, just not when x equals a, I get that
the limit of f of x minus f of a is equal to this limit over here. Well, we said this
limit was just zero. So my limit on the left is equal to zero.
of x minus a times f of x minus F of A over x minus a. And canceling these two copies
of x minus a, which is fine to do when x is near a, just not when x equals a, I get that
the limit of f of x minus f of a is equal to this limit over here. Well, we said this
limit was just zero. So my limit on the left is equal to zero.
And now I'm so close, I'd like to apply five the limit rule to this difference to break
it up into a difference of limits, but I can't quite do that, because I'm not sure yet that
the limit, as x goes to a of f of x exists, that's sort of part of what I'm trying to
prove as far as continuity. So instead, I think I'm gonna add to both sides, a limit
that I do know exists. And that's the limit
And now I'm so close, I'd like to apply five the limit rule to this difference to break
it up into a difference of limits, but I can't quite do that, because I'm not sure yet that
the limit, as x goes to a of f of x exists, that's sort of part of what I'm trying to
prove as far as continuity. So instead, I think I'm gonna add to both sides, a limit
that I do know exists. And that's the limit
of f of a.
of f of a.
Now, I do know that both of these two limits on the left side exists, so I can use the
limit rule about sums to rewrite this limit.
Now, I do know that both of these two limits on the left side exists, so I can use the
limit rule about sums to rewrite this limit.
Now, on the left side, I can cancel out my copies of f of a, whatever number that is,
and I get that the limit as x goes to a of f of x which has to exist by the the limit
rule for psalms that I applied
Now, on the left side, I can cancel out my copies of f of a, whatever number that is,
and I get that the limit as x goes to a of f of x which has to exist by the the limit
rule for psalms that I applied
above.
above.
That limit their past to equal the limit as x goes to a of f of a, well, f of a is just
some number, doesn't matter what X is doing f of a is it's just f of a, whatever that
is. So this limit on the right is just f of a, and look at that. That's exactly what it
means for a function to be continuous at x equals a, the limit as x goes to a of f of
x equals f of a. So f is continuous at x equals a, hella proof is complete. In this video,
we prove that if f is differentiable at x equals a, then f is continuous at x equals
a. This statement is equivalent to another statement known in logic as its contrapositive,
which says that if f is not continuous, at x equals a, then f is not differentiable at
x equals a. In this video, we'll learn a few rules for calculating derivatives, namely,
the power role, and the derivatives of sums, differences and constant multiples. These
rules will give us shortcuts for finding derivatives quickly, without needing to resort to the
old limit definition of derivative. In this video, we'll only do statements of the rules
and some examples, there won't be any proofs or justification for why the rules hold. Instead,
these proofs will be in a separate later video. Let's start with some basics. First of all,
if we have a constant C, and we will have the function f of x equals C. So if I graph
that, it's just going to be a straight horizontal line.
That limit their past to equal the limit as x goes to a of f of a, well, f of a is just
some number, doesn't matter what X is doing f of a is it's just f of a, whatever that
is. So this limit on the right is just f of a, and look at that. That's exactly what it
means for a function to be continuous at x equals a, the limit as x goes to a of f of
x equals f of a. So f is continuous at x equals a, hella proof is complete. In this video,
we prove that if f is differentiable at x equals a, then f is continuous at x equals
a. This statement is equivalent to another statement known in logic as its contrapositive,
which says that if f is not continuous, at x equals a, then f is not differentiable at
x equals a. In this video, we'll learn a few rules for calculating derivatives, namely,
the power role, and the derivatives of sums, differences and constant multiples. These
rules will give us shortcuts for finding derivatives quickly, without needing to resort to the
old limit definition of derivative. In this video, we'll only do statements of the rules
and some examples, there won't be any proofs or justification for why the rules hold. Instead,
these proofs will be in a separate later video. Let's start with some basics. First of all,
if we have a constant C, and we will have the function f of x equals C. So if I graph
that, it's just going to be a straight horizontal line.
So the derivative
So the derivative
df dx, ought to be zero, because the slope of the tangent line of this straight line
is just
df dx, ought to be zero, because the slope of the tangent line of this straight line
is just
zero.
zero.
Another simple example, is the derivative of the function f of x equals x. So again,
if I draw the graph, that's just going to be a straight line with slope one. And so
the tangent line for the straight line will also have slope one. And the derivative, f
prime of x has to be always equal to one. These two simple examples are actually special
cases of the power rule, which is one of the most useful rules for calculating
Another simple example, is the derivative of the function f of x equals x. So again,
if I draw the graph, that's just going to be a straight line with slope one. And so
the tangent line for the straight line will also have slope one. And the derivative, f
prime of x has to be always equal to one. These two simple examples are actually special
cases of the power rule, which is one of the most useful rules for calculating
derivatives.
derivatives.
So the power rule says that if you have the function, y equals x to the n, where n is
any real number, then you can find the derivative d y dx, simply by pulling that exponent and
down and multiplying it in the front and then reducing the exponent by one. So for example,
if you want to calculate the derivative of y equals x to the 15th, D y dX here is just
going to be 15 times x to the 15 minus one, or 14. The second example, f of x equals the
cube root of x might not immediately look like an example where we can apply the power
rule. But if we rewrite it, by putting the cube root in exponential tation as x to the
1/3, now we can apply the power rule, we bring the 1/3 down, multiplied on the front and
So the power rule says that if you have the function, y equals x to the n, where n is
any real number, then you can find the derivative d y dx, simply by pulling that exponent and
down and multiplying it in the front and then reducing the exponent by one. So for example,
if you want to calculate the derivative of y equals x to the 15th, D y dX here is just
going to be 15 times x to the 15 minus one, or 14. The second example, f of x equals the
cube root of x might not immediately look like an example where we can apply the power
rule. But if we rewrite it, by putting the cube root in exponential tation as x to the
1/3, now we can apply the power rule, we bring the 1/3 down, multiplied on the front and
reduce the
reduce the
exponent of 1/3 by one, or 1/3 minus one is negative two thirds. So we found the derivative
here using the power rule, we could rewrite it if we want to using exponent rules, as
one over 3x to the two thirds, either answers good.
exponent of 1/3 by one, or 1/3 minus one is negative two thirds. So we found the derivative
here using the power rule, we could rewrite it if we want to using exponent rules, as
one over 3x to the two thirds, either answers good.
In the third example, g of x is one over x to the 3.7. Again, we need to do a little
rewriting before we can apply the power rule. So I'm going to rewrite g of x as x to the
minus 3.7. Using exponent rules, now I can find dg dx by pulling down the negative 3.7
multiplying in the front, and now I have to reduce negative 3.7 by one, so I subtract
one that gives me x to the negative 4.7. Again, I can rewrite this if I wish as negative 3.7
over x to the 4.7. It's important to notice that in all these examples and in fact, in
any example where the power rule applies, the variable x is in the base. And the exponent
is just a constant, just a real number. The constant multiple rules says that if c is
just a constant real number, and f is a differentiable function, then the derivative of C times f
of x is just c times the derivative of f of x. In other words, when we take the derivative,
we can just pull a constant outside of the derivative sign. Let's use this rule in an
example. If we want to take the derivative of 5x cubed, that's the same thing as five
times the derivative of x cubed. And now using the power rule, we can bring down the three
and get 15x squared. f and g are differentiable functions, then the derivative of f of x plus
g of x is the derivative of f plus the derivative of g. Similarly, for a difference, if f and
g are differentiable functions, then the derivative of the difference is just the difference of
the derivatives. Now let's use all these rules together to calculate the derivative of this
polynomial. To find the y dx, we can use the sum and difference rule to calculate the derivative
of each term separately. Now using the constant multiple rule and the power rule, we can bring
out the seven, bring down the three getting x squared here, similarly, for the next term,
five times two times x to the one, four times the derivative of x, which is just one, and
the derivative of a constant two is just zero. So simplifying, we get 21x squared minus 10x
plus four, and notice that the derivative of the original polynomial is just another
polynomial of one less degree. In this video, we use some shortcuts to calculate the derivatives
of various functions, especially polynomials. If you're interested in seeing where these
rules come from, how they're derived from the limit definition of derivative, then look
for another video coming soon on proofs. This video is about identities involving trig functions
like sine and cosine. But I want to start with some examples that just involve
In the third example, g of x is one over x to the 3.7. Again, we need to do a little
rewriting before we can apply the power rule. So I'm going to rewrite g of x as x to the
minus 3.7. Using exponent rules, now I can find dg dx by pulling down the negative 3.7
multiplying in the front, and now I have to reduce negative 3.7 by one, so I subtract
one that gives me x to the negative 4.7. Again, I can rewrite this if I wish as negative 3.7
over x to the 4.7. It's important to notice that in all these examples and in fact, in
any example where the power rule applies, the variable x is in the base. And the exponent
is just a constant, just a real number. The constant multiple rules says that if c is
just a constant real number, and f is a differentiable function, then the derivative of C times f
of x is just c times the derivative of f of x. In other words, when we take the derivative,
we can just pull a constant outside of the derivative sign. Let's use this rule in an
example. If we want to take the derivative of 5x cubed, that's the same thing as five
times the derivative of x cubed. And now using the power rule, we can bring down the three
and get 15x squared. f and g are differentiable functions, then the derivative of f of x plus
g of x is the derivative of f plus the derivative of g. Similarly, for a difference, if f and
g are differentiable functions, then the derivative of the difference is just the difference of
the derivatives. Now let's use all these rules together to calculate the derivative of this
polynomial. To find the y dx, we can use the sum and difference rule to calculate the derivative
of each term separately. Now using the constant multiple rule and the power rule, we can bring
out the seven, bring down the three getting x squared here, similarly, for the next term,
five times two times x to the one, four times the derivative of x, which is just one, and
the derivative of a constant two is just zero. So simplifying, we get 21x squared minus 10x
plus four, and notice that the derivative of the original polynomial is just another
polynomial of one less degree. In this video, we use some shortcuts to calculate the derivatives
of various functions, especially polynomials. If you're interested in seeing where these
rules come from, how they're derived from the limit definition of derivative, then look
for another video coming soon on proofs. This video is about identities involving trig functions
like sine and cosine. But I want to start with some examples that just involve
quadratic functions. If I want to find the solutions to this equation, I can rewrite
it x squared minus 6x minus seven equals zero, factor it, x minus seven times x plus one
equals zero, set the factors equal to 0x minus seven equals zero, or x plus one equals zero.
And that gives me the solutions, x equals seven, or x equals negative one. Next, let's
look at this more complicated equation. I'm going to try to solve that for x by multiplying
out the right hand side. Next, our combined terms on the right hand side. So that gives
me x squared minus 6x on both sides, well, x squared minus 6x is equal to x squared minus
6x. That's true no matter what I plug in for x, and therefore, all values of x satisfy
this equation, we can say that the solution set is all real numbers. The second equation
is called an identity, because it holds for all values of the variable. The first equation,
on the other hand is not an identity, because it only holds for some values of x and not
all values. Please pause the video for a moment and try to decide which of the following three
equations or identities that is, which of these equations hold for all values of the
variable. To start out, you might want to test them by plugging in a few values of the
variable and see if the equation holds. The first equation is not an identity. It does
hold for some values of x. For example, if x equals zero, then sine of two times zero
is zero. Two times sine of zero is also zero. So it does hold when x is zero. However, when
x is say pi over two, then sine of two times pi over two, that's the same thing as sine
of pi, which is zero, but two times sine of pi over two, that's two times one, or two,
and zero is not equal to two. So the equation does not hold for x equals pi
quadratic functions. If I want to find the solutions to this equation, I can rewrite
it x squared minus 6x minus seven equals zero, factor it, x minus seven times x plus one
equals zero, set the factors equal to 0x minus seven equals zero, or x plus one equals zero.
And that gives me the solutions, x equals seven, or x equals negative one. Next, let's
look at this more complicated equation. I'm going to try to solve that for x by multiplying
out the right hand side. Next, our combined terms on the right hand side. So that gives
me x squared minus 6x on both sides, well, x squared minus 6x is equal to x squared minus
6x. That's true no matter what I plug in for x, and therefore, all values of x satisfy
this equation, we can say that the solution set is all real numbers. The second equation
is called an identity, because it holds for all values of the variable. The first equation,
on the other hand is not an identity, because it only holds for some values of x and not
all values. Please pause the video for a moment and try to decide which of the following three
equations or identities that is, which of these equations hold for all values of the
variable. To start out, you might want to test them by plugging in a few values of the
variable and see if the equation holds. The first equation is not an identity. It does
hold for some values of x. For example, if x equals zero, then sine of two times zero
is zero. Two times sine of zero is also zero. So it does hold when x is zero. However, when
x is say pi over two, then sine of two times pi over two, that's the same thing as sine
of pi, which is zero, but two times sine of pi over two, that's two times one, or two,
and zero is not equal to two. So the equation does not hold for x equals pi
over two.
over two.
Since it doesn't hold for all values of the variable, it's not an identity. The second
equation is an identity. You can build some evidence for this by plugging in numbers.
For example, cosine of zero plus pi, which is negative one is the same thing as negative
of cosine of zero. You can also check for example, that cosine of pi over six plus pi
is the same thing as negative cosine of pi over six. But even if we check a zillion examples,
that's just evidence, it's not a proof that the identity holds, we could have just gotten
lucky with the values we picked, we can build a little bit stronger evidence, by looking
at graphs, I'm going to put theta on the x axis in a pie graph, y equals cosine of theta
plus
Since it doesn't hold for all values of the variable, it's not an identity. The second
equation is an identity. You can build some evidence for this by plugging in numbers.
For example, cosine of zero plus pi, which is negative one is the same thing as negative
of cosine of zero. You can also check for example, that cosine of pi over six plus pi
is the same thing as negative cosine of pi over six. But even if we check a zillion examples,
that's just evidence, it's not a proof that the identity holds, we could have just gotten
lucky with the values we picked, we can build a little bit stronger evidence, by looking
at graphs, I'm going to put theta on the x axis in a pie graph, y equals cosine of theta
plus
pi,
pi,
that's just like the graph of cosine shifted over to the left by pi. On the other hand,
if I graph y equals negative cosine theta, that's the graph of cosine theta, reflected
across the x axis, which gives us the exact same graph. So graphing both sides gives us
strong evidence that this equation is an identity, it holds for all values of theta. Now, the
strongest evidence of all would be an algebraic proof, which we'll do later in the course,
once we have a formula for the cosine of a sum of two angles.
that's just like the graph of cosine shifted over to the left by pi. On the other hand,
if I graph y equals negative cosine theta, that's the graph of cosine theta, reflected
across the x axis, which gives us the exact same graph. So graphing both sides gives us
strong evidence that this equation is an identity, it holds for all values of theta. Now, the
strongest evidence of all would be an algebraic proof, which we'll do later in the course,
once we have a formula for the cosine of a sum of two angles.
In the meantime, let's look at equation C. It turns out equation C is an identity. And
we could build evidence for it again by plugging in values for x, or by graphing the left side
and the right side separately, and checking to see that the graphs coincided. But for
this example, I'm going to go ahead and do an algebraic verification.
In the meantime, let's look at equation C. It turns out equation C is an identity. And
we could build evidence for it again by plugging in values for x, or by graphing the left side
and the right side separately, and checking to see that the graphs coincided. But for
this example, I'm going to go ahead and do an algebraic verification.
In particular, I'm going to start with the left side of the equation, and rewrite things
and rewrite things until I get to the right side of the equation. The first thing I'll
rewrite
In particular, I'm going to start with the left side of the equation, and rewrite things
and rewrite things until I get to the right side of the equation. The first thing I'll
rewrite
is secant and tangent in terms of their constituent functions, sine and cosine. Since secant of
x is one over cosine x, and tangent of x is sine x over cosine x, I can rewrite this expression
as one over cosine x minus sine x times sine x over cosine x. I can clean up those fractions
and write this as one over cosine x minus sine squared x over cosine x. Now, I noticed
that I have two fractions with the same denominator. So I can pull them together as one minus sine
squared x over cosine x. Next, I'm going to rewrite the numerator one minus sine squared
x using the Pythagorean identity that says that cosine squared x plus sine squared x
equals one, and therefore, one minus sine squared x is equal to cosine squared x just
by subtracting sine squared x from both sides. So I can replace my numerator, one minus sine
squared x with cosine squared x. And canceling one cosine from the top and from the bottom,
that's the same thing as cosine of x, which is the right hand side that I was trying to
get to. So a combination of a bunch of algebra, and the Pythagorean identity allows me to
prove that this equation is true for all values of x,
is secant and tangent in terms of their constituent functions, sine and cosine. Since secant of
x is one over cosine x, and tangent of x is sine x over cosine x, I can rewrite this expression
as one over cosine x minus sine x times sine x over cosine x. I can clean up those fractions
and write this as one over cosine x minus sine squared x over cosine x. Now, I noticed
that I have two fractions with the same denominator. So I can pull them together as one minus sine
squared x over cosine x. Next, I'm going to rewrite the numerator one minus sine squared
x using the Pythagorean identity that says that cosine squared x plus sine squared x
equals one, and therefore, one minus sine squared x is equal to cosine squared x just
by subtracting sine squared x from both sides. So I can replace my numerator, one minus sine
squared x with cosine squared x. And canceling one cosine from the top and from the bottom,
that's the same thing as cosine of x, which is the right hand side that I was trying to
get to. So a combination of a bunch of algebra, and the Pythagorean identity allows me to
prove that this equation is true for all values of x,
it's an identity. The best way to prove that an equation is an identity
it's an identity. The best way to prove that an equation is an identity
is to use algebra and to use other identities, like the Pythagorean identity to rewrite one
side of the equation till it looks like the other side. The best way to prove the net
equation is not an identity is to plug in numbers that break the identity. That is make
the equation not true. Now if you're just trying to decide if an equation has an identity
or not, and not worried about proving it, then I recommend plugging in numbers, or graphing
the left and right sides to see if those graphs are the same. Recall that an identity is an
equation that holds for all values of the variable. This video states and proves three
identities called the Pythagorean identities. The first one is the familiar cosine squared
theta plus sine squared theta equals one. The second one says tan squared theta plus
one equals secant squared theta. And the third one goes cotangent squared theta plus one
equals cosecant squared theta. Let's start by proving that cosine squared theta plus
sine squared theta equals one. I'll do this by drawing the unit circle with a right triangle
inside it by the definition of sine and cosine, the x&y coordinates of this top point, r cosine
theta and sine theta, the high partners, my triangle is one, since that's the radius of
my unit circle. Now the length of the base of my triangle is the same thing as the x
coordinate of this point. So that's equal to cosine theta. The height of this triangle
is the same thing as the y coordinate of this point. So that's sine theta. Now the Pythagoras
theorem for right triangles, says this side length squared plus that side length squared
is equal to the hypothenar squared. So by the Pythagorean theorem, we have that cosine
theta squared plus sine theta squared equals one squared, I can rewrite that as cosine
squared theta plus sine squared theta equals one, since one squared is one, and cosine
squared theta is just a shorthand notation for cosine theta squared. That completes the
proof of the first Pythagorean identity, at least in the case, when the angle theta is
in the first quadrant. In the case, when the angle was in a different quadrant, you can
use symmetry to argue the same identity holds. But I won't give the details here. To prove
the next Pythagorean identity, tan squared theta plus one equals secant squared theta,
let's use the first without your an identity, which said that cosine squared theta plus
sine squared theta equals one, I'm going to divide both sides of this equation by cosine
squared theta. Now I'm going to rewrite the left side by breaking apart the fraction into
cosine squared theta over cosine squared theta plus sine squared theta over cosine squared
theta. Now cosine squared theta over cosine squared theta is just one. And I can rewrite
the next fraction as sine of theta over cosine of theta squared. That's because when I square
a fraction, I can just square the numerator and square the denominator. And sine squared
theta is shorthand for sine of theta squares. Similarly, for cosine squared theta. Now on
the other side of the equal sign, I can rewrite this fraction as one over cosine theta squared.
Again, that's because when I square the fraction, I just get the one squared, which is one divided
by the cosine theta squared, which is this. I'm almost done. sine theta over cosine theta
is the same thing as tangent theta. And one over cosine theta is the same thing as secant
theta. Using the shorthand notation, that says one plus tan squared theta equals sequencer
data, which, after rearranging is exactly the identity that we were looking for. The
proof of the third for that green identity is very similar. Once again, I'll start with
the identity
is to use algebra and to use other identities, like the Pythagorean identity to rewrite one
side of the equation till it looks like the other side. The best way to prove the net
equation is not an identity is to plug in numbers that break the identity. That is make
the equation not true. Now if you're just trying to decide if an equation has an identity
or not, and not worried about proving it, then I recommend plugging in numbers, or graphing
the left and right sides to see if those graphs are the same. Recall that an identity is an
equation that holds for all values of the variable. This video states and proves three
identities called the Pythagorean identities. The first one is the familiar cosine squared
theta plus sine squared theta equals one. The second one says tan squared theta plus
one equals secant squared theta. And the third one goes cotangent squared theta plus one
equals cosecant squared theta. Let's start by proving that cosine squared theta plus
sine squared theta equals one. I'll do this by drawing the unit circle with a right triangle
inside it by the definition of sine and cosine, the x&y coordinates of this top point, r cosine
theta and sine theta, the high partners, my triangle is one, since that's the radius of
my unit circle. Now the length of the base of my triangle is the same thing as the x
coordinate of this point. So that's equal to cosine theta. The height of this triangle
is the same thing as the y coordinate of this point. So that's sine theta. Now the Pythagoras
theorem for right triangles, says this side length squared plus that side length squared
is equal to the hypothenar squared. So by the Pythagorean theorem, we have that cosine
theta squared plus sine theta squared equals one squared, I can rewrite that as cosine
squared theta plus sine squared theta equals one, since one squared is one, and cosine
squared theta is just a shorthand notation for cosine theta squared. That completes the
proof of the first Pythagorean identity, at least in the case, when the angle theta is
in the first quadrant. In the case, when the angle was in a different quadrant, you can
use symmetry to argue the same identity holds. But I won't give the details here. To prove
the next Pythagorean identity, tan squared theta plus one equals secant squared theta,
let's use the first without your an identity, which said that cosine squared theta plus
sine squared theta equals one, I'm going to divide both sides of this equation by cosine
squared theta. Now I'm going to rewrite the left side by breaking apart the fraction into
cosine squared theta over cosine squared theta plus sine squared theta over cosine squared
theta. Now cosine squared theta over cosine squared theta is just one. And I can rewrite
the next fraction as sine of theta over cosine of theta squared. That's because when I square
a fraction, I can just square the numerator and square the denominator. And sine squared
theta is shorthand for sine of theta squares. Similarly, for cosine squared theta. Now on
the other side of the equal sign, I can rewrite this fraction as one over cosine theta squared.
Again, that's because when I square the fraction, I just get the one squared, which is one divided
by the cosine theta squared, which is this. I'm almost done. sine theta over cosine theta
is the same thing as tangent theta. And one over cosine theta is the same thing as secant
theta. Using the shorthand notation, that says one plus tan squared theta equals sequencer
data, which, after rearranging is exactly the identity that we were looking for. The
proof of the third for that green identity is very similar. Once again, I'll start with
the identity
cosine squared theta plus sine squared theta equals one, and this time, I'll divide both
sides by sine squared theta. I'll break up the fraction on the left. And now I'll rewrite
my fractions as cosine theta over sine theta squared plus one equals one over sine theta
squared. cosine over sine can be written As cotangent, and one over sine can be written
as cosecant. That gives me the identity that I'm looking for. We've now proved three trig
identities. The first one, we proved using the unit circle, and the Pythagorean Theorem.
The second and third identities, we proved by using the first identity, and a bit of
algebra. The sum and difference formulas are formulas for computing the sine of a sum of
two angles, the cosine of a sum of two angles, the sine of a difference of two angles, and
the cosine of a difference of two angles. Please pause the video for a moment to think
about this question. Is it true that the sine of A plus B is equal to the sine of A plus
the sine of B? No, it's not true. And we can see by an example, if we plug in say, A equals
pi over two and B equals pi, than the sine of pi over two plus pi, is the same thing
as a sine of three pi over two, which is negative one. Whereas the sine of pi over two plus
the sine of pi is equal to one plus zero, which is one, negative one is not equal to
one. So this equation does not hold for all values of a and b. There are a few values
of a and b for which it does hold. For example, if a is zero, and B is zero, but it's not
true in general, instead, we need more complicated formulas. It turns out that the sine of the
sum of two angles A plus B is given by sine of A cosine of B plus cosine of A, sine of
B. The cosine of A plus B is given by cosine A cosine B minus sine A sine Bay. I like to
remember these with a song, sine cosine cosine sine cosine cosine minus sine sine. Please
feel free to back up the video and sing along with me, I encourage you to memorize the two
formulas for the sine of a
cosine squared theta plus sine squared theta equals one, and this time, I'll divide both
sides by sine squared theta. I'll break up the fraction on the left. And now I'll rewrite
my fractions as cosine theta over sine theta squared plus one equals one over sine theta
squared. cosine over sine can be written As cotangent, and one over sine can be written
as cosecant. That gives me the identity that I'm looking for. We've now proved three trig
identities. The first one, we proved using the unit circle, and the Pythagorean Theorem.
The second and third identities, we proved by using the first identity, and a bit of
algebra. The sum and difference formulas are formulas for computing the sine of a sum of
two angles, the cosine of a sum of two angles, the sine of a difference of two angles, and
the cosine of a difference of two angles. Please pause the video for a moment to think
about this question. Is it true that the sine of A plus B is equal to the sine of A plus
the sine of B? No, it's not true. And we can see by an example, if we plug in say, A equals
pi over two and B equals pi, than the sine of pi over two plus pi, is the same thing
as a sine of three pi over two, which is negative one. Whereas the sine of pi over two plus
the sine of pi is equal to one plus zero, which is one, negative one is not equal to
one. So this equation does not hold for all values of a and b. There are a few values
of a and b for which it does hold. For example, if a is zero, and B is zero, but it's not
true in general, instead, we need more complicated formulas. It turns out that the sine of the
sum of two angles A plus B is given by sine of A cosine of B plus cosine of A, sine of
B. The cosine of A plus B is given by cosine A cosine B minus sine A sine Bay. I like to
remember these with a song, sine cosine cosine sine cosine cosine minus sine sine. Please
feel free to back up the video and sing along with me, I encourage you to memorize the two
formulas for the sine of a
sum of angles and the cosine of a sum of angles. Once you do, it's easy to figure out the sine
sum of angles and the cosine of a sum of angles. Once you do, it's easy to figure out the sine
and cosine of a difference of two angles. One way to do this is to think of sine of
and cosine of a difference of two angles. One way to do this is to think of sine of
A minus B as sine of A plus negative B. And then use the angle sum formula. So this works
out to sine cosine plus cosine, sine. And now, if I use the fact that cosine is even,
I know that cosine of negative B is cosine of B. And since sine is odd, sine of negative
b is negative sine of B. So I can rewrite this as sine of A cosine of B minus cosine
of A sine of B. Notice that this new formula for the difference is the same as the formula
for the sum is just that plus sign turned into a minus sign. We can do the same trick
for cosine of A minus B, that's cosine of A plus minus b, which is cosine A cosine minus
b minus sine of A sine of negative B. Again, using even an odd properties, this gives us
cosine A cosine B plus sine A sine B. Once again, the formula for the difference is almost
exactly like the for the song, just that minus sign has switched to
a plus sign. Now let's use the angle sum formula to find the exact value for the sign of 105
degrees. Now, 105 degrees is not a special angle on the unit circle, but I can write
it as the sum of two special angles. I can write it as 60 degrees plus 45 degrees. Therefore,
the sine of 105 degrees is the sine of 60 plus 45. And now by the angle some formula,
this is sine, cosine, cosine, sine. And I for my Unit Circle, I can figure out that
sine of 60 degrees is root three over two cosine of 45 degrees root two over two, cosine
of 60 degrees is one half, and sine of 45 degrees is root two over two. So this simplifies
to root six plus root two over four. For our last example, let's find the cosine of v plus
W, given the values of cosine v and cosine W, and the fact that v and w are angles in
the first quadrant. Remember, to compute the cosine of a sum, we can't just add together
the two cosines. That wouldn't even make sense in this case, because adding point nine and
point seven would give something bigger than one and the cosine of something's never bigger
than one. Instead, we have to use the angle sum formula for cosine. So that goes cosine
of v plus w equals cosine, cosine, minus sine, sine. Now, I've already know the cosine of
v and the cosine of W, so I could just plug those in. But I have to figure out the sine
of v and the sine of W from the given information. And one way to do that is to draw right triangles.
So here, I'm going to draw a right triangle with angle V,
A minus B as sine of A plus negative B. And then use the angle sum formula. So this works
out to sine cosine plus cosine, sine. And now, if I use the fact that cosine is even,
I know that cosine of negative B is cosine of B. And since sine is odd, sine of negative
b is negative sine of B. So I can rewrite this as sine of A cosine of B minus cosine
of A sine of B. Notice that this new formula for the difference is the same as the formula
for the sum is just that plus sign turned into a minus sign. We can do the same trick
for cosine of A minus B, that's cosine of A plus minus b, which is cosine A cosine minus
b minus sine of A sine of negative B. Again, using even an odd properties, this gives us
cosine A cosine B plus sine A sine B. Once again, the formula for the difference is almost
exactly like the for the song, just that minus sign has switched to a plus sign. Now let's
use the angle sum formula to find the exact value for the sign of 105 degrees. Now, 105
degrees is not a special angle on the unit circle, but I can write it as the sum of two
special angles. I can write it as 60 degrees plus 45 degrees. Therefore, the sine of 105
degrees is the sine of 60 plus 45. And now by the angle some formula, this is sine, cosine,
cosine, sine. And I for my Unit Circle, I can figure out that sine of 60 degrees is
root three over two cosine of 45 degrees root two over two, cosine of 60 degrees is one
half, and sine of 45 degrees is root two over two. So this simplifies to root six plus root
two over four. For our last example, let's find the cosine of v plus W, given the values
of cosine v and cosine W, and the fact that v and w are angles in the first quadrant.
Remember, to compute the cosine of a sum, we can't just add together the two cosines.
That wouldn't even make sense in this case, because adding point nine and point seven
would give something bigger than one and the cosine of something's never bigger than one.
Instead, we have to use the angle sum formula for cosine. So that goes cosine of v plus
w equals cosine, cosine, minus sine, sine. Now, I've already know the cosine of v and
the cosine of W, so I could just plug those in. But I have to figure out the sine of v
and the sine of W from the given information. And one way to do that is to draw right triangles.
So here, I'm going to draw a right triangle with angle V,
and another right triangle with angle W. Since I know that the cosine of V is point nine,
I can think of that as nine over 10.
and another right triangle with angle W. Since I know that the cosine of V is point nine,
I can think of that as nine over 10.
And I can think of that as adjacent over hypotenuse in my right triangle. So I'll decorate my
triangles adjacent side with the number nine and the hypotenuse with 10. Similarly, since
I know that the cosine of W is point seven, which is seven tenths, I can put a seven on
this adjacent side, and a 10 on this iPod news. Now, the Pythagorean Theorem, lets me
compute the length of the unlabeled side. So this one is going to be the square root
of 10 squared minus nine squared, that's going to be the square root of 19. And here I have
the square root of 10 squared minus seven squared. So that's the square root of 51.
I can now find the sign of V as the opposite over the hi partners. So that's the square
root of 19 over 10. And the sign of W will be the square root of 51 over 10. Because
we're assuming v and w are in the first quadrant, we know the values of sign need to be positive,
so we don't need to Jimmy around with positive or negative signs in our answers, we can just
leave them as is.
And I can think of that as adjacent over hypotenuse in my right triangle. So I'll decorate my
triangles adjacent side with the number nine and the hypotenuse with 10. Similarly, since
I know that the cosine of W is point seven, which is seven tenths, I can put a seven on
this adjacent side, and a 10 on this iPod news. Now, the Pythagorean Theorem, lets me
compute the length of the unlabeled side. So this one is going to be the square root
of 10 squared minus nine squared, that's going to be the square root of 19. And here I have
the square root of 10 squared minus seven squared. So that's the square root of 51.
I can now find the sign of V as the opposite over the hi partners. So that's the square
root of 19 over 10. And the sign of W will be the square root of 51 over 10. Because
we're assuming v and w are in the first quadrant, we know the values of sign need to be positive,
so we don't need to Jimmy around with positive or negative signs in our answers, we can just
leave them as is.
Now we're ready to plug into our formula. So we have that cosine of v plus w is equal
to point nine times point seven minus the square root of 19 over 10, times the square
root of 51 over 10.
Now we're ready to plug into our formula. So we have that cosine of v plus w is equal
to point nine times point seven minus the square root of 19 over 10, times the square
root of 51 over 10.
using a calculator, this works out to a decimal approximation of 0.3187. This video gave the
angle sum and difference formulas and use them to compute some values. To see a proof
for why the sum formulas hold, please watch my other video. This video gives formulas
for sine of two theta and cosine of two theta. Please pause the video for a moment and see
if you think this equation sine of two theta equals two sine theta is true or false. Remember
that true means always true for all values of theta were false means sometimes they're
always false. This equation is false, because it's not true for all values of theta. One
way to see this is graphically, if I graph y equals sine of two theta, that's like the
graph of sine theta, squished in horizontally by a factor of one half. On the other hand,
if I graph y equals two sine beta, that's like the graph of sine theta stretched vertically
by a factor of two. These two graphs are not the same. So instead, we need a more complicated
formula for sine of two theta. And that formula is sine of two theta is two sine theta, cosine
theta. It's not hard to see why that formula works based on the angle some formula. Recall
that sine of A plus B is equal to sine A cosine B plus cosine A sign Therefore, sine of two
theta, which is sine of theta plus theta is going to be sine theta, cosine theta, plus
cosine theta sine theta. simply plugging in theta for a and theta for B, in this angle,
some formula, will sine theta cosine theta is the same thing as cosine theta sine theta.
So I can rewrite this as twice sine theta cosine theta. That gives me this formula.
There's also a formula for cosine of two theta. And that formula is cosine squared theta minus
sine squared theta. Again, we can use the angle sum formula to see where this comes
from. cosine of A plus B is equal to cosine of A cosine of B minus sine A, sine B. So
if we want cosine of two theta, that's just cosine of theta plus theta, which is cosine
theta, cosine theta, minus sine theta, sine theta by plugging in beta for a and b, this
can be rewritten as cosine squared theta minus sine squared theta, which is exactly the formula
above. Now there are a couple other formulas for cosine of two theta that are also popular.
One of them is one minus two sine squared theta. And the other one is cosine of two
theta is two cosine squared theta minus one, you can get each of these two formulas from
the original one using the Pythagorean identity. We know that cosine squared theta plus sine
squared theta is one. So cosine squared theta is one minus sine squared theta. If I plug
that into my original formula, which I've copied here, so I'm plugging in, instead of
cosine squared, I'm gonna write one minus sine squared theta, I still have a nother
minus sine squared theta. So that's the same thing as one minus twice science grid data,
which is exactly what I'm looking for. Similarly, I can use the Pythagorean identity to write
sine squared theta as one minus cosine squared theta. Again, I'll take this equation and
copy it below. But this time, I'm going to plug in for sine squared right here. So that
gives me cosine of two theta is cosine squared theta minus the quantity one minus cosine
squared theta. That simplifies to two cosine squared theta minus one after distributing
the negative sign and combining like terms.
using a calculator, this works out to a decimal approximation of 0.3187. This video gave the
angle sum and difference formulas and use them to compute some values. To see a proof
for why the sum formulas hold, please watch my other video. This video gives formulas
for sine of two theta and cosine of two theta. Please pause the video for a moment and see
if you think this equation sine of two theta equals two sine theta is true or false. Remember
that true means always true for all values of theta were false means sometimes they're
always false. This equation is false, because it's not true for all values of theta. One
way to see this is graphically, if I graph y equals sine of two theta, that's like the
graph of sine theta, squished in horizontally by a factor of one half. On the other hand,
if I graph y equals two sine beta, that's like the graph of sine theta stretched vertically
by a factor of two. These two graphs are not the same. So instead, we need a more complicated
formula for sine of two theta. And that formula is sine of two theta is two sine theta, cosine
theta. It's not hard to see why that formula works based on the angle some formula. Recall
that sine of A plus B is equal to sine A cosine B plus cosine A sign Therefore, sine of two
theta, which is sine of theta plus theta is going to be sine theta, cosine theta, plus
cosine theta sine theta. simply plugging in theta for a and theta for B, in this angle,
some formula, will sine theta cosine theta is the same thing as cosine theta sine theta.
So I can rewrite this as twice sine theta cosine theta. That gives me this formula.
There's also a formula for cosine of two theta. And that formula is cosine squared theta minus
sine squared theta. Again, we can use the angle sum formula to see where this comes
from. cosine of A plus B is equal to cosine of A cosine of B minus sine A, sine B. So
if we want cosine of two theta, that's just cosine of theta plus theta, which is cosine
theta, cosine theta, minus sine theta, sine theta by plugging in beta for a and b, this
can be rewritten as cosine squared theta minus sine squared theta, which is exactly the formula
above. Now there are a couple other formulas for cosine of two theta that are also popular.
One of them is one minus two sine squared theta. And the other one is cosine of two
theta is two cosine squared theta minus one, you can get each of these two formulas from
the original one using the Pythagorean identity. We know that cosine squared theta plus sine
squared theta is one. So cosine squared theta is one minus sine squared theta. If I plug
that into my original formula, which I've copied here, so I'm plugging in, instead of
cosine squared, I'm gonna write one minus sine squared theta, I still have a nother
minus sine squared theta. So that's the same thing as one minus twice science grid data,
which is exactly what I'm looking for. Similarly, I can use the Pythagorean identity to write
sine squared theta as one minus cosine squared theta. Again, I'll take this equation and
copy it below. But this time, I'm going to plug in for sine squared right here. So that
gives me cosine of two theta is cosine squared theta minus the quantity one minus cosine
squared theta. That simplifies to two cosine squared theta minus one after distributing
the negative sign and combining like terms.
So I have one double angle formula for sine of two theta. And I have three versions of
the double angle formula for cosine of two theta.
So I have one double angle formula for sine of two theta. And I have three versions of
the double angle formula for cosine of two theta.
Now let's use these formulas in some examples. Let's find the cosine of two theta. If we
know that cosine theta is negative one over root 10, and theta terminates in quadrant
three, we have a choice of three formulas for cosine of two theta, I'm going to choose
the second one, because it only involves cosine if they had on the right side. And they already
know my value for cosine theta. Of course, I could use one of the other ones,
Now let's use these formulas in some examples. Let's find the cosine of two theta. If we
know that cosine theta is negative one over root 10, and theta terminates in quadrant
three, we have a choice of three formulas for cosine of two theta, I'm going to choose
the second one, because it only involves cosine if they had on the right side. And they already
know my value for cosine theta. Of course, I could use one of the other ones,
but then I'd have to work out the value of sine theta. So plugging in, I get cosine of
two theta is twice negative one over root n squared minus one, which simplifies to two
tenths minus one or negative eight tenths, negative four fifths. Finally, let's solve
the equation two cosine x plus sine of 2x equals zero. What makes this equation tricky
is that one of the trig functions has the argument of just x, but the other tree function
has the argument of 2x. So I want to use my double angle formula to rewrite sine of 2x.
I'll copy down the two cosine x, and now sine of 2x is equal to two sine x cosine x. At
this point, I see a way to factor my equation, I can factor out a two cosine x from both
of these two terms. That gives me one plus sine x and the product there is equal to zero.
That means that either two cosine x is equal to zero One plus sine x is equal to zero,
that simplifies to cosine x equals zero, or sine x is negative one. Using my unit circle,
I see that cosine of x is zero at pi over two, and three pi over two, while sine of
x is negative one at three pi over two, there's some redundancy here, but my solution set
is going to be pi over two plus multiples of two pi, and three pi over two plus multiples
of two pi. This video proved the double angle formulas, sine of two theta is two sine theta
cosine theta. and cosine of two theta is cosine squared theta minus sine squared theta. It
also proved to alternate versions of the equation for cosine of two theta. This video introduces
higher order derivatives and notation. We've seen that f prime of x denotes the derivative
of the function f of x, but f prime of x is also itself a function. So we can take its
derivative, that would be f prime prime of x, which is usually written instead as f double
prime of x. This is called the second derivative of f. And it means the derivative of the derivative,
we can also talk about the third derivative, f triple prime of x, which might sometimes
be written f to the three of x if you get tired of writing all those primes, and we
can talk about the nth derivative f, parentheses n of x, the parentheses here are important
to show that it's the nth derivative. The second, third and nth derivative are referred
to as higher order
but then I'd have to work out the value of sine theta. So plugging in, I get cosine of
two theta is twice negative one over root n squared minus one, which simplifies to two
tenths minus one or negative eight tenths, negative four fifths. Finally, let's solve
the equation two cosine x plus sine of 2x equals zero. What makes this equation tricky
is that one of the trig functions has the argument of just x, but the other tree function
has the argument of 2x. So I want to use my double angle formula to rewrite sine of 2x.
I'll copy down the two cosine x, and now sine of 2x is equal to two sine x cosine x. At
this point, I see a way to factor my equation, I can factor out a two cosine x from both
of these two terms. That gives me one plus sine x and the product there is equal to zero.
That means that either two cosine x is equal to zero One plus sine x is equal to zero,
that simplifies to cosine x equals zero, or sine x is negative one. Using my unit circle,
I see that cosine of x is zero at pi over two, and three pi over two, while sine of
x is negative one at three pi over two, there's some redundancy here, but my solution set
is going to be pi over two plus multiples of two pi, and three pi over two plus multiples
of two pi. This video proved the double angle formulas, sine of two theta is two sine theta
cosine theta. and cosine of two theta is cosine squared theta minus sine squared theta. It
also proved to alternate versions of the equation for cosine of two theta. This video introduces
higher order derivatives and notation. We've seen that f prime of x denotes the derivative
of the function f of x, but f prime of x is also itself a function. So we can take its
derivative, that would be f prime prime of x, which is usually written instead as f double
prime of x. This is called the second derivative of f. And it means the derivative of the derivative,
we can also talk about the third derivative, f triple prime of x, which might sometimes
be written f to the three of x if you get tired of writing all those primes, and we
can talk about the nth derivative f, parentheses n of x, the parentheses here are important
to show that it's the nth derivative. The second, third and nth derivative are referred
to as higher order
derivatives.
derivatives.
There are many alternative notations for derivatives stemming from their tangled and contentious
history in the 1600s. There are a few different notations for functions themselves. Most often
we write a function of something like f of x, but we can also use the variable y to refer
to the output of a function. When we're looking at the first derivative. We've been using
the notation f prime of x. But you might also see y prime, this means the same thing. Another
notation is df dx, is known as lug nuts denotation. After five minutes, you might see something
like dy dx of f of x. And you might see dy dx, another version of Lebanon's notation.
Sometimes you'll also see a capital D used to refer to the derivative. If we're looking
at the second derivative, we've seen that in Haitian f double prime of x, y double prime
is a similar notation, or we might write dy dx of df
There are many alternative notations for derivatives stemming from their tangled and contentious
history in the 1600s. There are a few different notations for functions themselves. Most often
we write a function of something like f of x, but we can also use the variable y to refer
to the output of a function. When we're looking at the first derivative. We've been using
the notation f prime of x. But you might also see y prime, this means the same thing. Another
notation is df dx, is known as lug nuts denotation. After five minutes, you might see something
like dy dx of f of x. And you might see dy dx, another version of Lebanon's notation.
Sometimes you'll also see a capital D used to refer to the derivative. If we're looking
at the second derivative, we've seen that in Haitian f double prime of x, y double prime
is a similar notation, or we might write dy dx of df
dx.
dx.
And the shorthand for that is d squared f dx squared. Similarly, we might write d squared
y dx squared using y in the place of F for the function.
And the shorthand for that is d squared f dx squared. Similarly, we might write d squared
y dx squared using y in the place of F for the function.
There's similar notations for third derivative, I'll jump ahead to nth derivatives. So that
would be F to the n of x, or y to the n, d to the n of f dx to the n, or D to the n of
y dx to the N. When using live minutes notation, we want to emphasize that we're evaluating
our derivative at a particular value of x, we might write something like at x equals
three, or at x equals a, using a vertical line. For better or for worse, you'll need
to become familiar with all of these alternative notations.
There's similar notations for third derivative, I'll jump ahead to nth derivatives. So that
would be F to the n of x, or y to the n, d to the n of f dx to the n, or D to the n of
y dx to the N. When using live minutes notation, we want to emphasize that we're evaluating
our derivative at a particular value of x, we might write something like at x equals
three, or at x equals a, using a vertical line. For better or for worse, you'll need
to become familiar with all of these alternative notations.
That's all for this video on higher order derivatives and notation. This video is about
the derivative of e to the x, one of my favorite functions ever simply because it has such
a great derivative. As you may recall, he is an irrational number whose decimal approximation
is something like 2.718 looks like it's repeating. But then it keeps going on forever, never
repeating never terminating.
That's all for this video on higher order derivatives and notation. This video is about
the derivative of e to the x, one of my favorite functions ever simply because it has such
a great derivative. As you may recall, he is an irrational number whose decimal approximation
is something like 2.718 looks like it's repeating. But then it keeps going on forever, never
repeating never terminating.
Its value notice is somewhere in between two and three. Here is a graph of y equals e to
the x.
Its value notice is somewhere in between two and three. Here is a graph of y equals e to
the x.
It says exponential function increase thing looks a lot like two to the x or three to
the x, not only is the graph of e to the x increasing, but it's increasing more and more
rapidly. So for negative values of x, the slope of this graph is positive but very close
to zero. Over here, when x equals zero, that slope is looks like approximately a slope
of one, we'll see that it is in fact exactly one. And as the x values increase, this tangent
lines get steeper and steeper. I'm going to state without proof, three really useful facts
about E. First, if you take the limit, as n goes to infinity of one plus one over N
raised to the nth power, that limit exists and equals E, you might have seen something
like that when you took precalculus, and looked at compound interest compounded over smaller
and smaller time periods. But even if you haven't seen it before, it's a really important
fact worth memorizing, you'll see it again later in the class. A second important formula
is that the limit as h goes to zero of e to the H minus one over h equals one. Now this
expression here on the left may remind you of a derivative, in fact, I can rewrite it
as the limit as h goes to zero of e to the zero plus h minus e to the zero since either
the zero is one over h, that's equal to one. And this expression right here on the left
is just the derivative of e to the x at x equals zero, according to the limit definition
of derivative. So this fact, is really saying that the derivative of e to the x at x equals
zero, that derivative is equal to one. So for the third fat,
It says exponential function increase thing looks a lot like two to the x or three to
the x, not only is the graph of e to the x increasing, but it's increasing more and more
rapidly. So for negative values of x, the slope of this graph is positive but very close
to zero. Over here, when x equals zero, that slope is looks like approximately a slope
of one, we'll see that it is in fact exactly one. And as the x values increase, this tangent
lines get steeper and steeper. I'm going to state without proof, three really useful facts
about E. First, if you take the limit, as n goes to infinity of one plus one over N
raised to the nth power, that limit exists and equals E, you might have seen something
like that when you took precalculus, and looked at compound interest compounded over smaller
and smaller time periods. But even if you haven't seen it before, it's a really important
fact worth memorizing, you'll see it again later in the class. A second important formula
is that the limit as h goes to zero of e to the H minus one over h equals one. Now this
expression here on the left may remind you of a derivative, in fact, I can rewrite it
as the limit as h goes to zero of e to the zero plus h minus e to the zero since either
the zero is one over h, that's equal to one. And this expression right here on the left
is just the derivative of e to the x at x equals zero, according to the limit definition
of derivative. So this fact, is really saying that the derivative of e to the x at x equals
zero, that derivative is equal to one. So for the third fat,
the third fact,
the third fact,
it talks about the derivative of e to the x in general. And that third fact is that
the derivative of the function, either the X is the function, e to the x, e to the x
is its own derivative. So this is a generalized version of the second fact. Because the second
fact is saying that the derivative at x equals zero is one, well, one is just the same thing
as e to the zero. So it saying the drought of either the exit at x equals zero is equal
to zero. And in general, the derivative of e to the x at any x is just e to the x. Now,
fact, one is frequently taken as the definition of E. Sometimes fact two instead is taken
as a definition of he, since he is the unique number with this, this property, the unique
number, you can plug in here and get this limit equal one, it's possible to prove that
fact one implies fact two and vice versa. But I won't do that here. It's also possible
to prove that fact two implies fact three about the derivative in general. And that's
pretty straightforward from the definition of derivative. So I will show you that argument.
So let's start out assuming fact to and try to prove fact three using the definition of
direct by the definition of derivative. The derivative of e to the x is the limit as h
goes to zero of e to the x plus h minus e to the x over h. If I factor out an E to the
X, from both terms on the numerator, I get the limit of e to the x times e to the H minus
one over h. Notice that e to the x times e to the H is in the x plus h by the exponent
rules. Now, either the X has nothing to do with H, so it's just a constant as far as
H is concerned. And I can pull it all the way out of the limit sign and rewrite this
limit. Now by factor two, which I'm assuming this limit here is just one, which means that
my derivative
it talks about the derivative of e to the x in general. And that third fact is that
the derivative of the function, either the X is the function, e to the x, e to the x
is its own derivative. So this is a generalized version of the second fact. Because the second
fact is saying that the derivative at x equals zero is one, well, one is just the same thing
as e to the zero. So it saying the drought of either the exit at x equals zero is equal
to zero. And in general, the derivative of e to the x at any x is just e to the x. Now,
fact, one is frequently taken as the definition of E. Sometimes fact two instead is taken
as a definition of he, since he is the unique number with this, this property, the unique
number, you can plug in here and get this limit equal one, it's possible to prove that
fact one implies fact two and vice versa. But I won't do that here. It's also possible
to prove that fact two implies fact three about the derivative in general. And that's
pretty straightforward from the definition of derivative. So I will show you that argument.
So let's start out assuming fact to and try to prove fact three using the definition of
direct by the definition of derivative. The derivative of e to the x is the limit as h
goes to zero of e to the x plus h minus e to the x over h. If I factor out an E to the
X, from both terms on the numerator, I get the limit of e to the x times e to the H minus
one over h. Notice that e to the x times e to the H is in the x plus h by the exponent
rules. Now, either the X has nothing to do with H, so it's just a constant as far as
H is concerned. And I can pull it all the way out of the limit sign and rewrite this
limit. Now by factor two, which I'm assuming this limit here is just one, which means that
my derivative
is e to the x, just like I wanted to show. Here's a slightly tricky example, asking you
to compute the derivative of a function that involves lots of ears and X's
is e to the x, just like I wanted to show. Here's a slightly tricky example, asking you
to compute the derivative of a function that involves lots of ears and X's
combined in lots of different ways. You'll need to use not only the rule for the derivative
of e to the x that we just talked about, but also the power rule and other rules. derivatives
that we've talked about
combined in lots of different ways. You'll need to use not only the rule for the derivative
of e to the x that we just talked about, but also the power rule and other rules. derivatives
that we've talked about
earlier.
earlier.
So please pause the video and try to compute this derivative yourself paying careful attention
to what's a variable and what's the constant.
So please pause the video and try to compute this derivative yourself paying careful attention
to what's a variable and what's the constant.
Okay, so we're taking the derivative here with respect to x, that's our variable. And
I'm taking the derivative of this entire expression, which I can split up as a sum of derivatives.
For the first term, I can just use the power rule, E is a constant coefficient, so I just
need to take down the exponent of two multiplied on the front, times x to the One Power. Now
for the second part,
Okay, so we're taking the derivative here with respect to x, that's our variable. And
I'm taking the derivative of this entire expression, which I can split up as a sum of derivatives.
For the first term, I can just use the power rule, E is a constant coefficient, so I just
need to take down the exponent of two multiplied on the front, times x to the One Power. Now
for the second part,
here, I do have my either the x function multiplied by two, so its derivative is just two times
the derivative of e to the x, which is either the X
here, I do have my either the x function multiplied by two, so its derivative is just two times
the derivative of e to the x, which is either the X
for my third part, I have just x times a constant e squared. So the derivative of x is one times
that constant. And so I just get e squared. Finally, to take the derivative of x to the
power of E squared, I can use the power rule because my variable is in the base, and I
have a constant e squared in my exponent. So using the power rule, I bring down the
E squared times that by x and subtract one from the exponent. This video states the happy
fact that the derivative of e to the x is just e to
for my third part, I have just x times a constant e squared. So the derivative of x is one times
that constant. And so I just get e squared. Finally, to take the derivative of x to the
power of E squared, I can use the power rule because my variable is in the base, and I
have a constant e squared in my exponent. So using the power rule, I bring down the
E squared times that by x and subtract one from the exponent. This video states the happy
fact that the derivative of e to the x is just e to
the x. This video prove some of the rules for taking derivatives.
the x. This video prove some of the rules for taking derivatives.
First, the constant row, it makes sense that the derivative of a constant real number has
to be zero, because the slope of a horizontal line
First, the constant row, it makes sense that the derivative of a constant real number has
to be zero, because the slope of a horizontal line
is zero.
is zero.
But we can also prove this fact, using the limit definition of derivative. The derivative
of any function is the limit as h goes to zero of the function of x plus h minus the
function at x divided by H. Well, here, our function is just a constant. So we're taking
the limit as h goes to zero of the constant minus the constant divided by h, which is
just the limit as h goes to zero of zero over h, which is just the limit of zero, which
is zero. Intuitively, it also makes sense that the derivative of the function y equals
x is got to be one, because the graph of y equals x is a straight line with slope one.
But again, we can prove this using the limit definition of derivative. So the derivative
of x is the limit as h goes to zero of x plus h minus x over h. Well, that simplifies to
the limit of h over h, since the Xs can't So in other words, the limit as h goes to
zero of one, which is one as wanted
But we can also prove this fact, using the limit definition of derivative. The derivative
of any function is the limit as h goes to zero of
the function of x plus h minus the function at x divided by H. Well, here, our function
is just a constant. So we're taking the limit as h goes to zero of the constant minus the
constant divided by h, which is just the limit as h goes to zero of zero over h, which is
just the limit of zero, which is zero. Intuitively, it also makes sense that the derivative of
the function y equals x is got to be one, because the graph of y equals x is a straight
line with slope one. But again, we can prove this using the limit definition of derivative.
So the derivative of x is the limit as h goes to zero of x plus h minus x over h. Well,
that simplifies to the limit of h over h, since the Xs can't So in other words, the
limit as h goes to zero of one, which is one as wanted
x,
x,
h to the n minus one, and then finally, a term of h to the N. That's the binomial expansion
of x plus h to the n. Now, we still have to subtract the x to the n that we had up here,
and we still have to divide this whole thing by H. Okay, that's looking kind of horribly
complicated.
h to the n minus one, and then finally, a term of h to the N. That's the binomial expansion
of x plus h to the n. Now, we still have to subtract the x to the n that we had up here,
and we still have to divide this whole thing by H. Okay, that's looking kind of horribly
complicated.
But
But
notice that the X to the ends can So notice that all of the remaining terms have an H
in them. So if we factor out that H, we get an x to the n minus one, plus a bunch of other
terms. And canceling the H's,
notice that the X to the ends can So notice that all of the remaining terms have an H
in them. So if we factor out that H, we get an x to the n minus one, plus a bunch of other
terms. And canceling the H's,
we get
we get
one term that doesn't have any ages in it, and another bunch of terms that all have H's
in them, as h goes to zero, all these other terms drop out because they go to zero. And
what we're left with is simply n times x to the n minus one, which is exactly what we
want for the power role. I think that's a pretty good proof if you're comfortable with
the binomial formula. But if you haven't seen the binomial formula before, that might leave
you feeling a little cold. So I'm going to offer you another proof using the other form
of the limit definition of derivative. So let me clear some space here. And I'll start
over using this definition, f prime at a is the limit as x goes to a of our function evaluated
at x. So that's x to the n minus our function evaluated at a, that's a to the n over x minus
a. Again, I'm going to need to rewrite things in order to evaluate this limit, since it's
currently in a zero over zero and determinant form. So I'm going to rewrite the top by factoring
out a copy of x minus a, which gives me x to the n minus one plus x to the n minus two
A plus x to the n minus three A squared, you see the pattern here. And I keep going until
I get to x a to the n minus two and finally a to the n minus one, that's still over x
minus a, you can verify this factoring formula, simply by multiplying out and checking that
you In fact, do get x to the n minus a to the n, after all your intermediate terms cancel.
Now that I've factored, I can cancel my x minus a, and simply evaluate my limit by plugging
in x equal to A to get a to the n minus one plus a to the n minus two A plus, and so on
each of these terms is equal to a to the n minus one. And there are a total of n terms,
since we got them from the terms above that started with x to the n minus one and did
with x to the zero. So that's n terms. So that means we've got a final sum of n times
a to the n minus one for a derivative f prime of A, which is exactly what we wanted to show.
Next, I'll prove the constant multiple rule that says that if c is a real number constant,
and f is a differentiable function, then the derivative of a constant times f is just the
constant times the derivative of f. Starting with the limit definition of derivative, I
have that the derivative of C times f of x is the limit as h goes to zero of C times
f of x plus h minus c times f of x over h. Now if I factor out the constant C, from both
of these terms, and actually I can pull it all the way out of the limit side, since the
constant has nothing to do with h. So now I get that was equal to the constant times
the limit as h goes to zero of f of x plus h minus f of x over h, which is just a constant
times the derivative of f, which is what we wanted to prove.
one term that doesn't have any ages in it, and another bunch of terms that all have H's
in them, as h goes to zero, all these other terms drop out because they go to zero. And
what we're left with is simply n times x to the n minus one, which is exactly what we
want for the power role. I think that's a pretty good proof if you're comfortable with
the binomial formula. But if you haven't seen the binomial formula before, that might leave
you feeling a little cold. So I'm going to offer you another proof using the other form
of the limit definition of derivative. So let me clear some space here. And I'll start
over using this definition, f prime at a is the limit as x goes to a of our function evaluated
at x. So that's x to the n minus our function evaluated at a, that's a to the n over x minus
a. Again, I'm going to need to rewrite things in order to evaluate this limit, since it's
currently in a zero over zero and determinant form. So I'm going to rewrite the top by factoring
out a copy of x minus a, which gives me x to the n minus one plus x to the n minus two
A plus x to the n minus three A squared, you see the pattern here. And I keep going until
I get to x a to the n minus two and finally a to the n minus one, that's still over x
minus a, you can verify this factoring formula, simply by multiplying out and checking that
you In fact, do get x to the n minus a to the n, after all your intermediate terms cancel.
Now that I've factored, I can cancel my x minus a, and simply evaluate my limit by plugging
in x equal to A to get a to the n minus one plus a to the n minus two A plus, and so on
each of these terms is equal to a to the n minus one. And there are a total of n terms,
since we got them from the terms above that started with x to the n minus one and did
with x to the zero. So that's n terms. So that means we've got a final sum of n times
a to the n minus one for a derivative f prime of A, which is exactly what we wanted to show.
Next, I'll prove the constant multiple rule that says that if c is a real number constant,
and f is a differentiable function, then the derivative of a constant times f is just the
constant times the derivative of f. Starting with the limit definition of derivative, I
have that the derivative of C times f of x is the limit as h goes to zero of C times
f of x plus h minus c times f of x over h. Now if I factor out the constant C, from both
of these terms, and actually I can pull it all the way out of the limit side, since the
constant has nothing to do with h. So now I get that was equal to the constant times
the limit as h goes to zero of f of x plus h minus f of x over h, which is just a constant
times the derivative of f, which is what we wanted to prove.
The difference rule can be proved just like the sum rule by writing out the definition
of derivative and regrouping terms. Or we could use kind of a sneaky shortcut, and put
together two of our previous roles. So if we think of f of x minus g of x as being f
of x plus minus one times g of x, then we can use the sum rule to rewrite this as a
sum of derivatives, and then use the constant multiplier rule to pull the constant of negative
one out, and then we have exactly what we wanted to prove. So in this video, we gave
the proof of the content multiple rule, the summon difference rules, and a proof of the
power rule when n is a positive integer. This video gives rules for calculating derivatives
of functions that are products, or quotients of other functions. In this video, you'll
find statements of the product rule and the quotient rule, and some examples. But no proofs,
proofs are in a separate video. Before we start, let's recall the Psalm and the difference
rules. If f and g are differentiable functions, then the derivative of the psalm f of x plus
g of x is just the sum of the derivatives. And a similar statement holds for differences.
The derivative of the difference is the difference of the derivatives. So does the same sort
of rule hold for products of functions, in other words, is the derivative of the product
equal to the product of the derivatives? Let's look at a simple example. To find out. For
example, if f of x is x, and g of x is x squared, then if we take the derivative of the product,
x times x squared, well, that's just the derivative of x cubed. We know how to do that with the
power rule. So it's 3x squared. On the other hand, if we look at the product of the derivatives,
we get one times 2x or just 2x. And these two things are not equal. So unfortunately,
the answer is no such a simple product rule does not hold. But not lose hope. There is
a product rule, it's just a little more complicated than the sum and difference rule. The product
rule says that  if f and g are differentiable functions, then
the derivative of the product f of x times g of x is equal to f of x times the derivative
of g of x plus the derivative of f of x times g of x. In other words, to take the derivative
of a product, we take the first function times the derivative of the second, plus the derivative
of the first times the second. Let's use this in an example. To take the derivative of the
square root of t times e to the t, we have to take the first function squared of t times
the derivative of the second function, plus the derivative of the first function
The difference rule can be proved just like the sum rule by writing out the definition
of derivative and regrouping terms. Or we could use kind of a sneaky shortcut, and put
together two of our previous roles. So if we think of f of x minus g of x as being f
of x plus minus one times g of x, then we can use the sum rule to rewrite this as a
sum of derivatives, and then use the constant multiplier rule to pull the constant of negative
one out, and then we have exactly what we wanted to prove. So in this video, we gave
the proof of the content multiple rule, the summon difference rules, and a proof of the
power rule when n is a positive integer. This video gives rules for calculating derivatives
of functions that are products, or quotients of other functions. In this video, you'll
find statements of the product rule and the quotient rule, and some examples. But no proofs,
proofs are in a separate video. Before we start, let's recall the Psalm and the difference
rules. If f and g are differentiable functions, then the derivative of the psalm f of x plus
g of x is just the sum of the derivatives. And a similar statement holds for differences.
The derivative of the difference is the difference of the derivatives. So does the same sort
of rule hold for products of functions, in other words, is the derivative of the product
equal to the product of the derivatives? Let's look at a simple example. To find out. For
example, if f of x is x, and g of x is x squared, then if we take the derivative of the product,
x times x squared, well, that's just the derivative of x cubed. We know how to do that with the
power rule. So it's 3x squared. On the other hand, if we look at the product of the derivatives,
we get one times 2x or just 2x. And these two things are not equal. So unfortunately,
the answer is no such a simple product rule does not hold. But not lose hope. There is
a product rule, it's just a little more complicated than the sum and difference rule. The product
rule says that if f and g are differentiable functions, then the derivative of the product
f of x times g of x is equal to f of x times the derivative of g of x plus the derivative
of f of x times g of x. In other words, to take the derivative of a product, we take
the first function times the derivative of the second, plus the derivative of the first
times the second. Let's use this in an example. To take the derivative of the square root
of t times e to the t, we have to take the first function squared of t times the derivative
of the second function, plus the derivative of the first function
times the second function. So that's the square root of t times the derivative of e to the
t is just e to the t.
times the second function. So that's the square root of t times the derivative of e to the
t is just e to the t.
And to find the derivative of the square root of t, it's going to be easier to write it
in exponential form. Now we can just use the power roll. Bring down the one half t to the
one half minus one is negative one half and I found the derivative. I'm going to clean
this up a little bit, and I'm done. quotient rule says that the derivative of a quotient
of two functions
And to find the derivative of the square root of t, it's going to be easier to write it
in exponential form. Now we can just use the power roll. Bring down the one half t to the
one half minus one is negative one half and I found the derivative. I'm going to clean
this up a little bit, and I'm done. quotient rule says that the derivative of a quotient
of two functions
is given By this quotient on the denominator, we have the denominator function g of x squared.
And on the numerator, we have g of x times the derivative
is given By this quotient on the denominator, we have the denominator function g of x squared.
And on the numerator, we have g of x times the derivative
of f of x
of f of x
minus f of x times the derivative of g of x. The way I remember this is this chant.
If you think of f of x as the high function, and g of x is a low function, you can say
this is low D high minus high D low
minus f of x times the derivative of g of x. The way I remember this is this chant.
If you think of f of x as the high function, and g of x is a low function, you can say
this is low D high minus high D low
over low low,
over low low,
we're low low means the low function squared. Let's start with a pretty simple example.
So this derivative, taking us back to z here,
we're low low means the low function squared. Let's start with a pretty simple example.
So this derivative, taking us back to z here,
we put the low low on the bottom, and then we go low,
we put the low low on the bottom, and then we go low,
D high.
D high.
z squared is to z minus high, D low the derivative of Z's cubed plus one is three z squared plus
zero, don't really need to write the zero there, I can simplify here a little bit to
z to the fourth plus two z minus three z to the fourth,
z squared is to z minus high, D low the derivative of Z's cubed plus one is three z squared plus
zero, don't really need to write the zero there, I can simplify here a little bit to
z to the fourth plus two z minus three z to the fourth,
over, I'm
over, I'm
not going to bother multiplying out this denominator, I think it looks simpler with it factored.
So when I cancel things on the numerator, I'm getting to z minus z to the fourth over
z cubed plus one squared as the derivative of my quotient. So in this video, we saw the
product rule and the quotient rule. I've written the rules here using the prime notation instead
of the dy dx notation, but you should check that the formulas are really the same as before.
For the proofs of these fabulously useful rules, you'll have to watch the next video.
In this video, I'll prove the product rule and the quotient rule, along with one more
related rule, called the reciprocal rule. First, the proof of the product rule. To find
the derivative of the product, f of x times g of x, I'm going to start as usual with the
limit definition of derivative. So the limit as h goes to zero of f of x plus h g of x
plus h minus f of x g of x, Oliver H. Now I'd like to make this expression here look
more like the expression above it, which is where I'm heading for. And to do this, I'm
going to use a classic trick of adding zero to my expression and kind of a devious way.
So I'm going to rewrite my expression, leaving the first term and the last term of the numerator
as they are, but inserting two new terms that cancel. So I'm subtracting the term, f of
x, g of x plus h, and then adding it back again. So they won't change the value of my
expression. This is not as pointless as it seems, because now we can factor out the common
factor of g of x plus h from the first two terms, and the common factor of f of x from
the next two terms. So I'm going to do that. And now I'll rewrite again, splitting up my
sum into two pieces here. You can use just algebra fractions to see that this expression
here and this expression here are the same. Notice that I'm taking the limit of this entire
expression here. Now my limit rules allow me to rewrite this limit as four separate
limits, provided that these four limits do in fact exist, which that tell you show you
in a moment that they do. So this first limit here is just equal to g of x, because G is
a continuous function. G is continuous because by assumption it's differentiable, so it has
to be continuous. This second limit here, you'll recognize as the definition of the
derivative of f, so that limit exists at equals d dx of f of x. The third limit, well, f of
x has nothing to do with age. So that limit is just f of x. And finally, the fourth limit
is the derivative of g. And we've done it. Well, modulus some minor rearrangement, you'll
see that this expression here, is exactly the same as this expression here, just with
the order of the terms switched around. Before we go on to prove the quotient rule, it'll
be really handy to prove the reciprocal rule, which states that the derivative of reciprocal,
one over f of x is given by negative the derivative of f of x divided by f of x squared. So to
prove this fact, let's start as usual, with the definition of derivative. The derivative
of one over f of x is the limit as h goes to zero of one over f of
not going to bother multiplying out this denominator, I think it looks simpler with it factored.
So when I cancel things on the numerator, I'm getting to z minus z to the fourth over
z cubed plus one squared as the derivative of my quotient. So in this video, we saw the
product rule and the quotient rule. I've written the rules here using the prime notation instead
of the dy dx notation, but you should check that the formulas are really the same as before.
For the proofs of these fabulously useful rules, you'll have to watch the next video.
In this video, I'll prove the product rule and the quotient rule, along with one more
related rule, called the reciprocal rule. First, the proof of the product rule. To find
the derivative of the product, f of x times g of x, I'm going to start as usual with the
limit definition of derivative. So the limit as h goes to zero of f of x plus h g of x
plus h minus f of x g of x, Oliver H. Now I'd like to make this expression here look
more like the expression above it, which is where I'm heading for. And to do this, I'm
going to use a classic trick of adding zero to my expression and kind of a devious way.
So I'm going to rewrite my expression, leaving the first term and the last term of the numerator
as they are, but inserting two new terms that cancel. So I'm subtracting the term, f of
x, g of x plus h, and then adding it back again. So they won't change the value of my
expression. This is not as pointless as it seems, because now we can factor out the common
factor of g of x plus h from the first two terms, and the common factor of f of x from
the next two terms. So I'm going to do that. And now I'll rewrite again, splitting up my
sum into two pieces here. You can use just algebra fractions to see that this expression
here and this expression here are the same. Notice that I'm taking the limit of this entire
expression here. Now my limit rules allow me to rewrite this limit as four separate
limits, provided that these four limits do in fact exist, which that tell you show you
in a moment that they do. So this first limit here is just equal to g of x, because G is
a continuous function. G is continuous because by assumption it's differentiable, so it has
to be continuous. This second limit here, you'll recognize as the definition of the
derivative of f, so that limit exists at equals d dx of f of x. The third limit, well, f of
x has nothing to do with age. So that limit is just f of x. And finally, the fourth limit
is the derivative of g. And we've done it. Well, modulus some minor rearrangement, you'll
see that this expression here, is exactly the same as this expression here, just with
the order of the terms switched around. Before we go on to prove the quotient rule, it'll
be really handy to prove the reciprocal rule, which states that the derivative of reciprocal,
one over f of x is given by negative the derivative of f of x divided by f of x squared. So to
prove this fact, let's start as usual, with the definition of derivative. The derivative
of one over f of x is the limit as h goes to zero of one over f of
x plus
x plus
h minus one over f of x over h. Now, these fractions here are just crying out to be combined
by finding a common denominator that common denominators f of x plus h times f of x. So
let me do that. I've just multiplied the first fraction by f of x over f of x and the second
fraction by f of x plus h over f of x plus h in order to rewrite with a common denominator.
So that gives me f of x minus f of x plus h divided by f of x plus h times f of x. And
instead of dividing this whole thing by H, I'll multiply it by one of our age, which
gives me another factor of H in the denominator. Now, this expression here is looking a lot
like the derivative of
h minus one over f of x over h. Now, these fractions here are just crying out to be combined
by finding a common denominator that common denominators f of x plus h times f of x. So
let me do that. I've just multiplied the first fraction by f of x over f of x and the second
fraction by f of x plus h over f of x plus h in order to rewrite with a common denominator.
So that gives me f of x minus f of x plus h divided by f of x plus h times f of x. And
instead of dividing this whole thing by H, I'll multiply it by one of our age, which
gives me another factor of H in the denominator. Now, this expression here is looking a lot
like the derivative of
f.
f.
It's just in the reverse order. So let me factor out a negative sign in order to let
me switch that order here. So this becomes f of x plus h minus f of x over h, and then
I've got the times one over f of x plus h times f of x. Now I can split this limit up
first, I'll factor out the negative sign. And then I'll write this product as a product
of two limits, which I can do provided the component limits exist. And I'll check that
these limits do exist. Let's see the first limit here. The first limit here is just the
derivative of
It's just in the reverse order. So let me factor out a negative sign in order to let
me switch that order here. So this becomes f of x plus h minus f of x over h, and then
I've got the times one over f of x plus h times f of x. Now I can split this limit up
first, I'll factor out the negative sign. And then I'll write this product as a product
of two limits, which I can do provided the component limits exist. And I'll check that
these limits do exist. Let's see the first limit here. The first limit here is just the
derivative of
f.
f.
And the second limit here exists because f is continuous, f is continuous since is differentiable.
So by continuity, as H is going to zero, since x plus h is approaching x, f of x plus h is
just approaching f of x. And I can rewrite this limit as one over f of x times f of x.
And so this is in other words, negative the derivative of f of x divided by f of x squared.
Now we've proved the reciprocal rule. Now we're in a great position to prove the quotient
rule with very little effort. So instead of going back to the definition of derivative,
this time, I'm just gonna think of the quotient f of x over g of x as a product of f of x
times the reciprocal
And the second limit here exists because f is continuous, f is continuous since is differentiable.
So by continuity, as H is going to zero, since x plus h is approaching x, f of x plus h is
just approaching f of x. And I can rewrite this limit as one over f of x times f of x.
And so this is in other words, negative the derivative of f of x divided by f of x squared.
Now we've proved the reciprocal rule. Now we're in a great position to prove the quotient
rule with very little effort. So instead of going back to the definition of derivative,
this time, I'm just gonna think of the quotient f of x over g of x as a product of f of x
times the reciprocal
of g of x.
of g of x.
And now, by the product rule, that's just the first function times the derivative of
the second plus the derivative of the first times the second. And by the quotient rule,
the derivative of this reciprocal is negative derivative of g over g of x squared. And I
still have this second term here, which I'm just going to write as derivative f of x divided
by g of x.
And now, by the product rule, that's just the first function times the derivative of
the second plus the derivative of the first times the second. And by the quotient rule,
the derivative of this reciprocal is negative derivative of g over g of x squared. And I
still have this second term here, which I'm just going to write as derivative f of x divided
by g of x.
Okay,
Okay,
so we're almost there. If we combine these two fractions, using a common denominator
of g of x squared, we just have to multiply this second fraction by g of x over g of x
to get that common denominator. Now we get negative f of x times the derivative of g
of x plus the derivative of f of x times g of x divided by g of x squared. And hopefully,
this bottom expression is the same as this top expression. And, yes, after rearranging
the terms it is. So that's the end of the proof of the quotient rule. So this video
gave proofs of the product rule, the reciprocal rule, and then the quotient rule. This video
is about two limits involving trig functions that turn out to be very useful. Namely, the
limit as theta goes to zero of sine theta
so we're almost there. If we combine these two fractions, using a common denominator
of g of x squared, we just have to multiply this second fraction by g of x over g of x
to get that common denominator. Now we get negative f of x times the derivative of g
of x plus the derivative of f of x times g of x divided by g of x squared. And hopefully,
this bottom expression is the same as this top expression. And, yes, after rearranging
the terms it is. So that's the end of the proof of the quotient rule. So this video
gave proofs of the product rule, the reciprocal rule, and then the quotient rule. This video
is about two limits involving trig functions that turn out to be very useful. Namely, the
limit as theta goes to zero of sine theta
over theta. And the limit as theta goes to zero of cosine theta minus one over theta.
These limits turn out to have really nice simple answers, as long as we keep theta in
radians,
over theta. And the limit as theta goes to zero of cosine theta minus one over theta.
These limits turn out to have really nice simple answers, as long as we keep theta in
radians,
not degrees. Let's consider the limit on the left first, that's the limit as theta goes
to zero of sine theta over theta. Notice that we can't just evaluate this limit by plugging
in zero for theta, because as theta goes to zero sine theta, and the numerator also goes
to zero, and theta itself goes to zero, so we end up with a zero over zero indeterminate
form. We can however, build up some evidence of what this limit by might be by using a
calculator and a table of values, or by looking at a graph. So here's the theta axis. And
here's the y axis. And you can see that as theta goes to zero from either the right or
the left, it's looking like the y value is going to one. The second limit here on the
right, is also zero over zero and determinant form. Since as theta goes to zero, cosine
theta goes to one, so cosine theta minus one goes to zero. But again, looking at the graph,
we have some evidence to suggest that as theta goes to zero, our expression is also going
to
not degrees. Let's consider the limit on the left first, that's the limit as theta goes
to zero of sine theta over theta. Notice that we can't just evaluate this limit by plugging
in zero for theta, because as theta goes to zero sine theta, and the numerator also goes
to zero, and theta itself goes to zero, so we end up with a zero over zero indeterminate
form. We can however, build up some evidence of what this limit by might be by using a
calculator and a table of values, or by looking at a graph. So here's the theta axis. And
here's the y axis. And you can see that as theta goes to zero from either the right or
the left, it's looking like the y value is going to one. The second limit here on the
right, is also zero over zero and determinant form. Since as theta goes to zero, cosine
theta goes to one, so cosine theta minus one goes to zero. But again, looking at the graph,
we have some evidence to suggest that as theta goes to zero, our expression is also going
to
zero.
zero.
these graphs provide strong evidence, but graphs can be misleading, and they're no substitute
for a rigorous proof. So for a pretty cool geometric and algebraic proof of these facts,
please see the proof video for this section. The fact that the limit as theta goes to zero
of sine theta over theta is one is really handy when you want to approximate sine theta.
Because intuitively, this is saying that sine theta is approximately equal to theta itself
when theta is near zero, because the ratio is approximately one. So if I want to approximate
sine of this value of theta, without a calculator, I can use that fact. And say that the sine
of 0.01769 is going to be approximately equal to 0.01769. This is an important time to remind
you that when we're doing these limits, we're assuming that theta is in radians. If it's
not in radians, we won't get this nice limit of one here. So that's our approximation.
And we can check it on a calculator, and I actually get an exact value of this number
up to 10 decimal places. So as you can see, this is a really good approximation. We can
use this same limit fat, again, in the next example, to calculate this complicated limit
as x goes to zero, the limit of tan of 7x over sine of forex. So when I see tangents
and signs and expression, I'm always tempted to rewrite things. And just in terms of sine
and cosine, so I'm going to do that first I'm going to rewrite tangent as as sine over
cosine. That still divided by sine of forex, and now I'm going to flip and multiply to
get sine of 7x over cosine of 7x times one over sine of 4x. Now intuitively, if x is
near zero, therefore 7x and 4x are also near zero, then sine of 7x is approximately equal
to 7x. And sine of 4x is approximately equal to 4x. So intuitively, this limit should be
pretty much the same thing as the limit as x goes to zero of 7x over cosine 7x times
4x. And canceling the access, this is just the same as seven fourths times the limit
of one of our cosine 7x. Since cosine of 7x is going to one, this should just be seven
fourths. So this is the intuitive approach, let me also give you a more rigorous approach.
So more rigorously, I'm going to rewrite this limit by multiplying by 7x over 7x. And by
multiplying by 4x, over 4x, that hasn't changed my expression, I'm just multiplying by one
and fancy forms. But this is really useful. Because if I regroup here, and write the sine
7x over the 7x, times the one over cosine 7x. Now I'm going to write the Forex over
the sign for x. And I'm still left with a 7x from the top and a 4x. From the bottom
here, I can cancel out those x's. And I can notice that this limit here, as x goes to
07, x is going to zero, so sine 7x over 7x is just going to be equal to one. And similarly
as x goes to zero, for x is going to zero. So the limit of 4x ever signed for x is the
reciprocal of one, it's also one. And finally, this limit in the middle here, as x goes to
07, x is going to zero, so cosine of 7x is going to one, and everything in the world
is going to one except the 7/4. So this limit is seven forth.
these graphs provide strong evidence, but graphs can be misleading, and they're no substitute
for a rigorous proof. So for a pretty cool geometric and algebraic proof of these facts,
please see the proof video for this section. The fact that the limit as theta goes to zero
of sine theta over theta is one is really handy when you want to approximate sine theta.
Because intuitively, this is saying that sine theta is approximately equal to theta itself
when theta is near zero, because the ratio is approximately one. So if I want to approximate
sine of this value of theta, without a calculator, I can use that fact. And say that the sine
of 0.01769 is going to be approximately equal to 0.01769. This is an important time to remind
you that when we're doing these limits, we're assuming that theta is in radians. If it's
not in radians, we won't get this nice limit of one here. So that's our approximation.
And we can check it on a calculator, and I actually get an exact value of this number
up to 10 decimal places. So as you can see, this is a really good approximation. We can
use this same limit fat, again, in the next example, to calculate this complicated limit
as x goes to zero, the limit of tan of 7x over sine of forex. So when I see tangents
and signs and expression, I'm always tempted to rewrite things. And just in terms of sine
and cosine, so I'm going to do that first I'm going to rewrite tangent as as sine over
cosine. That still divided by sine of forex, and now I'm going to flip and multiply to
get sine of 7x over cosine of 7x times one over sine of 4x. Now intuitively, if x is
near zero, therefore 7x and 4x are also near zero, then sine of 7x is approximately equal
to 7x. And sine of 4x is approximately equal to 4x. So intuitively, this limit should be
pretty much the same thing as the limit as x goes to zero of 7x over cosine 7x times
4x. And canceling the access, this is just the same as seven fourths times the limit
of one of our cosine 7x. Since cosine of 7x is going to one, this should just be seven
fourths. So this is the intuitive approach, let me also give you a more rigorous approach.
So more rigorously, I'm going to rewrite this limit by multiplying by 7x over 7x. And by
multiplying by 4x, over 4x, that hasn't changed my expression, I'm just multiplying by one
and fancy forms. But this is really useful. Because if I regroup here, and write the sine
7x over the 7x, times the one over cosine 7x. Now I'm going to write the Forex over
the sign for x. And I'm still left with a 7x from the top and a 4x. From the bottom
here, I can cancel out those x's. And I can notice that this limit here, as x goes to
07, x is going to zero, so sine 7x over 7x is just going to be equal to one. And similarly
as x goes to zero, for x is going to zero. So the limit of 4x ever signed for x is the
reciprocal of one, it's also one. And finally, this limit in the middle here, as x goes to
07, x is going to zero, so cosine of 7x is going to one, and everything in the world
is going to one except the 7/4. So this limit is seven forth.
In this video, we found that the limit as theta goes to zero of sine theta over theta
is equal to one. And the limit as theta goes to zero of cosine theta minus one over theta
is equal to zero. There's a nice proof of these facts in a later video for this section.
When you compose two functions, you apply the first function. And then you apply the
second function to the output of the first function. For example, the first function
might compute population size from time in years.
In this video, we found that the limit as theta goes to zero of sine theta over theta
is equal to one. And the limit as theta goes to zero of cosine theta minus one over theta
is equal to zero. There's a nice proof of these facts in a later video for this section.
When you compose two functions, you apply the first function. And then you apply the
second function to the output of the first function. For example, the first function
might compute population size from time in years.
So its input would be time in years, since a certain date, as output would be number
of people in the population. The second function g, might compute health care costs as a function
of population size. So it will take population size as input, and its output will be healthcare
costs. If you put these functions together, that is compose them, then you'll go all the
way from time in years to healthcare costs. This is your composition, g composed with
F. The composition of two functions, written g with a little circle, f of x is defined
as follows. g composed with f of x is G evaluated on f of x, we can think of it schematically
and so diagram, f acts on a number x and produces a number f of x, then g takes that output
f of x and produces a new number, g of f of x. Our composition of functions g composed
with F is the function that goes all the way from X to g of f of x. Let's work out some
examples where our functions are defined by tables of values. If we want to find g composed
with F of four, by definition, this means g of f of four. To evaluate this expression,
we always work from the inside out. So we start with the x value of four, and we find
f of four using the table of values for f of x. When x equals four, f of x is seven,
so we can replace F of four with the number seven. Now we need to evaluate g of seven,
seven becomes our new x value in our table of values for G, the x value of seven corresponds
to the G of X value of 10. So g of seven is equal to 10. We found that g composed with
F of four is equal to 10. If instead we want to find f composed with g of four, well, we
can rewrite that is f of g of four, and again, work from the inside out. Now we're trying
to find g of four, so four is our x value. And we use our table of values for G to see
that g of four is one. So we replaced by a four by one, and now we need to evaluate f
of one. Using our table for F values, f of one is eight. Notice that when we've computed
g of f of four, we got a different answer there when we computed F of G, F four. And
in general, g composed with F is not the same thing as f composed with g.
So its input would be time in years, since a certain date, as output would be number
of people in the population. The second function g, might compute health care costs as a function
of population size. So it will take population size as input, and its output will be healthcare
costs. If you put these functions together, that is compose them, then you'll go all the
way from time in years to healthcare costs. This is your composition, g composed with
F. The composition of two functions, written g with a little circle, f of x is defined
as follows. g composed with f of x is G evaluated on f of x, we can think of it schematically
and so diagram, f acts on a number x and produces a number f of x, then g takes that output
f of x and produces a new number, g of f of x. Our composition of functions g composed
with F is the function that goes all the way from X to g of f of x. Let's work out some
examples where our functions are defined by tables of values. If we want to find g composed
with F of four, by definition, this means g of f of four. To evaluate this expression,
we always work from the inside out. So we start with the x value of four, and we find
f of four using the table of values for f of x. When x equals four, f of x is seven,
so we can replace F of four with the number seven. Now we need to evaluate g of seven,
seven becomes our new x value in our table of values for G, the x value of seven corresponds
to the G of X value of 10. So g of seven is equal to 10. We found that g composed with
F of four is equal to 10. If instead we want to find f composed with g of four, well, we
can rewrite that is f of g of four, and again, work from the inside out. Now we're trying
to find g of four, so four is our x value. And we use our table of values for G to see
that g of four is one. So we replaced by a four by one, and now we need to evaluate f
of one. Using our table for F values, f of one is eight. Notice that when we've computed
g of f of four, we got a different answer there when we computed F of G, F four. And
in general, g composed with F is not the same thing as f composed with g.
Please
Please
pause the video and take a moment to compute the next two examples. We can replace f composed
with F of two by the equivalent expression, f of f of two. Working from the inside out,
we know that f of two is three, and f of three is six. If we want to find f composed with
g of six, rewrite that as f of g of six isn't the table for g, g of six is eight. But F
of eight, eight is not on the table as an x value for the for the f function. And so
there is no F of eight, this does not exist, we can say that six is not in the domain,
for F composed with g. Even though it was in the domain of g, we couldn't follow all
the way through and get a value for F composed with g of six. Next, let's turn our attention
to the composition of functions that are given by equations.
pause the video and take a moment to compute the next two examples. We can replace f composed
with F of two by the equivalent expression, f of f of two. Working from the inside out,
we know that f of two is three, and f of three is six. If we want to find f composed with
g of six, rewrite that as f of g of six isn't the table for g, g of six is eight. But F
of eight, eight is not on the table as an x value for the for the f function. And so
there is no F of eight, this does not exist, we can say that six is not in the domain,
for F composed with g. Even though it was in the domain of g, we couldn't follow all
the way through and get a value for F composed with g of six. Next, let's turn our attention
to the composition of functions that are given by equations.
p of x is x squared plus x and q of x is negative 2x. We want to find q composed with P of one.
p of x is x squared plus x and q of x is negative 2x. We want to find q composed with P of one.
As usual, I can rewrite this as Q of P of one and work from the inside out. P of one
is one squared plus one, so that's two. So this is the same thing as Q of two. But queue
of two is negative two times two or negative four. So this evaluates to negative four.
In this next example, we want to find q composed with P of some arbitrary x, or rewrite it
as usual as Q of p of x and work from the inside out. Well, p of x, we know the formula
for that, that's x squared plus x. So I can replace my P of x with that expression. Now,
I'm stuck with evaluating q on x squared plus x. Well queue of anything is negative two
times that thing. So q of x squared plus x is going to be negative two times the quantity
x squared plus x, what I've done is I've substituted in the whole expression x squared plus x,
where I saw the X in this formula for q of x, it's important to use the parentheses here,
so that we'll be multiplying negative two by the whole expression and not just by the
first piece, I can simplify this a bit as negative 2x squared minus 2x. And that's my
expression for Q composed with p of x. Notice that if I wanted to compute q composed with
P of one, which I already did in the first problem, I could just use this expression
now, negative two times one squared minus two and I get negative four, just like I did
before. Let's try another one. Let's try p composed with q of x. First I read right this
P of q of x. Working from the inside out, I can replace q of x with negative 2x. So
I need to compute P of negative 2x. Here's my formula for P. to compute P of this expression,
I need to plug in this expression everywhere I see an x in the formula for P. So that means
negative 2x squared plus negative 2x. Again, being careful to use parentheses to make sure
I plug in the entire expression in for x. let me simplify. This is 4x squared minus
2x. Notice that I got different expressions for Q of p of x. And for P of q of x. Once
again, we see that q composed with P is not necessarily equal to P composed with Q. Please
pause the video and try this last example yourself rewriting and working from the inside
out, we're going to replace p of x with its expression x squared plus x. And then we need
to evaluate p on x squared
As usual, I can rewrite this as Q of P of one and work from the inside out. P of one
is one squared plus one, so that's two. So this is the same thing as Q of two. But queue
of two is negative two times two or negative four. So this evaluates to negative four.
In this next example, we want to find q composed with P of some arbitrary x, or rewrite it
as usual as Q of p of x and work from the inside out. Well, p of x, we know the formula
for that, that's x squared plus x. So I can replace my P of x with that expression. Now,
I'm stuck with evaluating q on x squared plus x. Well queue of anything is negative two
times that thing. So q of x squared plus x is going to be negative two times the quantity
x squared plus x, what I've done is I've substituted in the whole expression x squared plus x,
where I saw the X in this formula for q of x, it's important to use the parentheses here,
so that we'll be multiplying negative two by the whole expression and not just by the
first piece, I can simplify this a bit as negative 2x squared minus 2x. And that's my
expression for Q composed with p of x. Notice that if I wanted to compute q composed with
P of one, which I already did in the first problem, I could just use this expression
now, negative two times one squared minus two and I get negative four, just like I did
before. Let's try another one. Let's try p composed with q of x. First I read right this
P of q of x. Working from the inside out, I can replace q of x with negative 2x. So
I need to compute P of negative 2x. Here's my formula for P. to compute P of this expression,
I need to plug in this expression everywhere I see an x in the formula for P. So that means
negative 2x squared plus negative 2x. Again, being careful to use parentheses to make sure
I plug in the entire expression in for x. let me simplify. This is 4x squared minus
2x. Notice that I got different expressions for Q of p of x. And for P of q of x. Once
again, we see that q composed with P is not necessarily equal to P composed with Q. Please
pause the video and try this last example yourself rewriting and working from the inside
out, we're going to replace p of x with its expression x squared plus x. And then we need
to evaluate p on x squared
plus x.
plus x.
That means we plug in
That means we plug in
x squared plus x,
x squared plus x,
everywhere we see an x in this formula, so that's x squared plus x quantity squared plus
x squared plus x. Once again, I can simplify by distributing out, that gives me x to the
fourth plus 2x cubed plus x squared plus x squared plus x, or x to the fourth plus 2x
cubed plus 2x squared plus x. In this last set of examples, we're asked to go backwards,
we're given a formula for a function of h of x. But we're supposed to rewrite h of x
as a composition of two functions, F and G. Let's think for a minute, which of these two
functions gets applied first, f composed with g of x, let's see, that means f of g of x.
And since we evaluate these expressions from the inside out, we must be applying g first,
and then F. In order to figure out what what f and g could be, I like to draw a box around
some thing inside my expression for H, so I'm going to draw a box around x squared plus
seven, then whatever's inside the box, that'll be my function, g of x, the first function
that gets applied, whatever happens to the box, in this case, taking the square root
sign, that becomes my outside function, my second function f. So here, we're gonna say
g of x is equal to x squared plus seven, and f of x is equal to the square root of x, let's
just check and make sure that this works. So I need to check that when I take the composition,
f composed with g, I need to get the same thing as my original h. So let's see, if I
do f composed with g of x, well, by definition, that's f of g of x.
everywhere we see an x in this formula, so that's x squared plus x quantity squared plus
x squared plus x. Once again, I can simplify by distributing out, that gives me x to the
fourth plus 2x cubed plus x squared plus x squared plus x, or x to the fourth plus 2x
cubed plus 2x squared plus x. In this last set of examples, we're asked to go backwards,
we're given a formula for a function of h of x. But we're supposed to rewrite h of x
as a composition of two functions, F and G. Let's think for a minute, which of these two
functions gets applied first, f composed with g of x, let's see, that means f of g of x.
And since we evaluate these expressions from the inside out, we must be applying g first,
and then F. In order to figure out what what f and g could be, I like to draw a box around
some thing inside my expression for H, so I'm going to draw a box around x squared plus
seven, then whatever's inside the box, that'll be my function, g of x, the first function
that gets applied, whatever happens to the box, in this case, taking the square root
sign, that becomes my outside function, my second function f. So here, we're gonna say
g of x is equal to x squared plus seven, and f of x is equal to the square root of x, let's
just check and make sure that this works. So I need to check that when I take the composition,
f composed with g, I need to get the same thing as my original h. So let's see, if I
do f composed with g of x, well, by definition, that's f of g of x.
Working from the inside out, I can replace g of x with its formula x squared plus seven.
So I need to evaluate f of x squared plus seven. That means I plug in x squared plus
seven, into the formula for for F. So that becomes the square root of x squared plus
seven, two, it works because it matches my original equation. So we found a correct answer
a correct way of breaking h down as a composition of two functions. But I do want to point out,
this is not the only correct answer. I'll write down my formula for H of X again, and
this time, I'll put the box in a different place, I'll just box the x squared. If I did
that, then my inside function, my first function, g of x would be x squared. And my second function
is what happens
Working from the inside out, I can replace g of x with its formula x squared plus seven.
So I need to evaluate f of x squared plus seven. That means I plug in x squared plus
seven, into the formula for for F. So that becomes the square root of x squared plus
seven, two, it works because it matches my original equation. So we found a correct answer
a correct way of breaking h down as a composition of two functions. But I do want to point out,
this is not the only correct answer. I'll write down my formula for H of X again, and
this time, I'll put the box in a different place, I'll just box the x squared. If I did
that, then my inside function, my first function, g of x would be x squared. And my second function
is what happens
to the box.
to the box.
So my f of x is what happens to the box, and the box gets added seven to it, and taking
the square root. So in other words, f of x is going to be the square root of x plus seven.
Again, I can check that this works. If I do f composed with g of x, that's f
So my f of x is what happens to the box, and the box gets added seven to it, and taking
the square root. So in other words, f of x is going to be the square root of x plus seven.
Again, I can check that this works. If I do f composed with g of x, that's f
of g of x.
of g of x.
So now g of x is x squared, so I'm taking f of x squared. When I plug in x squared for
x, I do in fact get the square root of x squared plus seven. So this is an alternative correct
solution. In this video, we learn to evaluate the composition of functions. by rewriting
it and working from the inside out. We also learn to break apart a complicated function
into a composition of two functions by boxing one piece of the function and letting the
first function applied in the composition. Let that be the inside of the box and the
second function applied in the composition be whatever happens to the box.
So now g of x is x squared, so I'm taking f of x squared. When I plug in x squared for
x, I do in fact get the square root of x squared plus seven. So this is an alternative correct
solution. In this video, we learn to evaluate the composition of functions. by rewriting
it and working from the inside out. We also learn to break apart a complicated function
into a composition of two functions by boxing one piece of the function and letting the
first function applied in the composition. Let that be the inside of the box and the
second function applied in the composition be whatever happens to the box.
This video is about solving rational equations. A rational equation like this one equation
that has rational expressions and that, in other words, an equation that has some variables
in the denominator. There are several different approaches for solving a rational equation,
but they all start by finding the least common denominator. In this example, the denominators
are x plus three and x, we can think of one as just having a denominator of one. Since
the denominators don't have any factors in common, I can find the least common denominator
just by multiplying them together. My next step is going to be clearing the denominator.
By this, I mean that I multiply both sides of my equation by this least common denominator,
x plus three times x, I multiply on the left side of the equation, and I multiply by the
same thing on the right side of the equation. Since I'm doing the same thing to both sides
of the equation, I don't change the the value of the equation. Multiplying the least common
denominator on both sides of the equation is equivalent to multiplying it by all three
terms in the equation, I can see this when I multiply out, I'll rewrite the left side
the same as before, pretty much. And then I'll distribute the right side to get x plus
three times x times one plus x plus three times x times one over x. So I've actually
multiplied the least common denominator by all three terms of my equation. Now I can
have a blast canceling things. The x plus three cancels with the x plus three on the
denominator. The here are nothing cancels out because there's no denominator, and here
are the x in the numerator cancels with the x in the denominator. So I can rewrite my
expression as x squared equals x plus three times x times one plus x plus three. Now I'm
going to simplify. So I'll leave the x squared alone on this side, I'll distribute out x
squared plus 3x plus x plus three, hey, look, the x squared is cancel on both sides. And
so I get zero equals 4x plus three, so 4x is negative three, and x is negative three
fourths. Finally, I'm going to plug in my answer to check. This is a good idea for any
kind of equation. But it's especially important for a rational equation because occasionally
for rational equations, you'll get what's called extraneous solution solutions that
don't actually work in your original equation because they make the denominator zero. Now,
in this example, I don't think we're going to get the extraneous equations, because negative
three fourths is not going to make any of these denominators zero, so it should work
out fine when I plug in. If I plug in, I get this, I can simplify the denominator here,
negative three fourths plus three, three is 12, for sets becomes nine fourths, and this
is one or flip and multiply to get minus four thirds. So here, I can simplify my complex
fraction, it ends up being negative three nights, and one minus four thirds is negative
1/3. So that all seems to check out. And so my final answer is x equals negative three
fourths. This next example looks a little trickier. And it is, but the same approach
will work. First off, find the least common denominator. So here, my denominators are
c minus five, c plus one, and C squared minus four c minus five, I'm going to factor that
as C minus five times c plus one. Now, my least common denominator needs to have just
enough factors to that each of these denominators divided into it. So I need the factor c minus
five, I need the factor c plus one. And now I've already got all the factors I need for
this denominator. So here is my least common denominator. Next step is to clear the denominators.
So I do this by multiplying both sides of the equation by my least common denominator.
In fact, I can just multiply each of the three terms by this least common denominator. I
went ahead and wrote my third denominator in factored form to make it easier to see
what cancels. Now canceling time dies, this dies, and both of those factors die.
This video is about solving rational equations. A rational equation like this one equation
that has rational expressions and that, in other words, an equation that has some variables
in the denominator. There are several different approaches for solving a rational equation,
but they all start by finding the least common denominator. In this example, the denominators
are x plus three and x, we can think of one as just having a denominator of one. Since
the denominators don't have any factors in common, I can find the least common denominator
just by multiplying them together. My next step is going to be clearing the denominator.
By this, I mean that I multiply both sides of my equation by this least common denominator,
x plus three times x, I multiply on the left side of the equation, and I multiply by the
same thing on the right side of the equation. Since I'm doing the same thing to both sides
of the equation, I don't change the the value of the equation. Multiplying the least common
denominator on both sides of the equation is equivalent to multiplying it by all three
terms in the equation, I can see this when I multiply out, I'll rewrite the left side
the same as before, pretty much. And then I'll distribute the right side to get x plus
three times x times one plus x plus three times x times one over x. So I've actually
multiplied the least common denominator by all three terms of my equation. Now I can
have a blast canceling things. The x plus three cancels with the x plus three on the
denominator. The here are nothing cancels out because there's no denominator, and here
are the x in the numerator cancels with the x in the denominator. So I can rewrite my
expression as x squared equals x plus three times x times one plus x plus three. Now I'm
going to simplify. So I'll leave the x squared alone on this side, I'll distribute out x
squared plus 3x plus x plus three, hey, look, the x squared is cancel on both sides. And
so I get zero equals 4x plus three, so 4x is negative three, and x is negative three
fourths. Finally, I'm going to plug in my answer to check. This is a good idea for any
kind of equation. But it's especially important for a rational equation because occasionally
for rational equations, you'll get what's called extraneous solution solutions that
don't actually work in your original equation because they make the denominator zero. Now,
in this example, I don't think we're going to get the extraneous equations, because negative
three fourths is not going to make any of these denominators zero, so it should work
out fine when I plug in. If I plug in, I get this, I can simplify the denominator here,
negative three fourths plus three, three is 12, for sets becomes nine fourths, and this
is one or flip and multiply to get minus four thirds. So here, I can simplify my complex
fraction, it ends up being negative three nights, and one minus four thirds is negative
1/3. So that all seems to check out. And so my final answer is x equals negative three
fourths. This next example looks a little trickier. And it is, but the same approach
will work. First off, find the least common denominator. So here, my denominators are
c minus five, c plus one, and C squared minus four c minus five, I'm going to factor that
as C minus five times c plus one. Now, my least common denominator needs to have just
enough factors to that each of these denominators divided into it. So I need the factor c minus
five, I need the factor c plus one. And now I've already got all the factors I need for
this denominator. So here is my least common denominator. Next step is to clear the denominators.
So I do this by multiplying both sides of the equation by my least common denominator.
In fact, I can just multiply each of the three terms by this least common denominator. I
went ahead and wrote my third denominator in factored form to make it easier to see
what cancels. Now canceling time dies, this dies, and both of those factors die.
cancel out the denominators, the whole point of multiplying by the least common denominator,
you're multiplying by something that's big enough to kill every single denominator so
you don't have to deal with denominators anymore. Now I'm going to simplify by multiplying out.
So I get, let's see, c plus one times four c, that's four c squared plus four c, now
I get minus just c minus five, and then over here, I get three c squared plus three, I
can rewrite the minus quantity c minus five as a minus c plus five. And now I can subtract
the three c squared from both sides to get just a C squared over here, and the four c
minus c, that becomes a three C. And finally, I can subtract the three from both sides to
get c squared plus three c plus two equals zero. got myself a quadratic equation that
looks like a nice one that factors. So this factors to C plus one times c plus two equals
zero. So either c plus one is zero, or C plus two is zero, so C equals negative one, or
C equals negative two. Now let's see, we need to still check our answers. Without even going
to the trouble of calculating anything, I can see that C equals negative one is not
going to work, because if I plug it in to this denominator here, I get a denominator
of zero, which doesn't make sense. So C equals minus one is an extraneous solution, it doesn't
actually satisfy my original equation. And so I can just cross it right out, C equals
negative two. I can go if I go ahead, and that doesn't make any of my denominators zero.
So if I haven't made any mistakes, it should satisfy my original equation, but, but I'll
just plug it in to be sure. And after some simplifying, I get a true statement. So my
final answer is C equals negative two. In this video, we solved a couple of rational
equations, using the method of finding the least common denominator, and then clearing
the denominator, we cleared the denominator by multiplying both sides of the equation
by the least common denominator or equivalently. multiplying each of the terms by that denominator.
There's another equivalent method that some people prefer, it still starts out the same,
we find the least common denominator, but then we write all the fractions over that
least common denominator. So in this example, we'd still use the least common denominator
of x plus three times x.
cancel out the denominators, the whole point of multiplying by the least common denominator,
you're multiplying by something that's big enough to kill every single denominator so
you don't have to deal with denominators anymore. Now I'm going to simplify by multiplying out.
So I get, let's see, c plus one times four c, that's four c squared plus four c, now
I get minus just c minus five, and then over here, I get three c squared plus three, I
can rewrite the minus quantity c minus five as a minus c plus five. And now I can subtract
the three c squared from both sides to get just a C squared over here, and the four c
minus c, that becomes a three C. And finally, I can subtract the three from both sides to
get c squared plus three c plus two equals zero. got myself a quadratic equation that
looks like a nice one that factors. So this factors to C plus one times c plus two equals
zero. So either c plus one is zero, or C plus two is zero, so C equals negative one, or
C equals negative two. Now let's see, we need to still check our answers. Without even going
to the trouble of calculating anything, I can see that C equals negative one is not
going to work, because if I plug it in to this denominator here, I get a denominator
of zero, which doesn't make sense. So C equals minus one is an extraneous solution, it doesn't
actually satisfy my original equation. And so I can just cross it right out, C equals
negative two. I can go if I go ahead, and that doesn't make any of my denominators zero.
So if I haven't made any mistakes, it should satisfy my original equation, but, but I'll
just plug it in to be sure. And after some simplifying, I get a true statement. So my
final answer is C equals negative two. In this video, we solved a couple of rational
equations, using the method of finding the least common denominator, and then clearing
the denominator, we cleared the denominator by multiplying both sides of the equation
by the least common denominator or equivalently. multiplying each of the terms by that denominator.
There's another equivalent method that some people prefer, it still starts out the same,
we find the least common denominator, but then we write all the fractions over that
least common denominator. So in this example, we'd still use the least common denominator
of x plus three times x.
But our next step would be to write each of these rational expressions over that common
denominator by multiplying the top and the bottom by the appropriate things. So one,
in order to get the common denominator of x plus 3x, I need to multiply the top and
the bottom by x plus three times x, whenever x, I need to multiply the top and the bottom
just by x plus three since that's what's missing from the denominator x. Now, if I simplify
a little bit, let's say this is x squared over that common denominator, and here I have
just x plus three times x over that denominator, and here I have x plus three over that common
denominator. Now add together my fractions on the right side, so they have a common denominator.
So this is x plus three times x plus x plus three. And now I have two fractions that have
that are equal, that have the same denominator, therefore, their numerators have to be equal
also. So the next step is to set the numerators equal. So I get x squared is x plus three
times x plus x plus three. And if you look back at the previous way, we solve this equation,
you'll recognize this equation. And so from here on we just continue as before. When choosing
between these two methods, I personally tend to prefer the clear the denominators method,
because it's a little bit less writing, you don't have to get rid of those denominators
earlier. You don't have to write them as many times, but some people find this one a little
bit easier to remember, a little easier to understand either of these methods is fine.
One last caution. Don't forget at the end, to check your solutions and eliminate any
extraneous solutions. These will be solutions that make the denominators of your original
equation. Go to zero. This video gives the derivative of sine cosine and other trig functions.
A graph of the function y equals sine x is given in blue here, we can estimate the shape
of the derivative of sine x by looking at the slopes of the tangent lines. Here, when
x equals zero, the tangent line has a positive slope of approximately one. As x increases
to pi over two, the slope of the tangent line is still positive, but decreases to zero.
Next, the slope turns negative more and more negative reaching a negative value of negative
one, before returning again to zero. Continuing like this, we see that the graph of the derivative
y equals sine prime of x looks like the graph of y equals cosine x below. Please pause the
video and do a similar exercise for the graph of y equals cosine of x below, that is use
the graph of y equals cosine x to estimate the shape of the graph of y equals cosine
prime of x. Notice that when x equals zero, the slope of the tangent line here is zero,
that slope turns negative, and then reaches zero again before turning positive. So the
graph of the derivative should look something like this. This new blue graph looks like
the vertical reflection of the blue graph above suggesting that the derivative of cosine
of x is equal to the negative of sine of x. So we have graphical evidence that the derivative
of sine x is equal to cosine of x, and the derivative of cosine of x is equal to negative
sine of x. For proofs of these facts, please see the separate proof video for this section.
Once we have the derivatives of sine and cosine, we have the power to compute the derivatives
of a lot of other trig functions as well.
But our next step would be to write each of these rational expressions over that common
denominator by multiplying the top and the bottom by the appropriate things. So one,
in order to get the common denominator of x plus 3x, I need to multiply the top and
the bottom by x plus three times x, whenever x, I need to multiply the top and the bottom
just by x plus three since that's what's missing from the denominator x. Now, if I simplify
a little bit, let's say this is x squared over that common denominator, and here I have
just x plus three times x over that denominator, and here I have x plus three over that common
denominator. Now add together my fractions on the right side, so they have a common denominator.
So this is x plus three times x plus x plus three. And now I have two fractions that have
that are equal, that have the same denominator, therefore, their numerators have to be equal
also. So the next step is to set the numerators equal. So I get x squared is x plus three
times x plus x plus three. And if you look back at the previous way, we solve this equation,
you'll recognize this equation. And so from here on we just continue as before. When choosing
between these two methods, I personally tend to prefer the clear the denominators method,
because it's a little bit less writing, you don't have to get rid of those denominators
earlier. You don't have to write them as many times, but some people find this one a little
bit easier to remember, a little easier to understand either of these methods is fine.
One last caution. Don't forget at the end, to check your solutions and eliminate any
extraneous solutions. These will be solutions that make the denominators of your original
equation. Go to zero. This video gives the derivative of sine cosine and other trig functions.
A graph of the function y equals sine x is given in blue here, we can estimate the shape
of the derivative of sine x by looking at the slopes of the tangent lines. Here, when
x equals zero, the tangent line has a positive slope of approximately one. As x increases
to pi over two, the slope of the tangent line is still positive, but decreases to zero.
Next, the slope turns negative more and more negative reaching a negative value of negative
one, before returning again to zero. Continuing like this, we see that the graph of the derivative
y equals sine prime of x looks like the graph of y equals cosine x below. Please pause the
video and do a similar exercise for the graph of y equals cosine of x below, that is use
the graph of y equals cosine x to estimate the shape of the graph of y equals cosine
prime of x. Notice that when x equals zero, the slope of the tangent line here is zero,
that slope turns negative, and then reaches zero again before turning positive. So the
graph of the derivative should look something like this. This new blue graph looks like
the vertical reflection of the blue graph above suggesting that the derivative of cosine
of x is equal to the negative of sine of x. So we have graphical evidence that the derivative
of sine x is equal to cosine of x, and the derivative of cosine of x is equal to negative
sine of x. For proofs of these facts, please see the separate proof video for this section.
Once we have the derivatives of sine and cosine, we have the power to compute the derivatives
of a lot of other trig functions as well.
And notice that a nice way to remember which of these answers have negative signs in them
is that the derivatives of the trig functions that start with a co
And notice that a nice way to remember which of these answers have negative signs in them
is that the derivatives of the trig functions that start with a co
always have a negative, and the root of the trig functions that don't have the ko are
positive. Now let's use these formulas in an example. g of x is a complicated expression
involving several trig functions as well as a constant m, and I have a couple choices
of how to proceed. I could try to rewrite all my trig functions in terms of sine and
cosine and simplify, or I could attack the derivative directly using the quotient rule.
I'm going to use the direct approach In this case, but sometimes you'll find that rewriting
will make things easier. So using the quotient rule on the denominator, I get the original
denominator squared. On the numerator, I get low D high to compute the derivative of x
cosine x, I need the product rule. So I get x times the derivative of cosine, which is
negative sine x, plus the derivative of x, which is just one times cosine of x. Now I
have to do a minus Hi, x cosine of x dillow. The derivative of M is just zero because M
is a constant, plus the derivative of cotangent which is negative cosecant squared of x. So
I found the derivative, I'm going to go ahead and simplify a little bit by multiplying out
then rewriting everything in terms of sine and cosine, and then multiplying the numerator
and denominator by sine squared of x, we have a somewhat simplified expression for the derivative,
you should memorize the derivatives of the trig functions will prove that the first two
formulas are correct in a separate proof video. In this video, I'll give proofs for the two
special trig limit. And I'll also prove that the derivative of sine is cosine. And the
derivative of cosine is minus sign. To prove that the limit of sine that over theta is
one as theta goes to zero, I'm going to start with a picture. In this picture, I have a
unit circle a circle of radius one, and I have two right triangles, a green triangle
and a smaller red triangle, both with angle theta. Now I'm going to argue in terms of
areas, if I want to compute the area of this sector that I've shaded in blue here, in other
words, that pie shaped piece, I can first compute the area of the circle, which is pi
times one squared for the radius. But since the sector has angle theta, and the full circle
has angle two pi, I need to multiply that area of the circle by the ratio theta over
two pi to represent the fraction of the area of the circle that's included in this sector.
So in other words, the area of the sector is just going to be theta over two, where
theta is given in the radians. Now if I want to compute the area of the little red triangle,
I can do one half times the base times the height. Now the base is going to be equal
to cosine theta, because I have a circle of radius one angle theta here, and the height
is going to be sine theta. Finally, the area of the green triangle is also one half times
the base times the height. But now the base is a full one unit, and the height is given
by tangent theta, since opposite, which is the height here over adjacent, which is one
has to equal tangent theta. Now if I put all those areas together, I know that the area
of the red triangle, alright is cosine theta sine theta over two has to be less than or
equal to the area of the blue sector, theta over two, which is less than or equal to the
area of the big green triangle, which is tan theta over two. Now I'm going to multiply
through this inequality by two and rewrite things in terms of sine and cosine to get
cosine theta sine theta is less than or equal to theta is less than or equal to sine theta
over cosine theta. Now I'm going to divide through my inequalities by sine theta, which
won't change the inequalities as long as theta is greater than zero, so that sine theta is
positive.
always have a negative, and the root of the trig functions that don't have the ko are
positive. Now let's use these formulas in an example. g of x is a complicated expression
involving several trig functions as well as a constant m, and I have a couple choices
of how to proceed. I could try to rewrite all my trig functions in terms of sine and
cosine and simplify, or I could attack the derivative directly using the quotient rule.
I'm going to use the direct approach In this case, but sometimes you'll find that rewriting
will make things easier. So using the quotient rule on the denominator, I get the original
denominator squared. On the numerator, I get low D high to compute the derivative of x
cosine x, I need the product rule. So I get x times the derivative of cosine, which is
negative sine x, plus the derivative of x, which is just one times cosine of x. Now I
have to do a minus Hi, x cosine of x dillow. The derivative of M is just zero because M
is a constant, plus the derivative of cotangent which is negative cosecant squared of x. So
I found the derivative, I'm going to go ahead and simplify a little bit by multiplying out
then rewriting everything in terms of sine and cosine, and then multiplying the numerator
and denominator by sine squared of x, we have a somewhat simplified expression for the derivative,
you should memorize the derivatives of the trig functions will prove that the first two
formulas are correct in a separate proof video. In this video, I'll give proofs for the two
special trig limit. And I'll also prove that the derivative of sine is cosine. And the
derivative of cosine is minus sign. To prove that the limit of sine that over theta is
one as theta goes to zero, I'm going to start with a picture. In this picture, I have a
unit circle a circle of radius one, and I have two right triangles, a green triangle
and a smaller red triangle, both with angle theta. Now I'm going to argue in terms of
areas, if I want to compute the area of this sector that I've shaded in blue here, in other
words, that pie shaped piece, I can first compute the area of the circle, which is pi
times one squared for the radius. But since the sector has angle theta, and the full circle
has angle two pi, I need to multiply that area of the circle by the ratio theta over
two pi to represent the fraction of the area of the circle that's included in this sector.
So in other words, the area of the sector is just going to be theta over two, where
theta is given in the radians. Now if I want to compute the area of the little red triangle,
I can do one half times the base times the height. Now the base is going to be equal
to cosine theta, because I have a circle of radius one angle theta here, and the height
is going to be sine theta. Finally, the area of the green triangle is also one half times
the base times the height. But now the base is a full one unit, and the height is given
by tangent theta, since opposite, which is the height here over adjacent, which is one
has to equal tangent theta. Now if I put all those areas together, I know that the area
of the red triangle, alright is cosine theta sine theta over two has to be less than or
equal to the area of the blue sector, theta over two, which is less than or equal to the
area of the big green triangle, which is tan theta over two. Now I'm going to multiply
through this inequality by two and rewrite things in terms of sine and cosine to get
cosine theta sine theta is less than or equal to theta is less than or equal to sine theta
over cosine theta. Now I'm going to divide through my inequalities by sine theta, which
won't change the inequalities as long as theta is greater than zero, so that sine theta is
positive.
And I get cosine theta is less than or equal to theta over sine theta is less than or equal
to one over cosine theta. Now this middle expression is the reciprocal of the expression
I want to take the limit of. So I'm going to go ahead and take limits. And since the
limits of the two expressions on the outside, both exist, and equal one by the sandwich
theorem, the limit of the expression on the inside has to exist an equal one as well.
Now I've cheated a little bit here. And I've really just taken the limit from the right
because I've assumed that theta is greater than zero. But you can check that if you say
that less than zero, so that sign that as negative, the limit from the left will also
equal one, the inequalities will flip around first, but you'll still get it use the sandwich
theorem to get a limit of one. And that's a cool geometric proof of this useful limit
from calculus. To show that the limit of cosine theta minus one over theta is zero, we can
actually rewrite this expression and reuse the limit that we just computed. So let me
write down my limit. And I'm going to multiply this expression by cosine theta plus one on
the numerator and the denominator. So I haven't changed the expression, I'm just multiply
that by one. Now, if I multiply my numerator out,
And I get cosine theta is less than or equal to theta over sine theta is less than or equal
to one over cosine theta. Now this middle expression is the reciprocal of the expression
I want to take the limit of. So I'm going to go ahead and take limits. And since the
limits of the two expressions on the outside, both exist, and equal one by the sandwich
theorem, the limit of the expression on the inside has to exist an equal one as well.
Now I've cheated a little bit here. And I've really just taken the limit from the right
because I've assumed that theta is greater than zero. But you can check that if you say
that less than zero, so that sign that as negative, the limit from the left will also
equal one, the inequalities will flip around first, but you'll still get it use the sandwich
theorem to get a limit of one. And that's a cool geometric proof of this useful limit
from calculus. To show that the limit of cosine theta minus one over theta is zero, we can
actually rewrite this expression and reuse the limit that we just computed. So let me
write down my limit. And I'm going to multiply this expression by cosine theta plus one on
the numerator and the denominator. So I haven't changed the expression, I'm just multiply
that by one. Now, if I multiply my numerator out,
I get cosine squared theta minus one. And from the trig identity, sine squared theta
plus cosine squared theta equals one, I know that cosine squared theta minus one has to
equal minus sine squared. So I can rewrite my limit as the limit of minus sine squared
theta over theta cosine theta plus one. And now I can regroup to write my sine theta over
theta, and my other copy of sine theta over cosine theta plus one, the limit of the first
expression is going to be negative one, because of the limit we just proved. And the limit
of the second expression is just zero over one plus one, or zero, and therefore, my entire
limit is just going to be negative one times zero, or zero, which is exactly what we want
it to prove. Now we can use these two limits that we've just proved to calculate the derivatives
of sine and cosine, using the limit definition of derivative and prove the results that were
stated previously. According to the limit definition of derivative, the derivative of
sine x is the limit as h goes to zero of sine of x plus h minus sine of x divided by H.
As usual, this is a zero for zero indeterminate form limits. So I'm going to need to rewrite
things to evaluate it. And I'm going to rewrite using the angle sum formula for sine, the
sine of x plus h is equal to sine x cosine H, cosine plus cosine x sine H. Now if I rearrange
things, and factor out a sine x from the first term, I can break up my limit into pieces
and compute every piece. So this is sine x times zero plus cosine x times one. And so
my final answer is cosine x as we wanted the proof that the derivative of cosine is minus
sine is very similar. So please stop the video and try it for yourself before proceeding.
Using the limit definition of derivative, we have that the derivative of cosine of x
is the limit as h goes to zero of cosine of x plus h minus cosine of x over h, we can
rewrite the cosine of x plus h using the angle sum formula as the cosine of x times the cosine
of H minus the sine of x times the sine of H. And then we still have the minus cosine
of x over h. As before, we're going to regroup things and factoring out the cosine x from
the first part, the same familiar limits just put together in different ways. So here, cosine
of x as h goes to zero is just cosine of x. This limit we know is zero. sine of x is just
saying sine of x, and sine of h over h is going to one which means that our final answer
is going to be negative sine of x times one or just negative sine of x, which is exactly
what we wanted. That's all for the proofs of these four useful calculus facts. rectilinear
motion or linear motion means the motion of an object along a straight line. For example,
a particle moving left and right, or a ball going up and down. In this video, we'll see
what the derivative and the second derivative tell us about the motion of an object constrained
to move along a straight line. In this example, A particle is moving up and down along a straight
line. And its position is given by this equation where the positive positions mean that the
particle is above its Baseline Position, whatever I'm calling position zero, and negative positions
mean the particle is below this Baseline Position. I'm asked to find s prime of T and S double
prime of t. So by deriving I get four t cubed minus 16 t squared plus 12 T for the first
derivative, and 12 t squared minus 32 t plus 12. For the second derivative,
I get cosine squared theta minus one. And from the trig identity, sine squared theta
plus cosine squared theta equals one, I know that cosine squared theta minus one has to
equal minus sine squared. So I can rewrite my limit as the limit of minus sine squared
theta over theta cosine theta plus one. And now I can regroup to write my sine theta over
theta, and my other copy of sine theta over cosine theta plus one, the limit of the first
expression is going to be negative one, because of the limit we just proved. And the limit
of the second expression is just zero over one plus one, or zero, and therefore, my entire
limit is just going to be negative one times zero, or zero, which is exactly what we want
it to prove. Now we can use these two limits that we've just proved to calculate the derivatives
of sine and cosine, using the limit definition of derivative and prove the results that were
stated previously. According to the limit definition of derivative, the derivative of
sine x is the limit as h goes to zero of sine of x plus h minus sine of x divided by H.
As usual, this is a zero for zero indeterminate form limits. So I'm going to need to rewrite
things to evaluate it. And I'm going to rewrite using the angle sum formula for sine, the
sine of x plus h is equal to sine x cosine H, cosine plus cosine x sine H. Now if I rearrange
things, and factor out a sine x from the first term, I can break up my limit into pieces
and compute every piece. So this is sine x times zero plus cosine x times one. And so
my final answer is cosine x as we wanted the proof that the derivative of cosine is minus
sine is very similar. So please stop the video and try it for yourself before proceeding.
Using the limit definition of derivative, we have that the derivative of cosine of x
is the limit as h goes to zero of cosine of x plus h minus cosine of x over h, we can
rewrite the cosine of x plus h using the angle sum formula as the cosine of x times the cosine
of H minus the sine of x times the sine of H. And then we still have the minus cosine
of x over h. As before, we're going to regroup things and factoring out the cosine x from
the first part, the same familiar limits just put together in different ways. So here, cosine
of x as h goes to zero is just cosine of x. This limit we know is zero. sine of x is just
saying sine of x, and sine of h over h is going to one which means that our final answer
is going to be negative sine of x times one or just negative sine of x, which is exactly
what we wanted. That's all for the proofs of these four useful calculus facts. rectilinear
motion or linear motion means the motion of an object along a straight line. For example,
a particle moving left and right, or a ball going up and down. In this video, we'll see
what the derivative and the second derivative tell us about the motion of an object constrained
to move along a straight line. In this example, A particle is moving up and down along a straight
line. And its position is given by this equation where the positive positions mean that the
particle is above its Baseline Position, whatever I'm calling position zero, and negative positions
mean the particle is below this Baseline Position. I'm asked to find s prime of T and S double
prime of t. So by deriving I get four t cubed minus 16 t squared plus 12 T for the first
derivative, and 12 t squared minus 32 t plus 12. For the second derivative,
S prime of t, which can also be written, D STD represents the instantaneous rate of change
of S of t, the position over time, well, the change in position over time is just the velocity.
And this can also be written as v of t, s double prime of t, the second derivative of
s with respect to t, can also be thought of as the derivative of the velocity function.
So that represents the rate of change of velocity over time, how fast the velocity is increasing
or decreasing. And that is called acceleration. And it can be written as a lefty. Like position,
velocity and acceleration can be both positive and negative. A positive velocity means the
position is increasing. So the particle is moving up, while a negative velocity means
the position is decreasing, so the particle is moving down. Of course, a velocity of zero
means the particles at rest, at least for that instant. from physics, we know that force
equals mass times acceleration. So if the acceleration is positive, then that means
the force is in the positive direction, it's like the particle is being pulled up. If on
the other hand, the acceleration is negative than the force is in the negative direction,
and it's like the particle is being pulled down. an acceleration of zero means there's
no force on the particle at that instant, and the velocity continues as is. Let's use
these ideas about velocity acceleration. And the following table of values to describe
the particles motion at time equals 1.5 seconds. At time 1.5 seconds, the position of the particle
is positive, so that means the particle is above its Baseline Position of zero. its velocity
is negative, so that means that its position is decreasing. In other words, the particle
is moving down. Its acceleration is negative acceleration is the derivative of velocity.
So a negative acceleration means the velocity is decreasing. Well, a negative velocity that's
decreasing is getting more and more negative. So in fact, the particle is moving down faster
and faster. This can be a little bit confusing, because even though the velocity is decreasing,
it's getting more and more negative, the speed, which is the absolute value of velocity is
increasing. We can also see what the particle is doing at 1.5 seconds by looking at this
graph, where the time is drawn on the x axis and position s of t is on the y axis. From
the graph, we can see that when t is zero, S of t is also zero. So the particle starts
at its Baseline Position of zero. At time 1.5 seconds, the particle is above this starting
position, but moving downwards. And since the slope of this graph is getting steeper
and steeper, we can conclude that the speed of the particle is increasing. Same thing
was we concluded from the table of values. Now let's do the same analysis when time is
2.5 seconds. s of 2.5 seconds is negative. So the particle is below its starting position.
Velocity s prime of t is also negative. So the particle is still going down. But now
the acceleration as double prime of t is positive. That means that the velocity Today is increasing,
well, a negative velocity that's increasing is getting less negative closer to zero. So
the particle must be slowing down. And in fact, the speed is decreasing. Again, the
graph agrees with this reasoning, at 2.5 seconds, our position is way down here, our graph is
decreasing, so the particles moving down, and the slope seems to be leveling off. So
the particle speed is decreasing. Even though it's velocity,
S prime of t, which can also be written, D STD represents the instantaneous rate of change
of S of t, the position over time, well, the change in position over time is just the velocity.
And this can also be written as v of t, s double prime of t, the second derivative of
s with respect to t, can also be thought of as the derivative of the velocity function.
So that represents the rate of change of velocity over time, how fast the velocity is increasing
or decreasing. And that is called acceleration. And it can be written as a lefty. Like position,
velocity and acceleration can be both positive and negative. A positive velocity means the
position is increasing. So the particle is moving up, while a negative velocity means
the position is decreasing, so the particle is moving down. Of course, a velocity of zero
means the particles at rest, at least for that instant. from physics, we know that force
equals mass times acceleration. So if the acceleration is positive, then that means
the force is in the positive direction, it's like the particle is being pulled up. If on
the other hand, the acceleration is negative than the force is in the negative direction,
and it's like the particle is being pulled down. an acceleration of zero means there's
no force on the particle at that instant, and the velocity continues as is. Let's use
these ideas about velocity acceleration. And the following table of values to describe
the particles motion at time equals 1.5 seconds. At time 1.5 seconds, the position of the particle
is positive, so that means the particle is above its Baseline Position of zero. its velocity
is negative, so that means that its position is decreasing. In other words, the particle
is moving down. Its acceleration is negative acceleration is the derivative of velocity.
So a negative acceleration means the velocity is decreasing. Well, a negative velocity that's
decreasing is getting more and more negative. So in fact, the particle is moving down faster
and faster. This can be a little bit confusing, because even though the velocity is decreasing,
it's getting more and more negative, the speed, which is the absolute value of velocity is
increasing. We can also see what the particle is doing at 1.5 seconds by looking at this
graph, where the time is drawn on the x axis and position s of t is on the y axis. From
the graph, we can see that when t is zero, S of t is also zero. So the particle starts
at its Baseline Position of zero. At time 1.5 seconds, the particle is above this starting
position, but moving downwards. And since the slope of this graph is getting steeper
and steeper, we can conclude that the speed of the particle is increasing. Same thing
was we concluded from the table of values. Now let's do the same analysis when time is
2.5 seconds. s of 2.5 seconds is negative. So the particle is below its starting position.
Velocity s prime of t is also negative. So the particle is still going down. But now
the acceleration as double prime of t is positive. That means that the velocity Today is increasing,
well, a negative velocity that's increasing is getting less negative closer to zero. So
the particle must be slowing down. And in fact, the speed is decreasing. Again, the
graph agrees with this reasoning, at 2.5 seconds, our position is way down here, our graph is
decreasing, so the particles moving down, and the slope seems to be leveling off. So
the particle speed is decreasing. Even though it's velocity,
which you can think of as speed with direction is increasing, simply because it's a negative
velocity that's getting less negative. Notice that in the first example, when velocity and
acceleration are both in the same direction, that is, they're both negative the particle
was speeding up. And the second example, were velocity and acceleration when the opposite
directions one positive one negative, the particle is slowing down.
which you can think of as speed with direction is increasing, simply because it's a negative
velocity that's getting less negative. Notice that in the first example, when velocity and
acceleration are both in the same direction, that is, they're both negative the particle
was speeding up. And the second example, were velocity and acceleration when the opposite
directions one positive one negative, the particle is slowing down.
This is true in general, when velocity acceleration had the same sign, that as they're both positive
or both negative, then the particle is speeding up.
This is true in general, when velocity acceleration had the same sign, that as they're both positive
or both negative, then the particle is speeding up.
And when velocity acceleration have opposite signs, then the particle is slowing down.
One way to think about this is in terms of force, forces in the same direction as acceleration.
So if velocity acceleration have the same sign, that means force is the same direction
as the particles already going, so it's making the particle speed up. But if velocity and
acceleration have opposite signs, then the force is going against the way that particles
moving, so it's causing it to slow down. Let's continue the same example with some more questions,
it'll be helpful to write down the velocity and acceleration functions that we calculated
earlier. I've also graphed position, velocity and acceleration here at the right. And before
you go on, it's a fun exercise to figure out which one is which, without even looking at
the equations just based on the shapes of the graphs. And where they're increasing where
they're decreasing where they're positive and where they're negative. Velocity is the
derivative of position. So velocity needs to be positive, where position is increasing.
The only pairs of functions that have this property are the blue one, that's positive,
when the red ones increasing, and the green function, which is positive, when the blue
one is increasing. Now acceleration, which is the derivative of velocity also needs to
be positive, when velocity is increasing. So the only way to correctly label the functions
with both of these relationships is to make the red one be position, the blue one be velocity,
and the green one be acceleration. This agrees with the equations that
And when velocity acceleration have opposite signs, then the particle is slowing down.
One way to think about this is in terms of force, forces in the same direction as acceleration.
So if velocity acceleration have the same sign, that means force is the same direction
as the particles already going, so it's making the particle speed up. But if velocity and
acceleration have opposite signs, then the force is going against the way that particles
moving, so it's causing it to slow down. Let's continue the same example with some more questions,
it'll be helpful to write down the velocity and acceleration functions that we calculated
earlier. I've also graphed position, velocity and acceleration here at the right. And before
you go on, it's a fun exercise to figure out which one is which, without even looking at
the equations just based on the shapes of the graphs. And where they're increasing where
they're decreasing where they're positive and where they're negative. Velocity is the
derivative of position. So velocity needs to be positive, where position is increasing.
The only pairs of functions that have this property are the blue one, that's positive,
when the red ones increasing, and the green function, which is positive, when the blue
one is increasing. Now acceleration, which is the derivative of velocity also needs to
be positive, when velocity is increasing. So the only way to correctly label the functions
with both of these relationships is to make the red one be position, the blue one be velocity,
and the green one be acceleration. This agrees with the equations that
we have over here.
we have over here.
The first question asks, When is the particle at rest, the particle is temporarily at rest
when the velocity is zero. In other words, S prime of t is zero. So plugging in the equation
for S prime of t, we can factor out a four T, and factor some more. To conclude the T
has to be 01, or three.
The first question asks, When is the particle at rest, the particle is temporarily at rest
when the velocity is zero. In other words, S prime of t is zero. So plugging in the equation
for S prime of t, we can factor out a four T, and factor some more. To conclude the T
has to be 01, or three.
This conclusion agrees with our graph of V of t, which has x intercepts at 01 and three,
and also agrees with our graph of position s of t. Since the particle stops for a moment,
it changed direction, when t equals 01. And three, the particle is moving up when velocity
is positive, and moving down when velocity is negative. Since we know from the previous
question, the velocity equals zero, when t equals 01. And three, we can look in between
those values to figure out whether the velocity is positive or negative, just by plugging
in values. So for example, when t is negative one, if I plug into the negative one to the
equation for velocity, I get a negative number. So velocity must be negative when t is less
than zero, between zero and one. If I plug in, for example, t equals one half, I get
a value of S prime of t or V of t of 2.5, which is a positive number. If I plug in a
value of t in between one and three, say t equals two, I get a value of v of t of negative
eight, which is a negative number. And finally, if I plug in a value of t greater than three,
so For, I get a positive answer for V of t. So from the sign chart, I see that V of t
is negative when t is between negative infinity and zero, and in between one and three, and
V of t is positive when t is between
This conclusion agrees with our graph of V of t, which has x intercepts at 01 and three,
and also agrees with our graph of position s of t. Since the particle stops for a moment,
it changed direction, when t equals 01. And three, the particle is moving up when velocity
is positive, and moving down when velocity is negative. Since we know from the previous
question, the velocity equals zero, when t equals 01. And three, we can look in between
those values to figure out whether the velocity is positive or negative, just by plugging
in values. So for example, when t is negative one, if I plug into the negative one to the
equation for velocity, I get a negative number. So velocity must be negative when t is less
than zero, between zero and one. If I plug in, for example, t equals one half, I get
a value of S prime of t or V of t of 2.5, which is a positive number. If I plug in a
value of t in between one and three, say t equals two, I get a value of v of t of negative
eight, which is a negative number. And finally, if I plug in a value of t greater than three,
so For, I get a positive answer for V of t. So from the sign chart, I see that V of t
is negative when t is between negative infinity and zero, and in between one and three, and
V of t is positive when t is between
zero and
zero and
one, and between three and infinity. Of course, I could have reached the same conclusion just
by looking at the graph of velocity and where it's above and below the x axis, or even by
looking at the graph of position and seeing where it's increasing and where it's decreasing.
To answer the next question, the particle will be speeding up when V of t and a of t
are both positive or both negative. And the particle will be slowing down when V of t
and a of t have opposite signs. So let's make a similar sign chart to figure out where a
of t is positive and negative. First, it'll be helpful to find out where a if t is zero.
So if I set zero equal to my S double prime, that's 12 t squared minus 32 t plus 12, I
could factor out
one, and between three and infinity. Of course, I could have reached the same conclusion just
by looking at the graph of velocity and where it's above and below the x axis, or even by
looking at the graph of position and seeing where it's increasing and where it's decreasing.
To answer the next question, the particle will be speeding up when V of t and a of t
are both positive or both negative. And the particle will be slowing down when V of t
and a of t have opposite signs. So let's make a similar sign chart to figure out where a
of t is positive and negative. First, it'll be helpful to find out where a if t is zero.
So if I set zero equal to my S double prime, that's 12 t squared minus 32 t plus 12, I
could factor out
a four
a four
and then use the quadratic equation to find the solution. Since this equation doesn't
factor easily, this simplifies to four thirds plus or minus the square root of seven over
three, which is approximately 0.45, and 2.22. Now I can build a similar sign chart for acceleration,
mark the places where acceleration is zero. and plug in values of t, say t equals zero
is the equation for acceleration, I get a positive answer here. When I plug in, say
t equals one, I get a negative answer here. And when I plug in, say, t equals three, I
get another positive answer here. Now, if I put this together with my velocity chart,
which changed sign at 01, and three, and went from negative to positive, to negative to
positive, I can try to figure out where velocity acceleration both have the same sign. It might
be helpful actually to shade in where acceleration is positive. And separately shade in where
velocity is positive. And then look for the places where both are shaded. So between zero
and 4.5, and greater than three. And then I can also look where both are unshaded that
looks like in between one and 2.22. So that's where they're both negative. And then I'll
know that V of t and a of t have opposite signs everywhere else. A slightly better answer,
we'll use exact values of four thirds plus or minus squared of seven thirds instead of
these decimal approximations. So let me write that down. So here is where the particle speeding
up. And here, it's where it's slowing down, we can check our work by looking at the graph
of position, the particle speeding up with the position graph is getting steeper and
steeper, that's the red graph is getting steeper and steeper here, here. And here, just like
we found algebraically. As our final example, let's look at net change in position and distance
traveled between one and four seconds for the same particle. At time, one second position
is five thirds, or about 1.67 millimeters at time for its position is given by 32 thirds,
or about 10.67 millimeters, all I'm doing is plugging one and four into this equation.
So the net change in position is just the difference of these two numbers. As a four
minus as of one, which is nine millimeters. At first glance, it might seem like the total
distance traveled between one and four seconds should also be nine millimeters, but actually,
it's a little more complicated. Because the particle switches direction during that time
period, it doesn't go straight from its position at one second to its position at four seconds.
Remember what the graph Position looked like the particle switches direction at one second
and at three seconds. So to find the total distance, we need the distance of travels
from one second to three seconds, plus the distance that travels from three seconds to
four seconds. Another way of thinking about this is that we need the absolute value of
s three minus s one, plus the absolute value of s four minus s three, we need these absolute
value signs because this difference in position will be negative instead of positive when
the particles moving down. Plugging in the t values into our equation, we get negative
27 minus five thirds, plus the absolute value of 32 thirds minus negative 27, which is a
total of 199 thirds, or 66.3 repeating millimeters, quite a bit more than the nine millimeter
difference in position. This video gave an in depth analysis of a particle moving up
and down along a straight
and then use the quadratic equation to find the solution. Since this equation doesn't
factor easily, this simplifies to four thirds plus or minus the square root of seven over
three, which is approximately 0.45, and 2.22. Now I can build a similar sign chart for acceleration,
mark the places where acceleration is zero. and plug in values of t, say t equals zero
is the equation for acceleration, I get a positive answer here. When I plug in, say
t equals one, I get a negative answer here. And when I plug in, say, t equals three, I
get another positive answer here. Now, if I put this together with my velocity chart,
which changed sign at 01, and three, and went from negative to positive, to negative to
positive, I can try to figure out where velocity acceleration both have the same sign. It might
be helpful actually to shade in where acceleration is positive. And separately shade in where
velocity is positive. And then look for the places where both are shaded. So between zero
and 4.5, and greater than three. And then I can also look where both are unshaded that
looks like in between one and 2.22. So that's where they're both negative. And then I'll
know that V of t and a of t have opposite signs everywhere else. A slightly better answer,
we'll use exact values of four thirds plus or minus squared of seven thirds instead of
these decimal approximations. So let me write that down. So here is where the particle speeding
up. And here, it's where it's slowing down, we can check our work by looking at the graph
of position, the particle speeding up with the position graph is getting steeper and
steeper, that's the red graph is getting steeper and steeper here, here. And here, just like
we found algebraically. As our final example, let's look at net change in position and distance
traveled between one and four seconds for the same particle. At time, one second position
is five thirds, or about 1.67 millimeters at time for its position is given by 32 thirds,
or about 10.67 millimeters, all I'm doing is plugging one and four into this equation.
So the net change in position is just the difference of these two numbers. As a four
minus as of one, which is nine millimeters. At first glance, it might seem like the total
distance traveled between one and four seconds should also be nine millimeters, but actually,
it's a little more complicated. Because the particle switches direction during that time
period, it doesn't go straight from its position at one second to its position at four seconds.
Remember what the graph Position looked like the particle switches direction at one second
and at three seconds. So to find the total distance, we need the distance of travels
from one second to three seconds, plus the distance that travels from three seconds to
four seconds. Another way of thinking about this is that we need the absolute value of
s three minus s one, plus the absolute value of s four minus s three, we need these absolute
value signs because this difference in position will be negative instead of positive when
the particles moving down. Plugging in the t values into our equation, we get negative
27 minus five thirds, plus the absolute value of 32 thirds minus negative 27, which is a
total of 199 thirds, or 66.3 repeating millimeters, quite a bit more than the nine millimeter
difference in position. This video gave an in depth analysis of a particle moving up
and down along a straight
line.
line.
A similar analysis could be done for a particle moving left and right along a straight line,
where a positive position means the particles on the right side, and a negative position
means the particles on the left side of his Baseline Position. Of course, the same analysis
can be done for other objects, not just particles. A typical application is to a ball being thrown
straight up and then falling down again. This video will give an economic application of
the derivative to a cost function. Suppose that the total cost of producing x tie dyed
t shirts is C of x.
A similar analysis could be done for a particle moving left and right along a straight line,
where a positive position means the particles on the right side, and a negative position
means the particles on the left side of his Baseline Position. Of course, the same analysis
can be done for other objects, not just particles. A typical application is to a ball being thrown
straight up and then falling down again. This video will give an economic application of
the derivative to a cost function. Suppose that the total cost of producing x tie dyed
t shirts is C of x.
I'm going to sketch a few graphs, and you try to decide which graph is the most reasonable
representation for C of x. Pause the video for a moment to think about it. All of these
candidate graphs that I've drawn have a nonzero y intercept, that's meant to reflect the idea
that there's some fixed startup cost and buying equipment before you can even get started.
Now, I would like to suggest that C of x should be an increasing function of x, because it's
going to cost more money to make more t shirts, you need more supplies and labor. So this
function is out. Now it's somewhat reasonable, I think that C of x might be a linear function
of x like it is here, if you've got the same cost per t shirt, whether you make 10 t shirts,
or 1000 t shirts, the slope in that case would represent the cost per t shirt. And the linear
function would mean that cost per t shirt is constant, no matter how many t shirts you're
making. But in reality, it's probably going to be cheaper to make 1000 t shirts than it
is to make just a few t shirts. And therefore the cost per t shirt, sure slope should be
going down as x increases. So this function right here is the one whose slope is going
down for larger access. And so I would say that this is the most reasonable representation
for C of x as a function of x. In other words, C of x should be an increasing function, but
C prime of x should be decreasing. C of 204 minus C of 200 represents the additional cost
for making 204 t shirts instead of 200. In formula, you might think of that as the cost
of making the last four t shirts. The ratio C of 200 for a minus C of 200 over four is
the average rate of change of C of x. The units are units of cost, which is probably
dollars per t shirt. And formula you might think of this as the additional cost per t
shirt of making the last four t shirts. C prime of 200 is the instantaneous rate of
change of C of x. c of x is known as the cost function. And C prime of x is called the marginal
cost, which is the rate at which cost is increasing per additional t shirt made. It might seem
a little bit weird to take the derivative of C of x since x can really only take on
integer values. But we can always approximate C of x with a function whose domain is all
real numbers. To make this a little more specific, let's use a cost function of C of x equals
500 plus 300 times the square root of x. In this example, x is supposed to be the number
of iPads that are produced, and C of x is the cost of producing them in dollars. Then
C of 401 minus C of 400. given by 500 plus 300, times the square root of 401 minus 500
plus 300 times the square root of 400. This simplifies to $7.50, rounded to the nearest
cent. This means that it costs an additional $7.50 to go from producing 400 iPads to 401
iPad. In this fictitious example,
I'm going to sketch a few graphs, and you try to decide which graph is the most reasonable
representation for C of x. Pause the video for a moment to think about it. All of these
candidate graphs that I've drawn have a nonzero y intercept, that's meant to reflect the idea
that there's some fixed startup cost and buying equipment before you can even get started.
Now, I would like to suggest that C of x should be an increasing function of x, because it's
going to cost more money to make more t shirts, you need more supplies and labor. So this
function is out. Now it's somewhat reasonable, I think that C of x might be a linear function
of x like it is here, if you've got the same cost per t shirt, whether you make 10 t shirts,
or 1000 t shirts, the slope in that case would represent the cost per t shirt. And the linear
function would mean that cost per t shirt is constant, no matter how many t shirts you're
making. But in reality, it's probably going to be cheaper to make 1000 t shirts than it
is to make just a few t shirts. And therefore the cost per t shirt, sure slope should be
going down as x increases. So this function right here is the one whose slope is going
down for larger access. And so I would say that this is the most reasonable representation
for C of x as a function of x. In other words, C of x should be an increasing function, but
C prime of x should be decreasing. C of 204 minus C of 200 represents the additional cost
for making 204 t shirts instead of 200. In formula, you might think of that as the cost
of making the last four t shirts. The ratio C of 200 for a minus C of 200 over four is
the average rate of change of C of x. The units are units of cost, which is probably
dollars per t shirt. And formula you might think of this as the additional cost per t
shirt of making the last four t shirts. C prime of 200 is the instantaneous rate of
change of C of x. c of x is known as the cost function. And C prime of x is called the marginal
cost, which is the rate at which cost is increasing per additional t shirt made. It might seem
a little bit weird to take the derivative of C of x since x can really only take on
integer values. But we can always approximate C of x with a function whose domain is all
real numbers. To make this a little more specific, let's use a cost function of C of x equals
500 plus 300 times the square root of x. In this example, x is supposed to be the number
of iPads that are produced, and C of x is the cost of producing them in dollars. Then
C of 401 minus C of 400. given by 500 plus 300, times the square root of 401 minus 500
plus 300 times the square root of 400. This simplifies to $7.50, rounded to the nearest
cent. This means that it costs an additional $7.50 to go from producing 400 iPads to 401
iPad. In this fictitious example,
if I want to compute C prime of 400, instead, I can see that C prime of x is equal to 300
times one half x to the minus one half. So C prime of 400 is going to be 300 times one
half times 400 to the negative one half, which simplifies to 300 over two times the square
root of 400, which is also 7.5, or $7.50. Per iPad. Up to rounding to the nearest cent,
these two answers are equal. And it makes sense that C prime of 400 should equal approximately
this difference. Since C prime of 400, the derivative is approximately equal to the average
rate of change going from 400 to 401, which is just this difference, divided by one. Once
again, C prime of 400 is called the marginal cost, and represents the rate at which the
cost function is increasing with each additional item. This video gave an example of the cost
function, and it's derivative, which is known as marginal cost. This video introduces logarithms.
logarithms are a way of writing exponents. The expression log base a of B equals c means
that a to the C equals b. In other words, log base a of B is the exponent that you raise
a to to get be. The number A is called the base of the logarithm. It's also called the
base when we write it in this exponential form. Some students find it helpful to remember
this relationship, log base a of B equals c means a to the C equals b, by drawing arrows,
if I want to compute C prime of 400, instead, I can see that C prime of x is equal to 300
times one half x to the minus one half. So C prime of 400 is going to be 300 times one
half times 400 to the negative one half, which simplifies to 300 over two times the square
root of 400, which is also 7.5, or $7.50. Per iPad. Up to rounding to the nearest cent,
these two answers are equal. And it makes sense that C prime of 400 should equal approximately
this difference. Since C prime of 400, the derivative is approximately equal to the average
rate of change going from 400 to 401, which is just this difference, divided by one. Once
again, C prime of 400 is called the marginal cost, and represents the rate at which the
cost function is increasing with each additional item. This video gave an example of the cost
function, and it's derivative, which is known as marginal cost. This video introduces logarithms.
logarithms are a way of writing exponents. The expression log base a of B equals c means
that a to the C equals b. In other words, log base a of B is the exponent that you raise
a to to get be. The number A is called the base of the logarithm. It's also called the
base when we write it in this exponential form. Some students find it helpful to remember
this relationship, log base a of B equals c means a to the C equals b, by drawing arrows,
a to the C equals b.
a to the C equals b.
Other students like to think of it in terms of asking a question, log base a of fee, asks,
What power do you raise a two in order to get b? Let's look at some examples. log base
two of eight is three, because two to the three equals eight. In general, log base two
of y is asking you the question, What power do you have to raise to two to get y? So for
example, log base two of 16 is four, because it's asking you the question to two what power
equals 16? And the answer is four. Please pause the video and try some of these other
examples. log base two of two is asking, What power do you raise to two to get to? And the
answer is one. Two to the one equals two. log base two of one half is asking two to
what power gives you one half? Well, to get one half, you need to raise two to a negative
power. So that would be two to the negative one. So the answer is negative one. log base
two of 1/8 means what power do we raise to two in order to get 1/8. Since one eight is
one over two cubed, we have to raise two to the negative three power to get one over two
cubed. So our exponent is negative three, and that's our answer to our log expression.
Finally, log base two of one is asking to to what power equals one, or anything raised
to the zero power gives us one, so this log expression evaluates to zero. Notice that
we can get positive negative and zero answers for our logarithm expressions. Please pause
the video and figure out what these logs evaluate to to work out log base 10 of another Again,
notice that a million is 10 to the sixth power. Now we're asking the question, What power
do we raise tend to to get a million? So that is what power do we raise 10? to to get 10
to the six? Well, of course, the answer is going to be six. Similarly, since point O
one is 10 to the minus three, this log expressions, the same thing as asking, what's the log base
10 of 10 to the minus three? Well, what power do you have to raise 10? to to get 10 to the
minus three? Of course, the answer is negative three. Log base 10 of zero is asking, What
power do we raise 10 to to get zero. If you think about it, there's no way to raise 10
to an exponent get zero. Raising 10 to a positive exponent gets us really big positive numbers.
Raising 10 to a negative exponent is like one over 10 to a power that's giving us tiny
fractions, but they're still positive numbers, we're never going to get zero. Even if we
raised 10 to the zero power, we'll just get one. So there's no way to get zero and the
log base 10 of zero does not exist. If you try it on your calculator using the log base
10 button, you'll get an error message. Same thing happens when we do log base 10 of negative
100. We're asking 10 to what power equals negative 100. And there's no exponent that
will work. And more generally, it's possible to take the log of numbers that are greater
than zero, but not for numbers that are less than or equal to zero. In other words, the
domain of the function log base a of x, no matter what base you're using, for a, the
domain is going to be all positive numbers. A few notes on notation. When you see ln of
x, that's called natural log, and it means the log base e of x, where he is that famous
number that's about 2.718. When you see log of x with no base at all, by convention, that
means log base 10 of x, and it's called the common log. Most scientific calculators have
buttons for natural log, and for common log. Let's practice rewriting expressions with
logs in them. log base three of one nine is negative two can be rewritten as the expression
three to the negative two equals 1/9. Log of 13 is shorthand for log base 10 of 13.
So that can be rewritten as 10 to the 1.11394 equals 13. Finally, in this last expression,
ln means natural log, or log base e, so I can rewrite this equation as log base e of
whenever E equals negative one. Well, that means the same thing as e to the negative
one equals one over e, which is true. Now let's go the opposite direction. We'll start
with exponential equations and rewrite them as logs. Remember that log base a of B equals
c means the same thing
Other students like to think of it in terms of asking a question, log base a of fee, asks,
What power do you raise a two in order to get b? Let's look at some examples. log base
two of eight is three, because two to the three equals eight. In general, log base two
of y is asking you the question, What power do you have to raise to two to get y? So for
example, log base two of 16 is four, because it's asking you the question to two what power
equals 16? And the answer is four. Please pause the video and try some of these other
examples. log base two of two is asking, What power do you raise to two to get to? And the
answer is one. Two to the one equals two. log base two of one half is asking two to
what power gives you one half? Well, to get one half, you need to raise two to a negative
power. So that would be two to the negative one. So the answer is negative one. log base
two of 1/8 means what power do we raise to two in order to get 1/8. Since one eight is
one over two cubed, we have to raise two to the negative three power to get one over two
cubed. So our exponent is negative three, and that's our answer to our log expression.
Finally, log base two of one is asking to to what power equals one, or anything raised
to the zero power gives us one, so this log expression evaluates to zero. Notice that
we can get positive negative and zero answers for our logarithm expressions. Please pause
the video and figure out what these logs evaluate to to work out log base 10 of another Again,
notice that a million is 10 to the sixth power. Now we're asking the question, What power
do we raise tend to to get a million? So that is what power do we raise 10? to to get 10
to the six? Well, of course, the answer is going to be six. Similarly, since point O
one is 10 to the minus three, this log expressions, the same thing as asking, what's the log base
10 of 10 to the minus three? Well, what power do you have to raise 10? to to get 10 to the
minus three? Of course, the answer is negative three. Log base 10 of zero is asking, What
power do we raise 10 to to get zero. If you think about it, there's no way to raise 10
to an exponent get zero. Raising 10 to a positive exponent gets us really big positive numbers.
Raising 10 to a negative exponent is like one over 10 to a power that's giving us tiny
fractions, but they're still positive numbers, we're never going to get zero. Even if we
raised 10 to the zero power, we'll just get one. So there's no way to get zero and the
log base 10 of zero does not exist. If you try it on your calculator using the log base
10 button, you'll get an error message. Same thing happens when we do log base 10 of negative
100. We're asking 10 to what power equals negative 100. And there's no exponent that
will work. And more generally, it's possible to take the log of numbers that are greater
than zero, but not for numbers that are less than or equal to zero. In other words, the
domain of the function log base a of x, no matter what base you're using, for a, the
domain is going to be all positive numbers. A few notes on notation. When you see ln of
x, that's called natural log, and it means the log base e of x, where he is that famous
number that's about 2.718. When you see log of x with no base at all, by convention, that
means log base 10 of x, and it's called the common log. Most scientific calculators have
buttons for natural log, and for common log. Let's practice rewriting expressions with
logs in them. log base three of one nine is negative two can be rewritten as the expression
three to the negative two equals 1/9. Log of 13 is shorthand for log base 10 of 13.
So that can be rewritten as 10 to the 1.11394 equals 13. Finally, in this last expression,
ln means natural log, or log base e, so I can rewrite this equation as log base e of
whenever E equals negative one. Well, that means the same thing as e to the negative
one equals one over e, which is true. Now let's go the opposite direction. We'll start
with exponential equations and rewrite them as logs. Remember that log base a of B equals
c means the same thing
as a to the C equals b,
as a to the C equals b,
the base stays the same in both expressions. So for this example, the base of three in
the exponential equation, that's going to be the same as the base in our log. Now I
just have to figure out what's in the argument of the log. And what goes on the other side
of the equal sign. Remember that the answer to a log is an exponent. So the thing that
goes in this box should be my exponent for my exponential equation. In other words, you
and I'll put the 9.78 as the argument of my log. This works because log base three of
9.78 equals u means the same thing as three to the U equals 9.78, which is just what we
started with. In the second example, the base of my exponential equation is E. So the base
of my log is going to be the answer to my log is an exponent. In this case, the exponent
3x plus seven. And the other expression, the four minus y becomes my argument of my log.
Let me check, log base e of four minus y equals 3x plus seven means e to 3x plus seven equals
four minus Y, which is just what I started with. I can also rewrite log base e as natural
log. This video introduced the idea of logs. And the fact that log base a of B equal c
means the same thing as a to the C equals b. So log base a of B is asking you the question,
What power exponent Do you raise a to in order to get b. In this video, we'll work out the
graph, so some log functions and also talk about their domains. For this first example,
let's graph a log function by hand by plotting some points. The function we're working with
is y equals log base two of x, I'll make a chart of x and y values. Since we're working
this out by hand, I want to pick x values for which it's easy to compute log base two
of x. So I'll start out with the x value of one. Because log base two of one is zero,
log base anything of one is 02 is another x value that's easy to compute log base two
of two, that's asking, What power do I raise to two to get to one? And the answer is one.
Power other powers of two are easy to work with. So for example, log base two of four
that saying what power do I raise to to to get four, so the answer is two. Similarly,
log base two of eight is three, and log base two of 16 is four. Let me also work with some
fractional values for X. If x is one half, then log base two of one half that saying
what power do I raise to two to get one half? Well, that needs a power of negative one.
It's also easy to compute by hand, the log base two of 1/4 and 1/8. log base two of 1/4
is negative two, since two to the negative two is 1/4. And similarly, log base two of
1/8 is negative three. I'll put some tick marks on my x and y axes. Please pause the
video and take a moment to plot these points. Let's see I have the point, one, zero, that's
here to one that's here, for two, that is here. And then eight, three, which is
the base stays the same in both expressions. So for this example, the base of three in
the exponential equation, that's going to be the same as the base in our log. Now I
just have to figure out what's in the argument of the log. And what goes on the other side
of the equal sign. Remember that the answer to a log is an exponent. So the thing that
goes in this box should be my exponent for my exponential equation. In other words, you
and I'll put the 9.78 as the argument of my log. This works because log base three of
9.78 equals u means the same thing as three to the U equals 9.78, which is just what we
started with. In the second example, the base of my exponential equation is E. So the base
of my log is going to be the answer to my log is an exponent. In this case, the exponent
3x plus seven. And the other expression, the four minus y becomes my argument of my log.
Let me check, log base e of four minus y equals 3x plus seven means e to 3x plus seven equals
four minus Y, which is just what I started with. I can also rewrite log base e as natural
log. This video introduced the idea of logs. And the fact that log base a of B equal c
means the same thing as a to the C equals b. So log base a of B is asking you the question,
What power exponent Do you raise a to in order to get b. In this video, we'll work out the
graph, so some log functions and also talk about their domains. For this first example,
let's graph a log function by hand by plotting some points. The function we're working with
is y equals log base two of x, I'll make a chart of x and y values. Since we're working
this out by hand, I want to pick x values for which it's easy to compute log base two
of x. So I'll start out with the x value of one. Because log base two of one is zero,
log base anything of one is 02 is another x value that's easy to compute log base two
of two, that's asking, What power do I raise to two to get to one? And the answer is one.
Power other powers of two are easy to work with. So for example, log base two of four
that saying what power do I raise to to to get four, so the answer is two. Similarly,
log base two of eight is three, and log base two of 16 is four. Let me also work with some
fractional values for X. If x is one half, then log base two of one half that saying
what power do I raise to two to get one half? Well, that needs a power of negative one.
It's also easy to compute by hand, the log base two of 1/4 and 1/8. log base two of 1/4
is negative two, since two to the negative two is 1/4. And similarly, log base two of
1/8 is negative three. I'll put some tick marks on my x and y axes. Please pause the
video and take a moment to plot these points. Let's see I have the point, one, zero, that's
here to one that's here, for two, that is here. And then eight, three, which is
here.
here.
And the fractional x values, one half goes with negative one, and 1/4 with negative two
1/8 with negative three.
And the fractional x values, one half goes with negative one, and 1/4 with negative two
1/8 with negative three.
And if I connect the dots, I get a graph that looks something like this. If I had smaller
and smaller fractions, I would keep getting more and more negative answers when I took
log base two of them, so my graph is getting more and more negative, my y values are getting
more and more negative as x is getting close to zero. Now I didn't draw any parts of the
graph over here with negative X values, I didn't put any negative X values on my chart.
That omission is no accident. Because if you try to take the log base two or base anything
of a negative number, like say negative four or something, there's no answer. This doesn't
exist because there's no power that you can raise to two to get a negative number. So
there are no points on the graph for negative X values. And similarly, there are no points
on the graph where x is zero, because you can't take log base two of zero, there's no
power you can raise to to to get zero. I want to observe some key features of this graph.
First of all, the domain is x values greater than zero. In interval notation, I can write
that as a round bracket because I don't want to include zero to infinity, the range is
going to be the y values, while they go all the way down into the far reaches of the negative
numbers. And the graph gradually increases y value is getting bigger and bigger. So the
range is actually all real numbers are an interval notation negative infinity to infinity.
Finally, I want to point out that this graph has a vertical asymptote at the y axis, that
is at the line x equals zero. I'll draw that on my graph with a dotted line. A vertical
asymptote is a line that our functions graph gets closer and closer to so this is The graph
of y equals log base two of x. But if I wanted to graph say, y equals log base 10 of x, it
would look very similar, it would still have a domain of X values greater than zero, a
range of all real numbers and a vertical asymptote at the y axis, it would still go through the
point one zero, but it would go through the point 10. One instead, because log base 10
of 10 is one, it would look pretty much the same, just a lot flatter over here. But even
though it doesn't look like it with the way I've drawn it, it still gradually goes up
to n towards infinity. In fact, the graph of y equals log base neaa of X for a bigger
than one looks pretty much the same, and has the same three properties. Now that we know
what the basic log graph looks like, we can plot at least rough graphs of other log functions
without plotting points. Here we have the graph of natural log of X plus five. And again,
I'm just going to draw a rough graph. If I did want to do a more accurate graph, I probably
would want to plot some points. But I know that roughly a log graph, if it was just like
y equals ln of x, that would look something like this, and it would go through the point
one zero,
And if I connect the dots, I get a graph that looks something like this. If I had smaller
and smaller fractions, I would keep getting more and more negative answers when I took
log base two of them, so my graph is getting more and more negative, my y values are getting
more and more negative as x is getting close to zero. Now I didn't draw any parts of the
graph over here with negative X values, I didn't put any negative X values on my chart.
That omission is no accident. Because if you try to take the log base two or base anything
of a negative number, like say negative four or something, there's no answer. This doesn't
exist because there's no power that you can raise to two to get a negative number. So
there are no points on the graph for negative X values. And similarly, there are no points
on the graph where x is zero, because you can't take log base two of zero, there's no
power you can raise to to to get zero. I want to observe some key features of this graph.
First of all, the domain is x values greater than zero. In interval notation, I can write
that as a round bracket because I don't want to include zero to infinity, the range is
going to be the y values, while they go all the way down into the far reaches of the negative
numbers. And the graph gradually increases y value is getting bigger and bigger. So the
range is actually all real numbers are an interval notation negative infinity to infinity.
Finally, I want to point out that this graph has a vertical asymptote at the y axis, that
is at the line x equals zero. I'll draw that on my graph with a dotted line. A vertical
asymptote is a line that our functions graph gets closer and closer to so this is The graph
of y equals log base two of x. But if I wanted to graph say, y equals log base 10 of x, it
would look very similar, it would still have a domain of X values greater than zero, a
range of all real numbers and a vertical asymptote at the y axis, it would still go through the
point one zero, but it would go through the point 10. One instead, because log base 10
of 10 is one, it would look pretty much the same, just a lot flatter over here. But even
though it doesn't look like it with the way I've drawn it, it still gradually goes up
to n towards infinity. In fact, the graph of y equals log base neaa of X for a bigger
than one looks pretty much the same, and has the same three properties. Now that we know
what the basic log graph looks like, we can plot at least rough graphs of other log functions
without plotting points. Here we have the graph of natural log of X plus five. And again,
I'm just going to draw a rough graph. If I did want to do a more accurate graph, I probably
would want to plot some points. But I know that roughly a log graph, if it was just like
y equals ln of x, that would look something like this, and it would go through the point
one zero,
with a vertical asymptote along the y axis. Now if I want to graph ln of x plus five,
that just shifts our graph by five units, it'll still have the same vertical asymptote,
since the vertical line shifted up by five units is still a vertical line. But instead
of going through one zero, it'll go through the point, one, five. So I'll draw a rough
sketch here. Let's compare our starting function y equals ln x, and the transformed version
y equals ln x plus five in terms of the domain, the range and the vertical asymptote. Our
original function y equals ln x had a domain of zero to infinity. Since adding five on
the outside affects the y values, and the domain is the x values, this transformation
doesn't change the domain. So the domain is still from zero to infinity. Now the range
of our original y equals ln x was from negative infinity to infinity. Shifting up by five
does affect the y values, and the range is talking about the y values. But since the
original range was all real numbers, if you add five to all set of all real numbers, you
still get the set of all real numbers. So in this case, the range doesn't change either.
And finally, we already saw that the original vertical asymptote of the y axis x equals
zero, when we shift that up by five units, it's still the vertical line x equals zero.
In this next example, we're starting with a log base 10 function. And since the plus
two is on the inside, that means we shift that graph left by two. So I'll draw our basic
log function. Here's our basic log function. So I'll think of that as y equals log of x
going through the point one, zero, here's its vertical asymptote. Now I need to shift
everything left by two. So my vertical asymptote shifts left, and now it's at the line x equals
negative two, instead of at x equals zero, and my graph, let's see my point, one zero
gets shifted to, let's see negative one zero, since I'm subtracting two from the axis, and
here's a rough sketch of the resulting graph. Let's compare the features of the two graphs
drawn here. We're talking about domains, the original kind of domain of from zero to infinity.
But now I've shifted that left. So I subtracted two from all my exercises. And here's my new
domain, which I can also verify just by looking at the picture. My range was originally from
negative infinity to infinity. Well, shifting left only affects the x value, so it doesn't
even affect the range. So my range is still negative infinity to infinity. My vertical
asymptote was originally at x equals zero. And since I subtract two from all my x values,
that shifts that to x equals negative two. In this last problem, I'm not going to worry
about drawing this graph. I'll just use algebra to compute its domain. So let's think about
What's the issue, when you're taking the logs of things? Well, you can't take the log of
a negative number or zero. So whatever is inside the argument of the log function, whatever
is being fed into log had better be greater than zero. So I'll write that down, we need
to minus 3x, to be greater than zero. Now it's a matter of solving an inequality. Two
has got to be greater than 3x. So two thirds is greater than x. In other words, x has to
be less than two thirds. So our domain is all the x values from negative infinity to
two thirds, not including two thirds, it's a good idea to memorize the basic shape of
the graph of a log function. It looks something like this, go through the point one zero,
and has a vertical asymptote on the y axis. Also, if you remember that you can't take
the log of a negative number, or zero, then that helps you quickly compute domains for
log functions. Whatever's inside the log function, you set that greater than zero, and solve.
This video is about combining logs and exponents. Please pause the video and take a moment to
use your calculator to evaluate the following four expressions.
with a vertical asymptote along the y axis. Now if I want to graph ln of x plus five,
that just shifts our graph by five units, it'll still have the same vertical asymptote,
since the vertical line shifted up by five units is still a vertical line. But instead
of going through one zero, it'll go through the point, one, five. So I'll draw a rough
sketch here. Let's compare our starting function y equals ln x, and the transformed version
y equals ln x plus five in terms of the domain, the range and the vertical asymptote. Our
original function y equals ln x had a domain of zero to infinity. Since adding five on
the outside affects the y values, and the domain is the x values, this transformation
doesn't change the domain. So the domain is still from zero to infinity. Now the range
of our original y equals ln x was from negative infinity to infinity. Shifting up by five
does affect the y values, and the range is talking about the y values. But since the
original range was all real numbers, if you add five to all set of all real numbers, you
still get the set of all real numbers. So in this case, the range doesn't change either.
And finally, we already saw that the original vertical asymptote of the y axis x equals
zero, when we shift that up by five units, it's still the vertical line x equals zero.
In this next example, we're starting with a log base 10 function. And since the plus
two is on the inside, that means we shift that graph left by two. So I'll draw our basic
log function. Here's our basic log function. So I'll think of that as y equals log of x
going through the point one, zero, here's its vertical asymptote. Now I need to shift
everything left by two. So my vertical asymptote shifts left, and now it's at the line x equals
negative two, instead of at x equals zero, and my graph, let's see my point, one zero
gets shifted to, let's see negative one zero, since I'm subtracting two from the axis, and
here's a rough sketch of the resulting graph. Let's compare the features of the two graphs
drawn here. We're talking about domains, the original kind of domain of from zero to infinity.
But now I've shifted that left. So I subtracted two from all my exercises. And here's my new
domain, which I can also verify just by looking at the picture. My range was originally from
negative infinity to infinity. Well, shifting left only affects the x value, so it doesn't
even affect the range. So my range is still negative infinity to infinity. My vertical
asymptote was originally at x equals zero. And since I subtract two from all my x values,
that shifts that to x equals negative two. In this last problem, I'm not going to worry
about drawing this graph. I'll just use algebra to compute its domain. So let's think about
What's the issue, when you're taking the logs of things? Well, you can't take the log of
a negative number or zero. So whatever is inside the argument of the log function, whatever
is being fed into log had better be greater than zero. So I'll write that down, we need
to minus 3x, to be greater than zero. Now it's a matter of solving an inequality. Two
has got to be greater than 3x. So two thirds is greater than x. In other words, x has to
be less than two thirds. So our domain is all the x values from negative infinity to
two thirds, not including two thirds, it's a good idea to memorize the basic shape of
the graph of a log function. It looks something like this, go through the point one zero,
and has a vertical asymptote on the y axis. Also, if you remember that you can't take
the log of a negative number, or zero, then that helps you quickly compute domains for
log functions. Whatever's inside the log function, you set that greater than zero, and solve.
This video is about combining logs and exponents. Please pause the video and take a moment to
use your calculator to evaluate the following four expressions.
Remember, that log base 10 on your calculator is the log button. While log base e on your
calculator is the natural log button, you should find that the log base 10 of 10 cubed
is three, the log base e of e to the 4.2 is 4.2 10 to the log base 10 of 1000 is 1000.
And eat the log base e of 9.6 is 9.6. In each case, the log and the exponential function
with the same base undo each other, and we're left with the exponent. In fact, it's true
that for any base a the log base a of a to the x is equal to x. The same sort of cancellation
happens if we do the exponential function in the log function with the same base in
the opposite order. For example, we take 10 to the power of log base 10 of 1000, the 10
to the power and the log base 10 undo each other, and we're left with the 1000s. This
happens for any base a a to the log base a of x is equal to x. We can describe this by
saying that an exponential function and a log function with the same base undo each
other. If you're familiar with the language of inverse functions, the exponential function
and log function are inverses. Let's see why these roles hold for the first log role. log
base a of a dx is asking the question, What power do we raise a two in order to get a
to the x? In other words, a to what power is a to the x? Well, the answer is clearly
x. And that's why log base a of a to the x equals x. For the second log rule, notice
that the log base a of x means the power we raise a two to get x. But this expression
is saying that we're supposed to raise a to that power. If we raise a to the power, we
need to raise a two to get x, then we'll certainly get x. Now let's use these two roles. In some
examples. If we want to find three to the log base three of 1.43 to the power and log
base three undo each other, so we're left with 1.4.
Remember, that log base 10 on your calculator is the log button. While log base e on your
calculator is the natural log button, you should find that the log base 10 of 10 cubed
is three, the log base e of e to the 4.2 is 4.2 10 to the log base 10 of 1000 is 1000.
And eat the log base e of 9.6 is 9.6. In each case, the log and the exponential function
with the same base undo each other, and we're left with the exponent. In fact, it's true
that for any base a the log base a of a to the x is equal to x. The same sort of cancellation
happens if we do the exponential function in the log function with the same base in
the opposite order. For example, we take 10 to the power of log base 10 of 1000, the 10
to the power and the log base 10 undo each other, and we're left with the 1000s. This
happens for any base a a to the log base a of x is equal to x. We can describe this by
saying that an exponential function and a log function with the same base undo each
other. If you're familiar with the language of inverse functions, the exponential function
and log function are inverses. Let's see why these roles hold for the first log role. log
base a of a dx is asking the question, What power do we raise a two in order to get a
to the x? In other words, a to what power is a to the x? Well, the answer is clearly
x. And that's why log base a of a to the x equals x. For the second log rule, notice
that the log base a of x means the power we raise a two to get x. But this expression
is saying that we're supposed to raise a to that power. If we raise a to the power, we
need to raise a two to get x, then we'll certainly get x. Now let's use these two roles. In some
examples. If we want to find three to the log base three of 1.43 to the power and log
base three undo each other, so we're left with 1.4.
If we want to find ln of e to the x, remember that ln means log base e, so we're taking
log base e of e to the x.
If we want to find ln of e to the x, remember that ln means log base e, so we're taking
log base e of e to the x.
Well, those functions undo each other and we're left with x. If we want to take 10 to
the log of three z, remember that log without a base written implies that the base is 10.
So really, we want to take 10 to the log base 10 of three z will tend to a power and log
base 10 undo each other. So we're left with a three z. Finally, does this last statement
hold is ln of 10 to the x equal to x? Well, ln means log base e. So we're taking log base
e of 10 to the x, notice that the base of the log and the base of the exponential function
are not the same thing. So they don't undo each other. And in fact, log base e of 10
to the x is not usually equal to x, we can check with one example, say if x equals one,
then log base e of 10 to the one, that's log base e of 10. And we can check on the calculator
that's equal to 2.3. In some more decimals, which is not the same thing as one. So this
statement is false, it does not hold. We need the basis to be the same for logs and exponent
to undo each other. In this video, we saw that logs and exponents with the same base
undo each other. Specifically, log base a of a to the x is equal to x and a to the log
base a of x is also equal to x
Well, those functions undo each other and we're left with x. If we want to take 10 to
the log of three z, remember that log without a base written implies that the base is 10.
So really, we want to take 10 to the log base 10 of three z will tend to a power and log
base 10 undo each other. So we're left with a three z. Finally, does this last statement
hold is ln of 10 to the x equal to x? Well, ln means log base e. So we're taking log base
e of 10 to the x, notice that the base of the log and the base of the exponential function
are not the same thing. So they don't undo each other. And in fact, log base e of 10
to the x is not usually equal to x, we can check with one example, say if x equals one,
then log base e of 10 to the one, that's log base e of 10. And we can check on the calculator
that's equal to 2.3. In some more decimals, which is not the same thing as one. So this
statement is false, it does not hold. We need the basis to be the same for logs and exponent
to undo each other. In this video, we saw that logs and exponents with the same base
undo each other. Specifically, log base a of a to the x is equal to x and a to the log
base a of x is also equal to x
for any values of x and any base a. This video is about rules or properties of logs. The
log rules are closely related to the exponent rules. So let's start by reviewing some of
the exponent rules.
for any values of x and any base a. This video is about rules or properties of logs. The
log rules are closely related to the exponent rules. So let's start by reviewing some of
the exponent rules.
To keep things simple, we'll write everything down with a base of two. Even though the exponent
rules hold for any base, we know that if we raise two to the zero power, we get one, we
have a product rule for exponents, which says that two to the M times two to the n is equal
to two to the m plus n. In other words, if we multiply two numbers, then we add the exponents.
We also have a quotient rule that says that two to the M divided by n to the n is equal
to two to the m minus n. In words, that says that if we divide two numbers, then we subtract
the exponents. Finally, we have a power rule that says if we take a power to a power, then
we multiply the exponents. Each of these exponent rules can be rewritten as a log rule. The
first rule, two to the zero equals one can be rewritten in terms of logs as log base
two of one equals zero. That's because log base two of one equals zero means two to the
zero equals one. The second rule, the product rule, can be rewritten in terms of logs by
saying log of x times y equals log of x plus log of y. I'll make these base two to agree
with my base that I'm using for my exponent rules. In words, that says the log of the
product is the sum of the logs. Since logs really represent exponents. This is saying
that when you multiply two numbers together, you add their exponents, which is just what
we said for the exponent version. The quotient rule for exponents can be rewritten in terms
of logs by saying the log of x divided by y is equal to the log of x minus the log of
y. In words, we can say that the log of the quotient is equal to the difference of the
logs. Since logs are really exponents, another way of saying the same thing is that when
you divide two numbers, you subtract their exponents. That's how we described the exponent
rule above. Finally, the power rule for exponents can be rewritten in terms of logs by saying
the log of x to the n is equal to n times log of x. Sometimes people describe this rule
by saying when you take the log of an expression with an exponent, you can bring down the exponent
and multiply. If we think of x as being some power of two, this is really saying when we
take a power to a power, we multiply their exponents. That's exactly how we described
the power rule above. It doesn't really matter if you multiply this exponent on the left
side, or on the right side, but it's more traditional to multiply it on the left side.
I've given these rules with the base of two, but they actually work for any base. To help
you remember them, please take a moment to write out the log rules using a base of a
you should get the following chart. Let's use the log rules to rewrite the following
expressions as a sum or difference of logs. In the first expression, we have a log base
10 of a quotient. So we can rewrite the log of the quotient
To keep things simple, we'll write everything down with a base of two. Even though the exponent
rules hold for any base, we know that if we raise two to the zero power, we get one, we
have a product rule for exponents, which says that two to the M times two to the n is equal
to two to the m plus n. In other words, if we multiply two numbers, then we add the exponents.
We also have a quotient rule that says that two to the M divided by n to the n is equal
to two to the m minus n. In words, that says that if we divide two numbers, then we subtract
the exponents. Finally, we have a power rule that says if we take a power to a power, then
we multiply the exponents. Each of these exponent rules can be rewritten as a log rule. The
first rule, two to the zero equals one can be rewritten in terms of logs as log base
two of one equals zero. That's because log base two of one equals zero means two to the
zero equals one. The second rule, the product rule, can be rewritten in terms of logs by
saying log of x times y equals log of x plus log of y. I'll make these base two to agree
with my base that I'm using for my exponent rules. In words, that says the log of the
product is the sum of the logs. Since logs really represent exponents. This is saying
that when you multiply two numbers together, you add their exponents, which is just what
we said for the exponent version. The quotient rule for exponents can be rewritten in terms
of logs by saying the log of x divided by y is equal to the log of x minus the log of
y. In words, we can say that the log of the quotient is equal to the difference of the
logs. Since logs are really exponents, another way of saying the same thing is that when
you divide two numbers, you subtract their exponents. That's how we described the exponent
rule above. Finally, the power rule for exponents can be rewritten in terms of logs by saying
the log of x to the n is equal to n times log of x. Sometimes people describe this rule
by saying when you take the log of an expression with an exponent, you can bring down the exponent
and multiply. If we think of x as being some power of two, this is really saying when we
take a power to a power, we multiply their exponents. That's exactly how we described
the power rule above. It doesn't really matter if you multiply this exponent on the left
side, or on the right side, but it's more traditional to multiply it on the left side.
I've given these rules with the base of two, but they actually work for any base. To help
you remember them, please take a moment to write out the log rules using a base of a
you should get the following chart. Let's use the log rules to rewrite the following
expressions as a sum or difference of logs. In the first expression, we have a log base
10 of a quotient. So we can rewrite the log of the quotient
as the difference of the logs. Now we still have the log of a product, I can rewrite the
log of a product as the sum of the logs
as the difference of the logs. Now we still have the log of a product, I can rewrite the
log of a product as the sum of the logs
So that is log of y plus log of z. When I put things together, I have to be careful,
because here I'm subtracting the entire log expression. So I need to subtract both terms
of this son.
So that is log of y plus log of z. When I put things together, I have to be careful,
because here I'm subtracting the entire log expression. So I need to subtract both terms
of this son.
I'll make sure I do that by putting them in parentheses. Now I can simplify a little bit
by distributing the negative sign. And here's my final answer. In my next expression, I
have the log of a product. So I can rewrite that as the sum of two logs. I can also use
my power rule to bring down the exponent T, and multiply it in the front. That gives me
the final expression log of five plus
I'll make sure I do that by putting them in parentheses. Now I can simplify a little bit
by distributing the negative sign. And here's my final answer. In my next expression, I
have the log of a product. So I can rewrite that as the sum of two logs. I can also use
my power rule to bring down the exponent T, and multiply it in the front. That gives me
the final expression log of five plus
t times log of two. One common mistake on this problem is to rewrite this expression
as t times log of five times two. In fact, those two expressions are not equal. Because
the T only applies to the two, not to the whole five times two, we can't just bring
it down in front using the power rule. After all, the power rule only applies to a single
expression raised to an exponent, and not to a product like this. And these next examples,
we're going to go the other direction. Here, we're given sums and differences of logs.
And we want to wrap them up into a single log expression. By look at the first two pieces,
that's a difference of logs. So I know I can rewrite it as the log of a quotient. Now I
have the sum of two logs. So I can rewrite that as the log of a product. I'll clean that
up a little bit and rewrite it as log base five of a times c over B. In my second example,
I can rewrite the sum of my logs as the log of a product. Now, I would like to rewrite
this difference of logs as the log of a quotient, but I can't do it yet, because of that factor
of two multiplied in front. But I can use the power rule backwards to put that two back
up in the exponent. So I'll do that first. So I will copy down the ln of x plus one times
x minus one, and rewrite this second term as ln of x squared minus one squared. Now
I have a straightforward difference of two logs, which I can rewrite as the log of a
quotient. I can actually simplify this some more. Since x plus one times x minus one is
the same thing as x squared minus one. I can cancel factors to get ln of one over x squared
minus one. In this video, we solve for rules for logs that are related to exponent rules.
First, we saw that the log with any base of one is equal to zero. Second, we saw the product
rule, the log of a product is equal to the sum of the logs. We saw the quotient rule,
the log of a quotient is the difference of the logs. And we saw the power rule. When
you take a log of an expression with an exponent in it, you can bring down the exponent and
multiply it. It's worth noticing that there's no log rule that helps you split up the log
of a song. In particular, the log of a psalm is not equal to the sum of the logs. If you
think about logs and exponent rules going together, this kind of makes sense, because
there's also no rule for rewriting the sum of two exponential expressions. Log roles
will be super handy. As we start to solve equations using locks. The chain rule is a
really useful method for finding the derivative of the composition of two functions. Let's
start with a brief review of composition. f composed with g means that we apply f to
the output of G as a diagram, this means we start with x and apply g first.
t times log of two. One common mistake on this problem is to rewrite this expression
as t times log of five times two. In fact, those two expressions are not equal. Because
the T only applies to the two, not to the whole five times two, we can't just bring
it down in front using the power rule. After all, the power rule only applies to a single
expression raised to an exponent, and not to a product like this. And these next examples,
we're going to go the other direction. Here, we're given sums and differences of logs.
And we want to wrap them up into a single log expression. By look at the first two pieces,
that's a difference of logs. So I know I can rewrite it as the log of a quotient. Now I
have the sum of two logs. So I can rewrite that as the log of a product. I'll clean that
up a little bit and rewrite it as log base five of a times c over B. In my second example,
I can rewrite the sum of my logs as the log of a product. Now, I would like to rewrite
this difference of logs as the log of a quotient, but I can't do it yet, because of that factor
of two multiplied in front. But I can use the power rule backwards to put that two back
up in the exponent. So I'll do that first. So I will copy down the ln of x plus one times
x minus one, and rewrite this second term as ln of x squared minus one squared. Now
I have a straightforward difference of two logs, which I can rewrite as the log of a
quotient. I can actually simplify this some more. Since x plus one times x minus one is
the same thing as x squared minus one. I can cancel factors to get ln of one over x squared
minus one. In this video, we solve for rules for logs that are related to exponent rules.
First, we saw that the log with any base of one is equal to zero. Second, we saw the product
rule, the log of a product is equal to the sum of the logs. We saw the quotient rule,
the log of a quotient is the difference of the logs. And we saw the power rule. When
you take a log of an expression with an exponent in it, you can bring down the exponent and
multiply it. It's worth noticing that there's no log rule that helps you split up the log
of a song. In particular, the log of a psalm is not equal to the sum of the logs. If you
think about logs and exponent rules going together, this kind of makes sense, because
there's also no rule for rewriting the sum of two exponential expressions. Log roles
will be super handy. As we start to solve equations using locks. The chain rule is a
really useful method for finding the derivative of the composition of two functions. Let's
start with a brief review of composition. f composed with g means that we apply f to
the output of G as a diagram, this means we start with x and apply g first.
Then we apply f to the output to get our final results I'm going to call G, the inner function,
and F the outer function. Because g looks like it's on the inside of f,
Then we apply f to the output to get our final results I'm going to call G, the inner function,
and F the outer function. Because g looks like it's on the inside of f,
in this standard notation, we can write h of x, which is the square root of sine of
x as the composition of two functions, by letting sine of x be the inner function, and
the square root function be the outer function, which I write as f of u equals the square
root of u. I like to do this sort of dissection of functions, by drawing a box around part
of the function, whatever is inside the box becomes my inner function, whatever we do
to the box becomes our outer function, in this case, taking the square root. This allows
us to write h of x as the composition, f of g of x, where f and g are the outer and inner
functions defined here. Please take a moment to write the next two functions as compositions
of functions, before you go on. A natural way to write k of x as a composition is to
let our inner function be tan of x plus seacon of x. The outer function describes what happens
to that box the inner function, it gets cubed and multiplied by five. There's several ways
to write the next example as a composition of functions. For example, we could take x
squared as our inner function, and then our outer function takes a to the sign of that
inner function. Alternatively, we could take the inner function to be sine of x squared.
And then the outer function
in this standard notation, we can write h of x, which is the square root of sine of
x as the composition of two functions, by letting sine of x be the inner function, and
the square root function be the outer function, which I write as f of u equals the square
root of u. I like to do this sort of dissection of functions, by drawing a box around part
of the function, whatever is inside the box becomes my inner function, whatever we do
to the box becomes our outer function, in this case, taking the square root. This allows
us to write h of x as the composition, f of g of x, where f and g are the outer and inner
functions defined here. Please take a moment to write the next two functions as compositions
of functions, before you go on. A natural way to write k of x as a composition is to
let our inner function be tan of x plus seacon of x. The outer function describes what happens
to that box the inner function, it gets cubed and multiplied by five. There's several ways
to write the next example as a composition of functions. For example, we could take x
squared as our inner function, and then our outer function takes a to the sign of that
inner function. Alternatively, we could take the inner function to be sine of x squared.
And then the outer function
has to be e to the power. It's also possible to write our function r of x as a composition
of three functions. An inner function of x squared, a middle function of sine and an
outermost function of e to the power, which are right as h of V equals e to the V.
has to be e to the power. It's also possible to write our function r of x as a composition
of three functions. An inner function of x squared, a middle function of sine and an
outermost function of e to the power, which are right as h of V equals e to the V.
When calculating the derivatives of complicated functions, it's really important to recognize
them as compositions of simpler functions. That way, we can build up the derivative in
terms of the simpler derivatives. And that's the idea behind the chain rule. The chain
rule tells us if we have two differentiable functions, then the derivative of the composition
f composed with g of x is equal to the derivative of the outer function evaluated on the inner
function times the derivative of the inner function. Sometimes the chain rule is written
instead, in lightness notation, that is the dydx notation. To see how this works, let's
let u equal g of x. And let's let y equal f of u. In other words, y is f of
When calculating the derivatives of complicated functions, it's really important to recognize
them as compositions of simpler functions. That way, we can build up the derivative in
terms of the simpler derivatives. And that's the idea behind the chain rule. The chain
rule tells us if we have two differentiable functions, then the derivative of the composition
f composed with g of x is equal to the derivative of the outer function evaluated on the inner
function times the derivative of the inner function. Sometimes the chain rule is written
instead, in lightness notation, that is the dydx notation. To see how this works, let's
let u equal g of x. And let's let y equal f of u. In other words, y is f of
g of x.
g of x.
Now do u dx is just another way of writing g prime of x, and d y d u is another way of
writing f prime of U. or in other words, f prime of g of x. Finally, if we write D y
dX, that means we're taking the derivative of f composed with g. So that's f composed
with g prime of x. Using this key, I can rewrite the expression above as the y dx equals d
y d u
Now do u dx is just another way of writing g prime of x, and d y d u is another way of
writing f prime of U. or in other words, f prime of g of x. Finally, if we write D y
dX, that means we're taking the derivative of f composed with g. So that's f composed
with g prime of x. Using this key, I can rewrite the expression above as the y dx equals d
y d u
times d u dx. These are the two alternative ways of writing the chain rule. Let's use
the chain rule to take the derivative of the square root of sine x.
times d u dx. These are the two alternative ways of writing the chain rule. Let's use
the chain rule to take the derivative of the square root of sine x.
Actually, I'm going to rewrite this as h of x equals sine x to the one half power to make
it easier to take derivatives. As a composition, we're thinking of the inner function as sine
x and the outer function as the one half power. So the chain rule tells us that to take h
prime
Actually, I'm going to rewrite this as h of x equals sine x to the one half power to make
it easier to take derivatives. As a composition, we're thinking of the inner function as sine
x and the outer function as the one half power. So the chain rule tells us that to take h
prime
of x we need to take the derivative of the outer function evaluated on the inner function
and then multiply that by the derivative of the inner function, we know that the derivative
of the inner function, sine x is just cosine x. And the derivative of the outer function
is one half times u to the negative one half. So h prime of x is then one half times u to
the negative one half. But that's evaluated on the inner function, sine of
of x we need to take the derivative of the outer function evaluated on the inner function
and then multiply that by the derivative of the inner function, we know that the derivative
of the inner function, sine x is just cosine x. And the derivative of the outer function
is one half times u to the negative one half. So h prime of x is then one half times u to
the negative one half. But that's evaluated on the inner function, sine of
x.
x.
And then we multiply that by cosine of x. Again, that's the derivative of the outer
function evaluated on the inner function times the derivative of the inner function. And
we found the derivative using the chain rule. For the next example, our inner function was
tan x plus seacon X and our outer function, f of u was five u cubed. So k prime of x is
15 times u squared. But that's the evaluated on the inner function 10x plus seacon x, then
we still need to multiply that by the derivative of the inner function 10x plus seacon x. So
we get the 15 tan x plus secant x squared times the derivative of tan x, which is secant
squared x, plus the derivative of secant x, which is secant x, tan x. And that's our chain
rule derivative. In this last example, we're thinking of the outermost function as being
e to the power and its inner function is sine of x squared. But sine of x squared itself
has an outer function of sine and an inner function of x squared. So to find r prime
of x, we first have to take the derivative of the outermost function, well, the derivative
of e to the power is just e to the power.
And then we multiply that by cosine of x. Again, that's the derivative of the outer
function evaluated on the inner function times the derivative of the inner function. And
we found the derivative using the chain rule. For the next example, our inner function was
tan x plus seacon X and our outer function, f of u was five u cubed. So k prime of x is
15 times u squared. But that's the evaluated on the inner function 10x plus seacon x, then
we still need to multiply that by the derivative of the inner function 10x plus seacon x. So
we get the 15 tan x plus secant x squared times the derivative of tan x, which is secant
squared x, plus the derivative of secant x, which is secant x, tan x. And that's our chain
rule derivative. In this last example, we're thinking of the outermost function as being
e to the power and its inner function is sine of x squared. But sine of x squared itself
has an outer function of sine and an inner function of x squared. So to find r prime
of x, we first have to take the derivative of the outermost function, well, the derivative
of e to the power is just e to the power.
And now we evaluate that on its inner function, sine of x squared. But now by the chain rule,
we have to multiply that by the derivative of the inner function, sine of x squared.
I'll copy down the E to the sine x squared. And I'll use the chain rule a second time
to find the derivative of sine x squared. Now the outer function is sine, and the derivative
of sine is cosine. I need to evaluate it on its inner function of x squared, and then
multiply that by the derivative of the inner function. After copying things down, I just
have to take the derivative of x squared, which is 2x
And now we evaluate that on its inner function, sine of x squared. But now by the chain rule,
we have to multiply that by the derivative of the inner function, sine of x squared.
I'll copy down the E to the sine x squared. And I'll use the chain rule a second time
to find the derivative of sine x squared. Now the outer function is sine, and the derivative
of sine is cosine. I need to evaluate it on its inner function of x squared, and then
multiply that by the derivative of the inner function. After copying things down, I just
have to take the derivative of x squared, which is 2x
by the power.
by the power.
This video introduced the chain rule, which says that the derivative of f composed with
g at x is equal to f prime at g of x times g prime at x, or equivalently, d y dx is equal
to d y d u times d u dx. This video gives some more examples and a justification of
the chain rule, and also includes a handy formula for the derivative of a to the x with
respect to x, where A is any positive number. In the next example, I want to show using
the chain rule that the derivative of five to the x is equal to ln five times five to
the x. First, I want to rewrite five to the x as easily ln five times x. And I can do
that because e to the ln five is equal to five. So e to the ln five to the x power is
equal to five to the x. But e to the ln five to the x power using exponent rules is just
e to the ln five times x. So if I want to take the derivative of five to the x, after
rewriting it as e to the ln five times x, let me think of the inner function as being
ln five times x. And I'm going to think of the outer function as being E to that power.
That's what I wanted to Make the derivative of. So by the chain rule, I can first take
the derivative of the outer function, derivative of e to the power is just e to the power,
and I evaluate it at its inner function. But then by the chain rule, I need to take the
derivative of the inner function, well, the derivative of ln five times x is just the
constant coefficient ln five. And that's my derivative. Now, I know that e to the ln five
times x is just five to the x. That's what we talked about before. So my final answer
is five to the x times ln five, or I guess I can rewrite that in the other order. That
is that there's nothing special about five in this example, I could have done this same
process with any a base positive base a. So I'm going to write that as a general principle
that the derivative of a to the x with respect to x is equal to ln a times a to the x. This
is a fact worth memorizing. I'll use this fact to compute the derivative of this complicated
expression, sine of 5x times the square root of two to the cosine 5x plus one. To find
dydx, I'll first use the product rule, since our expression is the product of two other
expressions. So D y dX is the first expression times the derivative of the second expression,
which I'll go ahead and write using an exponent instead of a square root sign, plus the derivative
of the first expression times the second expression. Now I'll need to use the chain rule to evaluate
the derivative here. My outermost function is the function that raises everything to
the one half power. So when I take the derivative, I can use the power rule, bring down the one
half, right to the cosine 5x plus one to the negative one half. But then by the chain rule,
I'm going to have to multiply by the derivative of the inner function, which is to to the
cosine 5x plus one, I'll just carry along the rest of my expression for now, what I
want to take the derivative of two to the cosine 5x plus one, I'm going to have to use
the chain rule again, thinking of my outer function as being two to the power plus one.
So let me copy things down on the next line. Now taking the derivative, the derivative
of one is just zero, so I'm really just taking the derivative of two to the cosine 5x. And
by my formula, this is going to be ln f two times two
This video introduced the chain rule, which says that the derivative of f composed with
g at x is equal to f prime at g of x times g prime at x, or equivalently, d y dx is equal
to d y d u times d u dx. This video gives some more examples and a justification of
the chain rule, and also includes a handy formula for the derivative of a to the x with
respect to x, where A is any positive number. In the next example, I want to show using
the chain rule that the derivative of five to the x is equal to ln five times five to
the x. First, I want to rewrite five to the x as easily ln five times x. And I can do
that because e to the ln five is equal to five. So e to the ln five to the x power is
equal to five to the x. But e to the ln five to the x power using exponent rules is just
e to the ln five times x. So if I want to take the derivative of five to the x, after
rewriting it as e to the ln five times x, let me think of the inner function as being
ln five times x. And I'm going to think of the outer function as being E to that power.
That's what I wanted to Make the derivative of. So by the chain rule, I can first take
the derivative of the outer function, derivative of e to the power is just e to the power,
and I evaluate it at its inner function. But then by the chain rule, I need to take the
derivative of the inner function, well, the derivative of ln five times x is just the
constant coefficient ln five. And that's my derivative. Now, I know that e to the ln five
times x is just five to the x. That's what we talked about before. So my final answer
is five to the x times ln five, or I guess I can rewrite that in the other order. That
is that there's nothing special about five in this example, I could have done this same
process with any a base positive base a. So I'm going to write that as a general principle
that the derivative of a to the x with respect to x is equal to ln a times a to the x. This
is a fact worth memorizing. I'll use this fact to compute the derivative of this complicated
expression, sine of 5x times the square root of two to the cosine 5x plus one. To find
dydx, I'll first use the product rule, since our expression is the product of two other
expressions. So D y dX is the first expression times the derivative of the second expression,
which I'll go ahead and write using an exponent instead of a square root sign, plus the derivative
of the first expression times the second expression. Now I'll need to use the chain rule to evaluate
the derivative here. My outermost function is the function that raises everything to
the one half power. So when I take the derivative, I can use the power rule, bring down the one
half, right to the cosine 5x plus one to the negative one half. But then by the chain rule,
I'm going to have to multiply by the derivative of the inner function, which is to to the
cosine 5x plus one, I'll just carry along the rest of my expression for now, what I
want to take the derivative of two to the cosine 5x plus one, I'm going to have to use
the chain rule again, thinking of my outer function as being two to the power plus one.
So let me copy things down on the next line. Now taking the derivative, the derivative
of one is just zero, so I'm really just taking the derivative of two to the cosine 5x. And
by my formula, this is going to be ln f two times two
to the power of cosine 5x. But of course, I have to use the chain rule
to the power of cosine 5x. But of course, I have to use the chain rule
and multiply by that the derivative of the inner function here, which is cosine 5x. Again,
I'm just going to carry the rest of the expression along with me for the ride for now. Now we're
taking the derivative of cosine 5x, I think of cosine as the outer function. And five
times x is the inner function. Similarly, down here, sign is the outer function, and
5x is the inner function. So I can complete my work by copying a lot of stuff down, and
now taking the derivative of cosine, which is minus sine, evaluated at its inner function,
times the derivative of the inner function 5x, which is just five. And I'm going to add
to that the derivative of sine of 5x. Well, the derivative of sine is cosine, evaluated
on its inner function times the derivative of the inner function 5x, which is just five
times the rest of the stuff. I'll do a modest amount of simplification, maybe bring the
constants out and combine any terms that I can. And that's the end of that complicated
example. And the next example, we'll try to find the derivative of a composition at the
value x equals one just based on a table of values. So the chain rule says that the derivative
of f composed with g is just going to be f prime evaluated g of x times g prime evaluated
x, but I want to do this whole process at x equals one. So that's just going to be f
prime at g of one times g prime of one. Well, g of one is two. So I really want f prime
at two, and f prime at two is 10. And g prime at one is negative five. So my answer is negative
50. I'm not going to give a rigorous proof of the chain rule. But I would like to give
a more informal explanation based on the limit definition of derivative. So I'm going to
write the derivative of f composed with g evaluated a point A, as the limit as x goes
to a of f composed with g of x minus f composed with g of A divided by x minus a. I'll rewrite
this slightly. And now we're going to multiply the top and the bottom by g of x minus g of
a, that doesn't change the value of expression, provided that g of x minus g of A is not zero.
That's the detail on sweeping under the rug here and why this is not a real proof, but
just a more informal explanation. Now if I rearrange things, and rewrite the limit of
the product as the product of the limits, my limit on the right here is just the derivative
of g. for the limit on the left, notice that as x goes to a, g of x has to go to g of a,
since G is differentiable on there for continuous function. So I can rewrite this and letting
say u be equal to g of x, I can rewrite this as the limit as u goes to g of a of f of u
minus f of g of A over u minus g of a. Now my expression on the left is just another
way of writing the derivative of f evaluated at g of a. And I've arrived at the expression
for the chain rule. Let me just emphasize again, this is just a pseudo proof, it's not
quite airtight, because g of x minus g of a might be zero. In this video, we saw some
more examples of the chain rule justification of it. And we saw that the derivative respect
to x
and multiply by that the derivative of the inner function here, which is cosine 5x. Again,
I'm just going to carry the rest of the expression along with me for the ride for now. Now we're
taking the derivative of cosine 5x, I think of cosine as the outer function. And five
times x is the inner function. Similarly, down here, sign is the outer function, and
5x is the inner function. So I can complete my work by copying a lot of stuff down, and
now taking the derivative of cosine, which is minus sine, evaluated at its inner function,
times the derivative of the inner function 5x, which is just five. And I'm going to add
to that the derivative of sine of 5x. Well, the derivative of sine is cosine, evaluated
on its inner function times the derivative of the inner function 5x, which is just five
times the rest of the stuff. I'll do a modest amount of simplification, maybe bring the
constants out and combine any terms that I can. And that's the end of that complicated
example. And the next example, we'll try to find the derivative of a composition at the
value x equals one just based on a table of values. So the chain rule says that the derivative
of f composed with g is just going to be f prime evaluated g of x times g prime evaluated
x, but I want to do this whole process at x equals one. So that's just going to be f
prime at g of one times g prime of one. Well, g of one is two. So I really want f prime
at two, and f prime at two is 10. And g prime at one is negative five. So my answer is negative
50. I'm not going to give a rigorous proof of the chain rule. But I would like to give
a more informal explanation based on the limit definition of derivative. So I'm going to
write the derivative of f composed with g evaluated a point A, as the limit as x goes
to a of f composed with g of x minus f composed with g of A divided by x minus a. I'll rewrite
this slightly. And now we're going to multiply the top and the bottom by g of x minus g of
a, that doesn't change the value of expression, provided that g of x minus g of A is not zero.
That's the detail on sweeping under the rug here and why this is not a real proof, but
just a more informal explanation. Now if I rearrange things, and rewrite the limit of
the product as the product of the limits, my limit on the right here is just the derivative
of g. for the limit on the left, notice that as x goes to a, g of x has to go to g of a,
since G is differentiable on there for continuous function. So I can rewrite this and letting
say u be equal to g of x, I can rewrite this as the limit as u goes to g of a of f of u
minus f of g of A over u minus g of a. Now my expression on the left is just another
way of writing the derivative of f evaluated at g of a. And I've arrived at the expression
for the chain rule. Let me just emphasize again, this is just a pseudo proof, it's not
quite airtight, because g of x minus g of a might be zero. In this video, we saw some
more examples of the chain rule justification of it. And we saw that the derivative respect
to x
of A to the X is equal to ln of a times a to the x. This video gives an explanation
for why the chain rule holds.
of A to the X is equal to ln of a times a to the x. This video gives an explanation
for why the chain rule holds.
I'm not going to give a rigorous proof of the chain rule. But I would like to give a
more informal explanation based on the limit definition of derivative.
I'm not going to give a rigorous proof of the chain rule. But I would like to give a
more informal explanation based on the limit definition of derivative.
So I'm going to write the derivative of f composed with g evaluated a point A, as the
limit as x goes to a of f composed with g of x minus f composed with g of A divided
by x minus a. I'll rewrite this slightly. And now we're going to multiply the top and
the bottom by g of x minus g of a, that doesn't change the value of expression, provided that
g of x minus g of A is not zero. That's the detail on sweeping under the rug here and
why this is not a real proof, but just a more informal explanation. Now if I rearrange things,
and rewrite the limit of the product as the product of the limits, my limit on the right
here is just the derivative of g. for the limit on the left, notice that as x goes to
a, g of x has to go to G evey, since G is differentiable on there for a continuous function,
so I can rewrite this and letting say u be equal to g of x, I can rewrite this as the
limit as u goes to g of a of f of u minus f of g of A
So I'm going to write the derivative of f composed with g evaluated a point A, as the
limit as x goes to a of f composed with g of x minus f composed with g of A divided
by x minus a. I'll rewrite this slightly. And now we're going to multiply the top and
the bottom by g of x minus g of a, that doesn't change the value of expression, provided that
g of x minus g of A is not zero. That's the detail on sweeping under the rug here and
why this is not a real proof, but just a more informal explanation. Now if I rearrange things,
and rewrite the limit of the product as the product of the limits, my limit on the right
here is just the derivative of g. for the limit on the left, notice that as x goes to
a, g of x has to go to G evey, since G is differentiable on there for a continuous function,
so I can rewrite this and letting say u be equal to g of x, I can rewrite this as the
limit as u goes to g of a of f of u minus f of g of A
over
over
u minus g of a. Now my expression on the left is just another way of writing the derivative
of f evaluated at G Ave. And I've arrived at the expression for the chain rule. Let
me just emphasize again, this is just a pseudo proof it's not quite airtight because g of
x minus g of a might be zero. That's all for the justification of the chain rule. For a
complete proof. Please see the textbook implicit differentiation is a technique for finding
the slopes of tangent lines for curves that are defined indirectly, and sometimes aren't
even functions. So far, we've developed a lot of techniques for finding derivatives
of functions defined explicitly, in terms of the equation y equals something. In this
section, we'll consider curves that are defined implicitly, in terms of any equation involving
x's and y's. So the points on this curve are the values of x and y that satisfy this equation.
As you can see, when you have implicitly defined curves, they are not necessarily functions.
And in fact, they can not only violate the vertical line test, but they can cross themselves,
or be broken up into several pieces or look like really cool pictures like this flower.
But small pieces of these curves do satisfy the vertical line test for small pieces, y
is a function of x. And that allows us to use our calculus techniques, especially the
chain rule, to compute derivatives for these implicitly defined curves.
u minus g of a. Now my expression on the left is just another way of writing the derivative
of f evaluated at G Ave. And I've arrived at the expression for the chain rule. Let
me just emphasize again, this is just a pseudo proof it's not quite airtight because g of
x minus g of a might be zero. That's all for the justification of the chain rule. For a
complete proof. Please see the textbook implicit differentiation is a technique for finding
the slopes of tangent lines for curves that are defined indirectly, and sometimes aren't
even functions. So far, we've developed a lot of techniques for finding derivatives
of functions defined explicitly, in terms of the equation y equals something. In this
section, we'll consider curves that are defined implicitly, in terms of any equation involving
x's and y's. So the points on this curve are the values of x and y that satisfy this equation.
As you can see, when you have implicitly defined curves, they are not necessarily functions.
And in fact, they can not only violate the vertical line test, but they can cross themselves,
or be broken up into several pieces or look like really cool pictures like this flower.
But small pieces of these curves do satisfy the vertical line test for small pieces, y
is a function of x. And that allows us to use our calculus techniques, especially the
chain rule, to compute derivatives for these implicitly defined curves.
As usual, the derivative dy dx represents the slope of a tangent line. For our first
example, let's find the equation of the tangent line for the lips 9x squared plus four y squared
equals 25. drawn below at the point, one, two. From the picture, it looks like the slope
of this tangent line should be about negative one, but let's use calculus to find it exactly.
So there are at least two ways we could proceed. First, we could solve for y, and then use
the same techniques that we've been using. So if we solve for y, we get for y squared
equals 25 minus 9x squared. So y squared is 25 minus 9x squared over four, which means
that y is plus or minus the square root of 25 minus 9x squared over four, or in other
words, plus or minus the square root of 25 minus 9x squared over two, the plus answer
is giving us the top half of the ellipse. And the minus answer is giving us this bottom.
Since the point one, two is on the top part of the ellipse, let's focus on the positive
version. And let's take the derivative. But first, let me rewrite one more time to put
it in a slightly easier form, instead of dividing by two, I'm going to think of multiplying
by the constant one half. And instead of taking the square root, I'm going to write that as
an exponent of one half here. So now if I want to take the y dx, I can pull out the
constant of one half. And now I'll start using the chain rule where my outer function
As usual, the derivative dy dx represents the slope of a tangent line. For our first
example, let's find the equation of the tangent line for the lips 9x squared plus four y squared
equals 25. drawn below at the point, one, two. From the picture, it looks like the slope
of this tangent line should be about negative one, but let's use calculus to find it exactly.
So there are at least two ways we could proceed. First, we could solve for y, and then use
the same techniques that we've been using. So if we solve for y, we get for y squared
equals 25 minus 9x squared. So y squared is 25 minus 9x squared over four, which means
that y is plus or minus the square root of 25 minus 9x squared over four, or in other
words, plus or minus the square root of 25 minus 9x squared over two, the plus answer
is giving us the top half of the ellipse. And the minus answer is giving us this bottom.
Since the point one, two is on the top part of the ellipse, let's focus on the positive
version. And let's take the derivative. But first, let me rewrite one more time to put
it in a slightly easier form, instead of dividing by two, I'm going to think of multiplying
by the constant one half. And instead of taking the square root, I'm going to write that as
an exponent of one half here. So now if I want to take the y dx, I can pull out the
constant of one half. And now I'll start using the chain rule where my outer function
is
is
taking things to the one half power, and my inner function is 25 minus 9x squared. So
I'll take the derivative of my outer function by bringing the one half down, taking the
inner function to the negative one half. Now I multiply by the derivative of the inner
function, which is negative 18x. If I simplify a little bit, I get D y dX is negative 18x
over four times 25 minus 9x squared to the one half power, or in other words, dy dx is
negative 9x over two times the square root of 25 minus 9x squared. This formula only
holds for the top half of the ellipse for the bottom half, we would need to use the
negative. Now I want to evaluate the derivative at the point one, two, so I'm going to take
dydx. When x equals one, I get negative nine over two times the square root of 25 minus
nine, which is negative nine eighths. Since I've found the slope of the tangent line,
and I know that point one, two is a point on the tangent line, I can now use the point
slope form to write down the equation of the tangent line. simplified this becomes y equals
negative nine is x plus nine eights plus two, or y equals negative nine, it's x plus 25.
It's now that we've solved the problem once using a familiar method. Let's go back to
the beginning and solve it again using a new method. method two is implicit differentiation.
The idea is that I'm going to take the derivative with respect to x of both sides. If my equation
without having to solve for y, I can rewrite the left side as nine times the derivative
of x squared plus four times the derivative of y squared. And the right side, the derivative
of a constant is zero. Going back to the left side, the derivative of x squared with respect
to x is 2x. Now for the derivative of y squared with respect to x, I'm going to need to use
the chain rule, I'm going to think of taking the squared power as my outside function,
I'm going to think of y itself as my inside function, my inside function of x. Even though
my entire curve is not a function, for small pieces of it, y is a function of x, so I can
get away with doing this, the derivative of my outside function y squared is to y, and
the derivative of my inside function, y as a function of x is just dydx. Now I can solve
for dy dx, which is going to tell me the slope of my tangent line. And so I get negative
18x from here, divided by eight y from here, which simplifies to negative nine fourths
times X over Y. Notice that the formula for my derivative dydx has both x's and y's in
it. Of course, for this problem, if I wanted to, I could solve for y in terms of x using
the original equation like I did in method one, and plug that in for y and get an expression
entirely in terms of x, which should be the same as the expression I got previously.
taking things to the one half power, and my inner function is 25 minus 9x squared. So
I'll take the derivative of my outer function by bringing the one half down, taking the
inner function to the negative one half. Now I multiply by the derivative of the inner
function, which is negative 18x. If I simplify a little bit, I get D y dX is negative 18x
over four times 25 minus 9x squared to the one half power, or in other words, dy dx is
negative 9x over two times the square root of 25 minus 9x squared. This formula only
holds for the top half of the ellipse for the bottom half, we would need to use the
negative. Now I want to evaluate the derivative at the point one, two, so I'm going to take
dydx. When x equals one, I get negative nine over two times the square root of 25 minus
nine, which is negative nine eighths. Since I've found the slope of the tangent line,
and I know that point one, two is a point on the tangent line, I can now use the point
slope form to write down the equation of the tangent line. simplified this becomes y equals
negative nine is x plus nine eights plus two, or y equals negative nine, it's x plus 25.
It's now that we've solved the problem once using a familiar method. Let's go back to
the beginning and solve it again using a new method. method two is implicit differentiation.
The idea is that I'm going to take the derivative with respect to x of both sides. If my equation
without having to solve for y, I can rewrite the left side as nine times the derivative
of x squared plus four times the derivative of y squared. And the right side, the derivative
of a constant is zero. Going back to the left side, the derivative of x squared with respect
to x is 2x. Now for the derivative of y squared with respect to x, I'm going to need to use
the chain rule, I'm going to think of taking the squared power as my outside function,
I'm going to think of y itself as my inside function, my inside function of x. Even though
my entire curve is not a function, for small pieces of it, y is a function of x, so I can
get away with doing this, the derivative of my outside function y squared is to y, and
the derivative of my inside function, y as a function of x is just dydx. Now I can solve
for dy dx, which is going to tell me the slope of my tangent line. And so I get negative
18x from here, divided by eight y from here, which simplifies to negative nine fourths
times X over Y. Notice that the formula for my derivative dydx has both x's and y's in
it. Of course, for this problem, if I wanted to, I could solve for y in terms of x using
the original equation like I did in method one, and plug that in for y and get an expression
entirely in terms of x, which should be the same as the expression I got previously.
But I don't really
But I don't really
need to do that in order to solve this problem. Instead, I can just plug in the x value of
one and the y value of two to get dy dx at x equals one, equal to negative nine fourths,
times one half or negative nine, eight, which we'll recognize as the same answer we got
before. So as before, we can compute the equation for the tangent line. And we'll again get
y equals negative nine 8x plus 25, eights. In this example, implicit differentiation
was a convenient way to find the derivative. But it was possible to solve for y and use
standard methods instead. But in many examples, like the next one, it's not possible to solve
for y directly. And so implicit differentiation is the only way to go. implicit differentiation
is definitely the key to finding y prime for this curve defined implicitly. So again, the
idea is to take the derivative of both sides with respect to x, I can break this up into
pieces. And now use the product rule for the first piece. So I get the first function x
cubed times the derivative of the second function y squared, the derivative of y squared is
to y, d y dx, don't forget the dydx there, because y is a function of x plus the derivative
of the first part 3x squared times the second part
need to do that in order to solve this problem. Instead, I can just plug in the x value of
one and the y value of two to get dy dx at x equals one, equal to negative nine fourths,
times one half or negative nine, eight, which we'll recognize as the same answer we got
before. So as before, we can compute the equation for the tangent line. And we'll again get
y equals negative nine 8x plus 25, eights. In this example, implicit differentiation
was a convenient way to find the derivative. But it was possible to solve for y and use
standard methods instead. But in many examples, like the next one, it's not possible to solve
for y directly. And so implicit differentiation is the only way to go. implicit differentiation
is definitely the key to finding y prime for this curve defined implicitly. So again, the
idea is to take the derivative of both sides with respect to x, I can break this up into
pieces. And now use the product rule for the first piece. So I get the first function x
cubed times the derivative of the second function y squared, the derivative of y squared is
to y, d y dx, don't forget the dydx there, because y is a function of x plus the derivative
of the first part 3x squared times the second part
y squared.
y squared.
Next, I need to take the derivative of sine x, y on the do use the chain rule here. So
the derivative of the outside sine is cosine. And now I need to take the derivative of the
inside x times y. And that's going to be a product rule application. So x times D y dX,
plus the derivative of x, which is just one times y. That all was just my left hand side.
But fortunately, my right hand side is easier. The derivative of x cubed with respect to
x is 3x squared. And the derivative of y cubed with respect to x is three y squared, dy dx.
Now I need to solve for the y dx. And since it's scattered all over the place in three
different places, I'm first going to distribute out to free it from these parentheses, and
then I'll try to move all the dydx is to the left side. So distributing out I get, I get
this expression. And now moving alternatives with dydx and add them to the left side and
all terms without dydx mm to the right side. I'm going to get this expression here. Now
I'm going to factor out the dy dx. I'm just using standard algebra techniques here. And
finally, I can just divide both sides by all this mess, too. isolate the dydx. And I found
my derivative using implicit differentiation. This video talked about using implicit differentiation
to find the slopes of tangent lines for curves defined implicitly, the main two steps, were
first to take the derivative of both sides with respect to x. And then to solve for dy
dx. This video is about finding the derivatives of exponential functions, we've already seen
that the derivative of the exponential function, e to the x is just itself, e to the x. But
what's the derivative of an exponential function with a different base, like five to the x,
one way to find the derivative of an exponential function like five to the X is to write five
as e to a power. So five, is the same thing as e to the ln five, where ln is the natural
log or the log base. See, this makes sense because ln five, which is the same thing as
log base e of five, means the power that we raise E to to get five. So now if we take
e to the ln five, that means we raise e to the power that we raise E to to get five?
Well, when you raise E to that power, you get five. All right, if five is the same thing
as e to the ln five, then that means if we take five to the x, that's the same thing
as e to the ln five raised to the x power BI properties of exponents, when I take a
power to a power, I multiply the exponents. so this can be written as e to the ln five
times x. Now I want to take the derivative with respect to x of five to the x. So by
my rewriting trick, that's the same thing as taking the derivative with respect to x
of e to the ln five times x. Now we know how to calculate this using the chain rule, we
can think of e to the power as our outside function, and ln five times x as our inside
function. So now by the chain rule, I take the derivative of the outside function e to
the power, and that's just gives me e to the power evaluated on the inside function. So
I stick ln five times x as my inside function, and by the chain rule, I multiply that by
the derivative of the inside function, ln five times x. Five is a constant. So let me
copy over first part, the derivative of a constant times x is just the constant. Let
me rewrite this a little bit. So e to the ln five times x is the same thing as e to
the ln five to the x power, just like before, because the exponent rules say, when I take
a power to a power, I multiply the exponent, and remember, is the LL five, it's just a
fancy way of writing five. So I've got five to the x times ln five as the derivative with
respect to x of five to the x. The same argument works not just for a base five exponential
function, but for any base exponential function. So if I take the respect to x of A to the
X for any number A, I'm going to get
Next, I need to take the derivative of sine x, y on the do use the chain rule here. So
the derivative of the outside sine is cosine. And now I need to take the derivative of the
inside x times y. And that's going to be a product rule application. So x times D y dX,
plus the derivative of x, which is just one times y. That all was just my left hand side.
But fortunately, my right hand side is easier. The derivative of x cubed with respect to
x is 3x squared. And the derivative of y cubed with respect to x is three y squared, dy dx.
Now I need to solve for the y dx. And since it's scattered all over the place in three
different places, I'm first going to distribute out to free it from these parentheses, and
then I'll try to move all the dydx is to the left side. So distributing out I get, I get
this expression. And now moving alternatives with dydx and add them to the left side and
all terms without dydx mm to the right side. I'm going to get this expression here. Now
I'm going to factor out the dy dx. I'm just using standard algebra techniques here. And
finally, I can just divide both sides by all this mess, too. isolate the dydx. And I found
my derivative using implicit differentiation. This video talked about using implicit differentiation
to find the slopes of tangent lines for curves defined implicitly, the main two steps, were
first to take the derivative of both sides with respect to x. And then to solve for dy
dx. This video is about finding the derivatives of exponential functions, we've already seen
that the derivative of the exponential function, e to the x is just itself, e to the x. But
what's the derivative of an exponential function with a different base, like five to the x,
one way to find the derivative of an exponential function like five to the X is to write five
as e to a power. So five, is the same thing as e to the ln five, where ln is the natural
log or the log base. See, this makes sense because ln five, which is the same thing as
log base e of five, means the power that we raise E to to get five. So now if we take
e to the ln five, that means we raise e to the power that we raise E to to get five?
Well, when you raise E to that power, you get five. All right, if five is the same thing
as e to the ln five, then that means if we take five to the x, that's the same thing
as e to the ln five raised to the x power BI properties of exponents, when I take a
power to a power, I multiply the exponents. so this can be written as e to the ln five
times x. Now I want to take the derivative with respect to x of five to the x. So by
my rewriting trick, that's the same thing as taking the derivative with respect to x
of e to the ln five times x. Now we know how to calculate this using the chain rule, we
can think of e to the power as our outside function, and ln five times x as our inside
function. So now by the chain rule, I take the derivative of the outside function e to
the power, and that's just gives me e to the power evaluated on the inside function. So
I stick ln five times x as my inside function, and by the chain rule, I multiply that by
the derivative of the inside function, ln five times x. Five is a constant. So let me
copy over first part, the derivative of a constant times x is just the constant. Let
me rewrite this a little bit. So e to the ln five times x is the same thing as e to
the ln five to the x power, just like before, because the exponent rules say, when I take
a power to a power, I multiply the exponent, and remember, is the LL five, it's just a
fancy way of writing five. So I've got five to the x times ln five as the derivative with
respect to x of five to the x. The same argument works not just for a base five exponential
function, but for any base exponential function. So if I take the respect to x of A to the
X for any number A, I'm going to get
a to the x times ln A. Now, you might be wondering, what if I use the same roll on our old favorite
e to the x. So our base here is E. That means I should get e to the x times ln e? Wait a
sec, ln E, that's log base e of E, that's asking what power? Do I raise II today get
he? Well, the answer there is one. And so the derivative respect to x of e to the x
by this new rule we have is e to the x, it agrees with our old rule. I want to draw your
attention to the difference between two expressions. And the first expression, dy dx of a to the
x, the variable that we're taking the derivative with respect to is in the exponent. So for
this exponential function, or we use the derivative rule that we just found, dy dx of a dx is
eight of the x times ln A. On the other hand, if we take respect to x of x to the A, where
the variable x that we're taking the derivative with respect to is in the base, then we don't
need this exponential rule. In fact, it doesn't even apply, although we have here is the power
rule, right? dy dx of x cubed would be 3x squared dy dx of x to the seventh would be
7x to the sixth and enjoy Add x of x to the a is just a times x to the A minus one by
the power role. So it's important to pay attention to where the variable is when you're taking
a derivative. In this video, we found that the derivative with respect to x of five to
the x is given by ln five times five to the x. And in general, the derivative of respect
to x of A to the X is going to be ln a times a to the x. This gives us a general formula
for the derivative of exponential functions. The main goal of this video is to figure out
the derivatives of logarithmic functions, functions like y equals ln x, or y equals
log base A x for any positive base a, I want to find the derivative of log base a of x.
In other words, I want to find the derivative of y, where y is log base a of x. By the definition
of logarithms, log base a of x equals y means that a to the Y power is equal to x. And that's
useful because now I can take the derivative of both sides and use implicit differentiation.
Recall of the derivative of a to the power is ln A times A to the power. But since why
we're thinking of as a function of x, I have to multiply that by dou y dx by the chain
rule. The right hand side here is just one. Solving for dydx, I get one over ln A times
A to the Y. But since age the y is equal to x, I can rewrite that as the y dx equals one
over ln A times x. So the derivative of log base a of x for any base a is one over ln
of A times x. And in particular, the derivative of natural log of x is one over ln of E times
x. But since ln of E is just one, that saying that the derivative of ln x is one over x,
this is a very handy fact. And this more general derivative is also worth memorizing. While
we're talking about the derivative of the natural log of x, let's look at the derivative
of the natural log of the absolute value of x. The function y equals ln of absolute value
of x is of course closely related to the function y equals ln of x, the difference being that
the domain for ln x is just x values greater than zero, whereas the domain of ln of absolute
value of x is all X's not equal to zero. The graphs are also related. When you look at
the graph of y equals ln absolute value of x, it looks like you're seeing double since
the absolute value of x is equal to x, when x is greater than or equal to zero and negative
x when x is less than zero, ln of the absolute value of x is going to be equal to ln x when
x is greater than or equal to zero, and ln of negative x when x is less than zero. If
I consider the derivative of ln of absolute value of x, I can think of taking the derivative
of each piece separately. We just saw that the derivative of ln x is one over x. So the
derivative of ln of minus x is going to be one over minus x times the derivative of minus
x, which is minus one by the chain rule. Notice that this second expression simplifies to
one over x. So the derivative of ln of absolute value of x is equal to one over x, whether
x is positive or negative. This formula will come in handy later when we start doing integrals.
In this video, we found that the derivative of ln x is equal to one over x, kind of a
nice derivative. And more generally, the derivative of log base a of x is one over ln A times
x. We've seen previously that the derivative of x to a constant A is equal to a times x
to the A minus one. This is the power rule. We've also seen that the derivative of a positive
number a raised to the x power is equal to ln A times A to the X So we know how to take
the derivative when the variable x is in the base, or when it's in the exponent. But what
if the variables in both the base and the exponent, how do we take the derivative of
x to the x. To differentiate functions like this, we'll need to use the technique of logarithmic
differentiation. To find the derivative of x to the x, I'm going to set y equal to x
to the x. Now we want to find dy dx. Since we don't know how to compute dydx directly,
so let's take the natural log of both sides. taking the log is often a handy trick when
you have a variable in the exponent that you don't know how to deal with, because the properties
of logs allow us to bring that exponent down and multiply it. Now we have y implicitly
defined in terms of x, so let's use implicit differentiation, we'll take the derivative
of both sides with respect to x. And now we should have no trouble taking the derivatives
because we've gotten rid of the awkward exponential expression. So the derivative on the left
of ln y is one over y times dy dx. And the derivative on the right using the product
rule is x times one over x plus one times ln x.
a to the x times ln A. Now, you might be wondering, what if I use the same roll on our old favorite
e to the x. So our base here is E. That means I should get e to the x times ln e? Wait a
sec, ln E, that's log base e of E, that's asking what power? Do I raise II today get
he? Well, the answer there is one. And so the derivative respect to x of e to the x
by this new rule we have is e to the x, it agrees with our old rule. I want to draw your
attention to the difference between two expressions. And the first expression, dy dx of a to the
x, the variable that we're taking the derivative with respect to is in the exponent. So for
this exponential function, or we use the derivative rule that we just found, dy dx of a dx is
eight of the x times ln A. On the other hand, if we take respect to x of x to the A, where
the variable x that we're taking the derivative with respect to is in the base, then we don't
need this exponential rule. In fact, it doesn't even apply, although we have here is the power
rule, right? dy dx of x cubed would be 3x squared dy dx of x to the seventh would be
7x to the sixth and enjoy Add x of x to the a is just a times x to the A minus one by
the power role. So it's important to pay attention to where the variable is when you're taking
a derivative. In this video, we found that the derivative with respect to x of five to
the x is given by ln five times five to the x. And in general, the derivative of respect
to x of A to the X is going to be ln a times a to the x. This gives us a general formula
for the derivative of exponential functions. The main goal of this video is to figure out
the derivatives of logarithmic functions, functions like y equals ln x, or y equals
log base A x for any positive base a, I want to find the derivative of log base a of x.
In other words, I want to find the derivative of y, where y is log base a of x. By the definition
of logarithms, log base a of x equals y means that a to the Y power is equal to x. And that's
useful because now I can take the derivative of both sides and use implicit differentiation.
Recall of the derivative of a to the power is ln A times A to the power. But since why
we're thinking of as a function of x, I have to multiply that by dou y dx by the chain
rule. The right hand side here is just one. Solving for dydx, I get one over ln A times
A to the Y. But since age the y is equal to x, I can rewrite that as the y dx equals one
over ln A times x. So the derivative of log base a of x for any base a is one over ln
of A times x. And in particular, the derivative of natural log of x is one over ln of E times
x. But since ln of E is just one, that saying that the derivative of ln x is one over x,
this is a very handy fact. And this more general derivative is also worth memorizing. While
we're talking about the derivative of the natural log of x, let's look at the derivative
of the natural log of the absolute value of x. The function y equals ln of absolute value
of x is of course closely related to the function y equals ln of x, the difference being that
the domain for ln x is just x values greater than zero, whereas the domain of ln of absolute
value of x is all X's not equal to zero. The graphs are also related. When you look at
the graph of y equals ln absolute value of x, it looks like you're seeing double since
the absolute value of x is equal to x, when x is greater than or equal to zero and negative
x when x is less than zero, ln of the absolute value of x is going to be equal to ln x when
x is greater than or equal to zero, and ln of negative x when x is less than zero. If
I consider the derivative of ln of absolute value of x, I can think of taking the derivative
of each piece separately. We just saw that the derivative of ln x is one over x. So the
derivative of ln of minus x is going to be one over minus x times the derivative of minus
x, which is minus one by the chain rule. Notice that this second expression simplifies to
one over x. So the derivative of ln of absolute value of x is equal to one over x, whether
x is positive or negative. This formula will come in handy later when we start doing integrals.
In this video, we found that the derivative of ln x is equal to one over x, kind of a
nice derivative. And more generally, the derivative of log base a of x is one over ln A times
x. We've seen previously that the derivative of x to a constant A is equal to a times x
to the A minus one. This is the power rule. We've also seen that the derivative of a positive
number a raised to the x power is equal to ln A times A to the X So we know how to take
the derivative when the variable x is in the base, or when it's in the exponent. But what
if the variables in both the base and the exponent, how do we take the derivative of
x to the x. To differentiate functions like this, we'll need to use the technique of logarithmic
differentiation. To find the derivative of x to the x, I'm going to set y equal to x
to the x. Now we want to find dy dx. Since we don't know how to compute dydx directly,
so let's take the natural log of both sides. taking the log is often a handy trick when
you have a variable in the exponent that you don't know how to deal with, because the properties
of logs allow us to bring that exponent down and multiply it. Now we have y implicitly
defined in terms of x, so let's use implicit differentiation, we'll take the derivative
of both sides with respect to x. And now we should have no trouble taking the derivatives
because we've gotten rid of the awkward exponential expression. So the derivative on the left
of ln y is one over y times dy dx. And the derivative on the right using the product
rule is x times one over x plus one times ln x.
This simplifies to one over y d y dx equals one plus ln x. So D y dX is going to equal
y times one plus ln x. and replacing y with x to the x, I have dy dx is x to the x times
one plus ln x. This technique of taking the log of both sides differentiating and solving
for dydx is known as logarithmic differentiation. And it's enormously useful whenever you have
variables in both the base and the exponent. Here's another example where our variable
is in both the base and the exponent. So as before, I'm going to set y equal to the expression
that I want to differentiate and compute dydx. First, I'll take the log of both sides. Use
my log rules to bring my exponent down and multiply it and take the derivative of both
sides with respect to x. On the left, I get one over y d y dx. And on the right, I get
one of our x times the derivative of ln tangent x, which is one over tangent x times the derivative
of tangent x, or secant squared x, continuing with the product rule, and didn't take the
derivative of one of our x, that's going to be the derivative of x to the minus one, which
is minus one times x to the minus
This simplifies to one over y d y dx equals one plus ln x. So D y dX is going to equal
y times one plus ln x. and replacing y with x to the x, I have dy dx is x to the x times
one plus ln x. This technique of taking the log of both sides differentiating and solving
for dydx is known as logarithmic differentiation. And it's enormously useful whenever you have
variables in both the base and the exponent. Here's another example where our variable
is in both the base and the exponent. So as before, I'm going to set y equal to the expression
that I want to differentiate and compute dydx. First, I'll take the log of both sides. Use
my log rules to bring my exponent down and multiply it and take the derivative of both
sides with respect to x. On the left, I get one over y d y dx. And on the right, I get
one of our x times the derivative of ln tangent x, which is one over tangent x times the derivative
of tangent x, or secant squared x, continuing with the product rule, and didn't take the
derivative of one of our x, that's going to be the derivative of x to the minus one, which
is minus one times x to the minus
two
two
times ln tangent of x. Simplifying the right hand side, I get one over x times one over
sine x over cosine x times one over cosine squared x minus ln 10x over x squared. rewriting,
I can flip and multiply to get one over x times cosine x over sine x times one over
cosine squared x minus the second term, canceling one copy of cosine and rewriting in terms
of cosecant and secant, I get this expression, I still have to solve for dy dx.
times ln tangent of x. Simplifying the right hand side, I get one over x times one over
sine x over cosine x times one over cosine squared x minus ln 10x over x squared. rewriting,
I can flip and multiply to get one over x times cosine x over sine x times one over
cosine squared x minus the second term, canceling one copy of cosine and rewriting in terms
of cosecant and secant, I get this expression, I still have to solve for dy dx.
So multiplying both sides by y, I get the following. And since y was equal to 10x to
the one over x, I can rewrite everything in terms of x. The technique of logarithmic differentiation
is most useful when taking the derivative of an expression that has a variable in both
the base and the exponent, like in this example, but sometimes it's also handy just as a way
to take the derivative of a complicated product and quotient like in this example. Now, we
could take the derivative here just by using the quotient rule on the product rule, but
it's a little easier to take the log of both sides. And the reason is that when we take
the log of a product, we get a sum and the log of a quotient is a difference and sums
and quotients are a lot easier to deal with. So in this example, the log of y is equal
to ln of x plus ln of cosine of x minus ln of x squared plus x to the fifth power, I
can even bring that fifth power down, because that's another one of my log roles. Now, it's
much more straightforward to take the log of both sides. On the left, I have one over
y dydx, as usual, and on the right, the derivative of ln x is one over x, the derivative of ln
cosine x is one over cosine of x times negative sine of x. And the derivative of ln x squared
plus x is one over x squared plus x times 2x plus one. I'll solve for dydx and get y
times one over x minus sine x over cosine x, that's the same as tangent x minus five
times 2x plus one over x squared plus x. Now I can just rewrite y in terms of x and I'll
be done. Again, I didn't have to use logarithmic differentiation. To find this derivative,
I could have just used the product rule in the quotient rule, but logarithmic differentiation
made it computationally much easier. In this video, we learned how to take the derivative
of expressions that have a variable both in the base and in the exponent. And the idea
was first to set y equal to the expression we want to derive. Next, to take the natural
log of both sides. Next, to derive both sides. And finally, to solve for dy dx. This process
is called logarithmic differentiation. The inverse of a function undoes what the function
does, so the inverse of tying your shoes would be to untie them. And the inverse of the function
that adds two to a number would be the function that subtracts two from a number. This video
introduces inverses and their properties. Suppose f of x is the function defined by
this chart. In other words, f of two is three, f of three is five, f of four is six, and
f of five is one, the inverse function for F written f superscript. Negative 1x undoes
what f does. Since f takes two to three, F inverse takes three back to two. So we write
this f superscript, negative one of three is two. Similarly, since f takes three to
five, F inverse takes five to three. And since f takes four to six, f inverse of six is four.
And since f takes five to one, f inverse of one is five. I'll use these numbers to fill
in the chart. Notice that the chart of values when y equals f of x and the chart of values
when y equals f inverse of x are closely related. They share the same numbers, but the x values
for f of x correspond to the y values for f inverse of x, and the y values for f of
x correspond to the x values for f inverse of x. That leads us to the first key fact
inverse functions reverse the roles of y and x. I'm going to plot the points for y equals
f of x in blue.
So multiplying both sides by y, I get the following. And since y was equal to 10x to
the one over x, I can rewrite everything in terms of x. The technique of logarithmic differentiation
is most useful when taking the derivative of an expression that has a variable in both
the base and the exponent, like in this example, but sometimes it's also handy just as a way
to take the derivative of a complicated product and quotient like in this example. Now, we
could take the derivative here just by using the quotient rule on the product rule, but
it's a little easier to take the log of both sides. And the reason is that when we take
the log of a product, we get a sum and the log of a quotient is a difference and sums
and quotients are a lot easier to deal with. So in this example, the log of y is equal
to ln of x plus ln of cosine of x minus ln of x squared plus x to the fifth power, I
can even bring that fifth power down, because that's another one of my log roles. Now, it's
much more straightforward to take the log of both sides. On the left, I have one over
y dydx, as usual, and on the right, the derivative of ln x is one over x, the derivative of ln
cosine x is one over cosine of x times negative sine of x. And the derivative of ln x squared
plus x is one over x squared plus x times 2x plus one. I'll solve for dydx and get y
times one over x minus sine x over cosine x, that's the same as tangent x minus five
times 2x plus one over x squared plus x. Now I can just rewrite y in terms of x and I'll
be done. Again, I didn't have to use logarithmic differentiation. To find this derivative,
I could have just used the product rule in the quotient rule, but logarithmic differentiation
made it computationally much easier. In this video, we learned how to take the derivative
of expressions that have a variable both in the base and in the exponent. And the idea
was first to set y equal to the expression we want to derive. Next, to take the natural
log of both sides. Next, to derive both sides. And finally, to solve for dy dx. This process
is called logarithmic differentiation. The inverse of a function undoes what the function
does, so the inverse of tying your shoes would be to untie them. And the inverse of the function
that adds two to a number would be the function that subtracts two from a number. This video
introduces inverses and their properties. Suppose f of x is the function defined by
this chart. In other words, f of two is three, f of three is five, f of four is six, and
f of five is one, the inverse function for F written f superscript. Negative 1x undoes
what f does. Since f takes two to three, F inverse takes three back to two. So we write
this f superscript, negative one of three is two. Similarly, since f takes three to
five, F inverse takes five to three. And since f takes four to six, f inverse of six is four.
And since f takes five to one, f inverse of one is five. I'll use these numbers to fill
in the chart. Notice that the chart of values when y equals f of x and the chart of values
when y equals f inverse of x are closely related. They share the same numbers, but the x values
for f of x correspond to the y values for f inverse of x, and the y values for f of
x correspond to the x values for f inverse of x. That leads us to the first key fact
inverse functions reverse the roles of y and x. I'm going to plot the points for y equals
f of x in blue.
Next, I'll plot the points for y equals f inverse of x in red. Pause the video for a
moment and see what kind of symmetry you observe in this graph. How are the blue points related
to the red points, you might have noticed that the blue points and the red points are
mirror images over the mirror line, y equals x. So our second key fact is that the graph
of y equals f inverse of x can be obtained from the graph of y equals f of x by reflecting
over the line y equals x. This makes sense, because inverses, reverse the roles of y and
x. In the same example, let's compute f inverse of f of two. This open circle means composition.
In other words, we're computing f inverse of f of two. We compute this from the inside
out. So that's f inverse of three. Since F of two is three, and f inverse of three, we
see is two. Similarly, we can compute f of f inverse of three. And that means we take
f of f inverse of three. Since f inverse of three is two, that's the same thing as computing
F of two, which is three. Please pause the video for a moment and compute these other
compositions. You should have found that in every case, if you take f inverse of f of
a number, you get back to the very same number you started with. And similarly, if you take
f of f inverse of any number, you get back to the same number you started with. So in
general, f inverse of f of x is equal to x, and f of f inverse of x is also equal to x.
This is the mathematical way of saying that F and envir f inverse undo each other. Let's
look at a different example. Suppose that f of x is x cubed. Pause the video for a moment
and guess what the inverse of f should be? Remember, F inverse undoes the work that F
does. You might have guessed that f inverse of x is going to be the cube root function.
And we can check that this is true by looking at f of f inverse of x, that's F of the cube
root of function, which means the cube root function cubed, which gets us back to x. Similarly,
if we compute f inverse of f of x, that's the cube root of x cubed. And we get back
to excellence again. So the cube root function really is the inverse of the cubing function.
When we compose the two functions, we get back to the number that we started with. It'd
be nice to have a more systematic way of finding inverses of functions besides guessing and
checking. One method uses the fact that inverses, reverse the roles of y and x. So if we want
to find the inverse of the function, f of x equals five minus x over 3x, we can write
it as y equals five minus x over 3x. Reverse the roles of y and x to get x equals five
minus y over three y, and then solve for y. To solve for y, let's multiply both sides
by three y. Bring all terms with wisened to the left side, and alternans without y's and
then to the right side, factor out the y and divide to isolate y. This gives us f inverse
of x as five over 3x plus one. Notice that our original function f and our inverse function,
f inverse are both rational functions, but they're not the reciprocals of each other.
And then General, f inverse of x is not usually equal to one over f of x. This can be confusing,
because when we write two to the minus one, that does mean one of our two, but f to the
minus one of x means the inverse function and not the reciprocal. It's natural to ask
if all functions have inverse functions, that is for any function you might encounter. Is
there always a function that it is its inverse?
Next, I'll plot the points for y equals f inverse of x in red. Pause the video for a
moment and see what kind of symmetry you observe in this graph. How are the blue points related
to the red points, you might have noticed that the blue points and the red points are
mirror images over the mirror line, y equals x. So our second key fact is that the graph
of y equals f inverse of x can be obtained from the graph of y equals f of x by reflecting
over the line y equals x. This makes sense, because inverses, reverse the roles of y and
x. In the same example, let's compute f inverse of f of two. This open circle means composition.
In other words, we're computing f inverse of f of two. We compute this from the inside
out. So that's f inverse of three. Since F of two is three, and f inverse of three, we
see is two. Similarly, we can compute f of f inverse of three. And that means we take
f of f inverse of three. Since f inverse of three is two, that's the same thing as computing
F of two, which is three. Please pause the video for a moment and compute these other
compositions. You should have found that in every case, if you take f inverse of f of
a number, you get back to the very same number you started with. And similarly, if you take
f of f inverse of any number, you get back to the same number you started with. So in
general, f inverse of f of x is equal to x, and f of f inverse of x is also equal to x.
This is the mathematical way of saying that F and envir f inverse undo each other. Let's
look at a different example. Suppose that f of x is x cubed. Pause the video for a moment
and guess what the inverse of f should be? Remember, F inverse undoes the work that F
does. You might have guessed that f inverse of x is going to be the cube root function.
And we can check that this is true by looking at f of f inverse of x, that's F of the cube
root of function, which means the cube root function cubed, which gets us back to x. Similarly,
if we compute f inverse of f of x, that's the cube root of x cubed. And we get back
to excellence again. So the cube root function really is the inverse of the cubing function.
When we compose the two functions, we get back to the number that we started with. It'd
be nice to have a more systematic way of finding inverses of functions besides guessing and
checking. One method uses the fact that inverses, reverse the roles of y and x. So if we want
to find the inverse of the function, f of x equals five minus x over 3x, we can write
it as y equals five minus x over 3x. Reverse the roles of y and x to get x equals five
minus y over three y, and then solve for y. To solve for y, let's multiply both sides
by three y. Bring all terms with wisened to the left side, and alternans without y's and
then to the right side, factor out the y and divide to isolate y. This gives us f inverse
of x as five over 3x plus one. Notice that our original function f and our inverse function,
f inverse are both rational functions, but they're not the reciprocals of each other.
And then General, f inverse of x is not usually equal to one over f of x. This can be confusing,
because when we write two to the minus one, that does mean one of our two, but f to the
minus one of x means the inverse function and not the reciprocal. It's natural to ask
if all functions have inverse functions, that is for any function you might encounter. Is
there always a function that it is its inverse?
In fact, the answer is no. See, if you can come up with an example of a function that
does not have an inverse function. The word function here is key. Remember that a function
is a relationship between x values and y values, such that for each x value in the domain,
there's only one corresponding y value. One example of a function that does not have an
inverse function is the function f of x equals x squared. To see that, the inverse of this
function is not a function. Note that for the x squared function, the number two and
the number negative two, both go to number four. So if I had an inverse, he would have
to send four to both two and negative two. The inverse would not be a function, it might
be easier to understand the problem, when you look at a graph of y equals x squared.
Recall that inverse functions reverse the roles of y and x and flip the graph over the
line y equals x. But when I flipped the green graph over the line y equals x, I get this
red graph. This red graph is not the graph of a function, because it violates the vertical
line test. The reason that violates the vertical line test is because the original green function
violates the horizontal line test, and has 2x values with the same y value. In general,
a function f has an inverse function if and only if the graph of f satisfies the horizontal
line test, ie every horizontal line intersects the graph. In most one point, pause the video
for a moment and see which of these four graphs satisfy the horizontal line test. In other
words, which of the four corresponding functions would have an inverse function, you may have
found that graphs A and B, violate the horizontal line test. So their functions would not have
inverse functions. But graph C and D satisfy the horizontal line test. So these graphs
represent functions that do have inverses. functions that satisfy the horizontal line
test are sometimes called One to One functions. Equivalent way of function is one to one,
if for any two different x values, x one and x two, the y value is f of x one and f of
x two are different numbers. Sometimes, as I said, f is one to one, if, whenever f of
x one is equal to f of x two, then x one has to equal x two. As our last example, let's
try to find P inverse of x, where p of x is the square root of x minus two drawn here.
If we graph P inverse on the same axis as p of x, we get the following graph, simply
by flipping over the line y equals x. If we try to solve the problem, algebraically, we
can write y equal to a squared of x minus two, reverse the roles of y and x and solve
for y by squaring both sides and adding two. Now if we were to graph y equals x squared
plus two, that would look like a parabola, it would look like the red graph was already
drawn together with another arm on the left side. But we know that our actual inverse
function consists only of this right arm, we can specify this algebraically by making
the restriction that x has to be bigger than or equal to zero. This corresponds to the
fact that on the original graph, for the square root of x, y was only greater than or equal
to zero. Looking more closely at the domain and range of P and P inverse, we know that
the domain of P is all values of x such that x minus two is greater than or equal to zero.
Since we can't take the square root of a negative number. This corresponds to x values being
greater than or equal to two, or an interval notation, the interval from two to infinity.
The range of P, we can see from the graph is our y value is greater than or equal to
zero, or the interval from zero to infinity.
In fact, the answer is no. See, if you can come up with an example of a function that
does not have an inverse function. The word function here is key. Remember that a function
is a relationship between x values and y values, such that for each x value in the domain,
there's only one corresponding y value. One example of a function that does not have an
inverse function is the function f of x equals x squared. To see that, the inverse of this
function is not a function. Note that for the x squared function, the number two and
the number negative two, both go to number four. So if I had an inverse, he would have
to send four to both two and negative two. The inverse would not be a function, it might
be easier to understand the problem, when you look at a graph of y equals x squared.
Recall that inverse functions reverse the roles of y and x and flip the graph over the
line y equals x. But when I flipped the green graph over the line y equals x, I get this
red graph. This red graph is not the graph of a function, because it violates the vertical
line test. The reason that violates the vertical line test is because the original green function
violates the horizontal line test, and has 2x values with the same y value. In general,
a function f has an inverse function if and only if the graph of f satisfies the horizontal
line test, ie every horizontal line intersects the graph. In most one point, pause the video
for a moment and see which of these four graphs satisfy the horizontal line test. In other
words, which of the four corresponding functions would have an inverse function, you may have
found that graphs A and B, violate the horizontal line test. So their functions would not have
inverse functions. But graph C and D satisfy the horizontal line test. So these graphs
represent functions that do have inverses. functions that satisfy the horizontal line
test are sometimes called One to One functions. Equivalent way of function is one to one,
if for any two different x values, x one and x two, the y value is f of x one and f of
x two are different numbers. Sometimes, as I said, f is one to one, if, whenever f of
x one is equal to f of x two, then x one has to equal x two. As our last example, let's
try to find P inverse of x, where p of x is the square root of x minus two drawn here.
If we graph P inverse on the same axis as p of x, we get the following graph, simply
by flipping over the line y equals x. If we try to solve the problem, algebraically, we
can write y equal to a squared of x minus two, reverse the roles of y and x and solve
for y by squaring both sides and adding two. Now if we were to graph y equals x squared
plus two, that would look like a parabola, it would look like the red graph was already
drawn together with another arm on the left side. But we know that our actual inverse
function consists only of this right arm, we can specify this algebraically by making
the restriction that x has to be bigger than or equal to zero. This corresponds to the
fact that on the original graph, for the square root of x, y was only greater than or equal
to zero. Looking more closely at the domain and range of P and P inverse, we know that
the domain of P is all values of x such that x minus two is greater than or equal to zero.
Since we can't take the square root of a negative number. This corresponds to x values being
greater than or equal to two, or an interval notation, the interval from two to infinity.
The range of P, we can see from the graph is our y value is greater than or equal to
zero, or the interval from zero to infinity.
Similarly, based on the graph, we see the domain of P inverse is x values greater than
or equal to zero, the interval from zero to infinity. And the range of P inverse is Y
values greater than or equal to two, or the interval from two to infinity. If you look
closely at these domains and ranges, you'll notice that the domain of P corresponds exactly
to the range of P inverse, and the range of P corresponds to the domain of P inverse.
This makes sense because inverse functions reverse the roles of y and x. The domain of
f inverse of x is the x values for F inverse, which corresponds to the y values or the range
of F. The range of f inverse is the y values for F inverse, which correspond to the x values
or the domain of f. In this video, we discussed
Similarly, based on the graph, we see the domain of P inverse is x values greater than
or equal to zero, the interval from zero to infinity. And the range of P inverse is Y
values greater than or equal to two, or the interval from two to infinity. If you look
closely at these domains and ranges, you'll notice that the domain of P corresponds exactly
to the range of P inverse, and the range of P corresponds to the domain of P inverse.
This makes sense because inverse functions reverse the roles of y and x. The domain of
f inverse of x is the x values for F inverse, which corresponds to the y values or the range
of F. The range of f inverse is the y values for F inverse, which correspond to the x values
or the domain of f. In this video, we discussed
five key properties of inverse functions. inverse functions reverse the roles of y and
x. The graph of y equals f inverse of x is the graph of y equals f of x reflected over
the line y equals x. When we compose F with F inverse, we get the identity function y
equals x. And similarly, when we compose f inverse with F, that brings x to x. In other
words, F and F inverse undo each other. The function f of x has an inverse function if
and only if the graph of y equals f of x satisfies the horizontal line test. And finally, the
domain of f is the range of f inverse. And the range of f is the domain of f inverse.
These properties of inverse functions will be important when we study exponential functions
and their inverses logarithmic functions. This video defines the standard inverse trig
functions, sine inverse cosine inverse and tan inverse. In this crazy looking graph,
please focus first on the thin black line. This is a graph of y equals sine x. The graph
of the inverse of a function can be found by flipping the graph of the original function
over the line y equals x. I've drawn the flipped graph with this blue dotted line. But you'll
notice that the blue dotted line is not the graph of a function, because it violates the
vertical line test. So in order to get a function, that's the inverse of y equals sine x, we
need to restrict the domain of sine of x, we'll restrict it to this piece that's drawn
with a thick black line. If I invert that piece, by flipping it over the line y equals
x, I get the piece drawn with a red dotted line here. And that piece does satisfy the
vertical line test. So it is in fact a function. of the regular sine x has domain from negative
infinity to infinity, or restricted sine x has domain from negative pi over two to pi
over two. It's it's range is still from negative one to one, just like the regular sine x.
Because I've taken the biggest possible piece of the graph whose flipped version is still
a function. The inverse sine function is often written as arc sine of x. And since inverting
a function reverses the roles of y at x, it reverses the domain and the range. So arc
sine of x, the inverse function has domain from negative one to one, and range from negative
pi over two to pi over two, which seems plausible from the graph. Now an inverse function undoes
the work of a function. So if the function sine takes angle theta, two numbers, x, then
the inverse sine, or arc sine takes numbers x, two angles theta. For example, since sine
of pi over two is one, arc sine of one is pi over two. And in general, the output of
arc sine of x is the angle between negative pi over two and pi over two whose side is
x. y is equal to arc sine x means that x is equal to sine of y. But since there are many
angles, y who sine is x, right, they all differ by multiples of two pi. We specify also, that
y is between negative pi over two and pi over two. That was the whole point of doing this
domain restriction in order to get a well defined inverse value. There's an alternate
notation for inverse sine. Sometimes it's written as sine to the negative one of x.
But this notation can be confusing, so be careful. In particular, sine to the negative
one of x does not equal one over sine of x.
One over sine of x the reciprocal function is another word for cosecant The backs, but
sign to the negative one of x is another word for arc sine of x, the inverse sine function,
which is not the same thing as the reciprocal function. Let's go through the same process
to build an inverse cosine function, we start with a graph of cosine of x, we flip it over
the line y equals x to get the blue dotted line. But the blue dotted line is not a function.
So we go back and restrict the domain for our original cosine of x to just be between
zero and pi. The resulting red graph now satisfies the vertical line test. So it's a proper inverse
function. Our restricted cosine has domain from zero to pi, and range from negative one
to one. And so our inverse function, arc cosine has domain from negative one to one, and range
from zero to pi. Since cosine takes us from angles to numbers, arc cosine takes us from
numbers back to angles. For example, cosine of pi over four is the square root of two
over two. So arc cosine of the square root of two over two is equal to pi over four.
arc cosine of x is the angle between zero and pi, whose cosine is x. In other words,
y equals r cosine of x means that x is equal to cosine of y, and y is between zero and
pi. Since otherwise, there'd be lots of possible answers for an angle y whose cosine is x.
The alternative notation for arc cosine is cosine inverse. And again, we have to be careful.
cosine to the negative 1x is not the same thing as one over cosine of x. one over cosine
of x is also called secant of x. cosine to the negative 1x means arc cosine, the inverse
function, and these two things are not the same. Finally, let's look take a look at inverse
tangent function. Here's a graph of tangent and black, these vertical lines aren't really
part of the function, they're just vertical asymptotes. So in order to get actual function,
when we flip over the line y equals x, we take just one piece of the tangent function.
Here we've taken the piece marked in black, we flip that over the line y equals x, we
get this piece in red, which is actually a function because it satisfies the vertical
line test. Now, you might ask, would it be possible to pick a different piece of the
tangent function to invert? And the answer is yes, we could do that. And on another planet,
maybe mathematicians do that. But on our planet, we use the convention that we pick this piece
of tangent to invert, which is kind of a convenient choice since it's centered here around the
origin. In the previous two examples, our choice of restricted domain for sine and for
cosine was also a convention that led to a conveniently defined inverse function. In
any case, based on our choice, our restricted tan function has domain from negative pi over
two to pi over two. We don't include the endpoints in that interval, because our restricted tan
function has vertical asymptotes, that negative pi over two pi over two, so it's not defined
there. The range of our restricted tan function is from negative infinity to infinity. Therefore,
arc tan of X has domain from negative infinity to infinity and range from negative pi over
two to pi over two. Once again, tangent is taking us from angles to numbers. So arc tan
is taking us from numbers to angles. For example, tangent of pi over four is one, and therefore,
arc tan of one is pi over four. So arc tan of x means the angle between negative pi over
two and pi over two whose tangent is x.
y is equal to arc tan x means that x is equal to tangent of y and the Why is between negative
pi over two and pi over two. The inverse tan function can also be written as 10 to the
minus one of x. And once again, tan inverse of x means the inverse trig function, arc
tan of x. And it's not equal to one over tan of x, which is called cotangent of x. And
that's all for this video on the three basic inverse trig functions. sine inverse x, also
known as arc sine of x, cosine inverse x, also known as arc cosine x, and tan inverse
x, also known as arc tan x. In this video, we'll use implicit differentiation to find
the derivatives of the inverse trig functions. First, the inverse sine function. Recall that
y equals sine inverse of x means that y is the angle in radians whose sine is x. In other
words, we can write x equals sine of y as an almost equivalent statements, I say almost
equivalent, because there are lots of different y's whose side is x, lots of different angles
can have the same resulting side. And for the inverse sine function, we specify that
that angle y has to be between pi over two and negative pi over two, that's just the
convention. Be careful not to mistake sine inverse of x, which is an inverse function,
and one over sine x, which is a reciprocal function. These are not the same thing. This
negative one does not mean reciprocal here, it means inverse function. There is another
notation for inverse sine, which is arc sine. So arc sine of x is the same thing as sine
inverse of x, we want to find the derivative of sine inverse of x, in other words, the
derivative of y with respect to x, where y is sine inverse of x, I'm going to rewrite
this equation here as x equals sine y, and then use implicit differentiation. So taking
the derivative of both sides with respect to x, I have derivative of x is equal to the
derivative of sine y. In other words, one is equal to cosine of y times dy dx. Solving
for dydx, I have that dydx is one over cosine of y. Now I found the derivative, but it's
not in a super useful form, because there's still a Y and expression, I'd rather have
it all in terms of x. Well, I could rewrite this as dy dx is one over cosine of sine inverse
of x, since after all, y is equal to sine inverse of x, but that's still not a super
useful form, because it's difficult to evaluate this. So instead of doing this, I'm going
to look at a right triangle. I want to label my triangle with y and x. Since y is my angle,
I'll put it here. And since sine of y is x, and sine is opposite over hypotenuse, I can
label my opposite side with x and my hypothesis with one. From this, I can figure out the
length of my remaining side, it's going to have to be the square root of one minus x
squared by the Pythagorean Theorem. Now I can compute cosine of y just from the triangle.
Cosine of y is adjacent overhype hotness, so that's the square root of one minus x squared
over one, or just the square root of one minus x squared. I've been implicitly assuming that
y is a positive angle between zero and pi over two when I've been drawing this triangle,
but you can check that the same formula also works. If y is a negative angle, think of
it going down here on the unit circle instead of up here.
So now that I have a formula for cosine y in terms of x, I can go back to my derivative
and substitute and I get dy dx is one over the square root of one minus x squared. In
other words, I found a formula for With the derivative of inverse sine of x, we can carry
out a similar process to find the derivative of inverse cosine y equals cosine inverse
x means that x is equal to cosine of y. And by convention, y lies between zero and pi.
To find the derivative of arc cosine of x, arc cosine is just an alternative notation
for cosine inverse, I can write y equals arc cosine of x, and equivalently x equals the
cosine of y. And then I want to find dydx. Using implicit differentiation, please pause
the video and try it for yourself before going on. So starting with the equation x equals
cosine y, we're going to take the derivative of both sides with respect to x. The derivative
of x is one, and the derivative of cosine y is negative sine y, dy dx. So d y dx is
equal to negative one over sine y. As before, I can draw and label a right triangle, the
angle is y. And now I know that x is cosine of y. So I'm going to put x on the adjacent
side and one on my partner's leaving the square root of one minus x squared on the opposite
side, which means that sine of y, which is opposite overhype hotness, is equal to the
square root of one minus x squared. And so dy dx is going to be negative one over the
square root of one minus x squared. And I have my formula for the derivative of arc
cosine of x. inverse tangent can be handled very similarly. And again, you may want to
try it for yourself before watching the video. Y equals inverse tangent of x means that x
is tangent of y. And the convention is that y is supposed to lie between negative pi over
two and pi over two. Proceeding as before, we write y equals our tan of x and x equals
tan of y, take the derivative of both sides. So we get one equals secant squared of y d
y dx. Solving for dydx we have one over seacon square root of y. And using our right triangle
as before, we can label the angle of as y. Since I know that tangent y is x, and tangent
is opposite over adjacent, I'm going to label the opposite side x and the adjacent side
one, which gives us a hypothesis of the square root of one plus x squared. Now we know that
secant of y is one over cosine of y. So that's going to be hi partners over adjacent. So
that's the square root of one plus x squared over one. And so secant squared of y is just
the square of this, which is one plus x squared. Now I can substitute into my formula for dy
dx, and I get dy dx is one over one plus x squared, which gives me a nice formula for
the derivative of inverse tangent. The other inverse trig functions cotesia inverse, seacon
inverse and cosequin inverse, have derivatives that can be computed. Similarly, the following
table summarizes these results. In some books, you may see absolute value signs around the
X for the formulas for inverse secant and inverse cosecant. Of course, when x is positive,
this makes no difference. And when x is negative, this discrepancy comes from differences in
the convention for the range of y for these inverse trig functions seeking inverse and
cosecant numbers.
Notice that the derivatives of the inverse trig functions that start with CO, all have
negative signs in front of them and are the negatives of the corresponding inverse trig
functions without the CO that makes it easier to remember them. You should memorize these
formulas. Let's do one example using the formulas that we just found. Let's compute The derivative
of tan inverse of A plus x over a minus x, we'll want to use the formula for the derivative
of tan inverse x. Now, to compute dydx, for our function, we can use the chain rule, the
outside function is tan inverse, whose derivative is one over one plus the inside function squared,
we'll need to multiply that by the derivative of the inside function. I'll just copy over
the first part and take the derivative of a plus x over a minus x using the quotient
rule. So I put the denominator on the bottom and square it. And then I take low times d,
hi, the derivative of a plus x with respect to x is just one minus high D low, the derivative
of a minus x with respect to x is negative one. I'll simplify my numerator, a minus x,
the negative one and the negative sign here cancel, so I get plus a plus x. On the denominator,
I have one plus a plus x squared over a minus x squared, multiplied by a minus x squared.
canceling in the numerator, I get to a and distributing and the denominator I get a minus
x squared plus a plus x squared. If I expand out the denominator, the same simplifies to
two a over two A squared plus 2x squared, or just a over a squared plus x squared, which
is a pretty nice derivative. So now you know the derivatives of the inverse trig functions.
And you also know how to find them using implicit differentiation, if you ever forget them.
When two or more quantities are related by an equation, then their rates of change over
time are also related. That's the idea behind related rates. And this video gives an example
of related rates involving distances. A tornado is 20 miles west of us, heading due east towards
Phillips Hall at a rate of 40 miles per hour. you hop on your bike and ride due south at
a speed of 12 miles per hour. How fast is the distance between you and the tornado changing
after 15 minutes. In a related rates problem, it's always a good idea to draw a picture
first, that can help you uncover the geometry of the problem and see how quantities are
related. In this problem, we have a right triangle. Because the tornadoes traveling
due east and the bicycles traveling due south at right angles. Let's assign variables to
the quantities of interest. I'll call the distance between the tornado and Phillips
Hall a. Although it starts at 20 miles, it varies with time, and therefore it's a good
idea to assign it a letter a variable. I'll use B to refer to the distance between Philips
Hall and the bicycle, a quantity that also varies with time. And I'll let c stand for
the distance between the tornado and the bicycle. The problem asks us to find how fast this
distance is changing. In other words, DC dt. The next step is to write down equations that
relate the quantities of interest. In this problem, we know by the Pythagorean Theorem,
that a squared plus b squared equals c squared. We're interested in how fast the distance
between you and the tornado is changing. That's a rate of change. And the rate at which the
bicycle is traveling and the tornado is moving. These are also rates of change. In order to
work these rates of change into the problem, I'm going to take the derivative of both sides
of this equation with respect to time. That's the third step. So I'm going to take DDT of
a squared plus b squared. And that's equal to DDT of C squared. Notice that I'm thinking
of a B and C as functions of T here, since they vary over time, on the left side, I get
to a times dA DT using the chain rule,
plus two B DB dt. And on the right side, I get to see DC dt. Now I can use the information
given to me in the problem to plug in numbers and solve for the quantity of interest DC
DT since the tornado was moving at a rate of 40 miles per hour The distance between
the tornado and Philip's Hall is decreasing at 40 miles per hour. In other words, da dt
is negative 40. That negative sign is important here, and comes from the fact that the distance
is decreasing. Since the bicycle is moving at 12 miles per hour, the distance between
Philips Hall and the bicycle is increasing at a rate of 12 miles per hour. So DBT is
positive 12. The quantity is a, b and c are constantly changing. But at the time of interest,
t equals 15 minutes or in hours, 0.25 hours, we can figure out what a B and C are. The
tornado is starts 20 miles away, but it's moving at a rate of 40 miles per hour. So
after a quarter of an hour, it's gone 10 miles, that means after a quarter of an hour, it's
only 10 miles away. And so at the time of point two, five hours, a equals 10. The bike
is moving at 12 miles per hour. So after a quarter of a mile, it's gone three miles.
And so at this time, b equals three. Now using the same equation we started with, we can
plug in a and b and solve for C, we know that c squared is going to be 10 squared plus three
squared. So C is going to be the square root of 109. Plugging in the numbers into this
equation, we get two times 10 times negative 40 plus two times three times 12 equals two
times the square root of 109 times DC dt. So DC dt is going to be negative 800 plus
72 over two times a squared of 109, which is approximately negative 35. In other words,
the distance between the tornado and us is decreasing at 35 miles per hour, the tornado
is gaining on us quickly.
These same steps will get you through a variety of related rates problems. A couple of cautionary
notes. Don't plug in numbers to send. Any quantities that vary with time should be written
as variables, so you can properly take the derivative with respect to time. In addition,
be careful to use negative numbers for negative rates of change. That is, for quantities that
are decreasing, we wouldn't have gotten the right answer if we hadn't have used a negative
40 for the rate of change of the distance here. In this video, we solved the related
rates problem, and found that riding a bicycle may not be the best way to escape a tornado.
In this classic related rates problem, water's flowing into a cone shaped tank, and we have
to figure out how fast the water is rising. Water flows into a tank at a rate of three
cubic meters per minute, the tank is shaped like a cone with a height of four meters,
and a radius of five meters at the top, we're supposed to find the rate at which the water
level is rising in the tank. When the water height is two meters. we've drawn our picture.
Now let's label some quantities of interest. It's fine to use numbers for the quantities
that stay fixed throughout the problem. Like the dimensions of the tank. For any quantities
that are varying with time, I need to use letters variables to represent those quantities.
So the height of the water is varying throughout the problem. I'll call that H. And it might
be handy to also talk about the radius of the part of the cone that's filled with water.
I'll call that our ultimately I want to find the rate at which the water level is rising.
So that's DHD T. Next, I want to write down equations that relate the quantities of interest.
From geometry, I know that the volume of a cone is 1/3 times the area of the base times
the height. So the volume of water in the cone is going to be 1/3 times pi r squared
times h since h is the height of the piece of the code that contains water. And pi r
squared is the area of that circular base for that piece of account. I'm calling it
the base even though it's at the top. There's one more equation that's going to be handy
here that comes from similar triangles. From similar triangles, we know the ratio of sides
for the little triangle here is the same as the ratio of sides for the big triangle. In
other words, we know that our over h is going to be equal to five over four. I can use this
relationship to eliminate one of the variables in this equation. Let's think for a minute
which one we want to eliminate. Since we're ultimately interested in finding DHD T, we
need to keep the variable h in here. But since we don't have any information about how are
you changing, it's a good idea to get rid of the R. So let's solve for r here. And we
get r equals five fourths times h, and plug that back into our volume equation. So we
get v equals 1/3 pi times five fourths h squared times h. Or in other words, V equals 25/48
pi h cubed. Now we're gonna derive both sides of the equation with respect to time t, to
get rates of change into the problem. Remember that we're thinking of the volume of water
and the height of water as functions of time t, we get dv dt equals 2548. It's pi times
three h squared DHD t. Now let's plug in numbers and solve for the quantity of interest DHD
T. From our problem, we know that water's flowing into the tank at a rate of three cubic
meters per minute. So dv dt is three, we're asked to find the rate at which the water
level is rising when the water height is two meters. So that's when h is two. Plugging
in those values and solving for a DHD T, we get d h dt is equal to three divided by 2540.
It's pi times three times two squared, which is 12 over 25 pi meters per second, are about
point one five meters per second.
This video solve the related rates problem involving volume, and use the trick of finding
similar triangles to eliminate one variable. In this video, we'll do a related rates problem
involving rotation and angles. a lighthouse that's half a mile west of shore, has a rotating
light that makes two revolutions per minute in the counterclockwise direction, the shore
runs north south, and there's a cave directly east of the lighthouse. How fast is the beam
of light moving along the shore at a point one mile north of the cave. we've drawn a
picture. Now let's label it with variables for all the quantities that are changing with
time, the distance between the lighthouse and the cave that's fixed. So we don't have
to put a variable for that. But the distance between the cave and the point on the shore
where the light is hitting, that's varying. So I'll call that say x. Since we want to
know how fast the beam of light is moving, we're going to want to know how that distance
x is changing. In other words, we want to calculate dx dt, when x is one. The high partners
of this right triangle made by the beam of light is also changing with time as as the
angle here between the beam of light and the East West line, I'll call that angle theta.
And the angle up here, I suppose is also changing and call that fee. This angle is the right
angle between the East West line and the north south line. So that doesn't change, it's always
90 degrees. Next, we want to write down equations to relate the quantities of interest. Whenever
I see a right triangle in a problem, I'm tempted to write down the Pythagorean theorem, which
in this case would say one half squared plus x squared equals h squared. But in this particular
problem, it doesn't look like that's going to help us much because of a tiger in theory
and would relate x and h. But we don't have any information about how H is changing. The
only rate of change information given to us is this two revolutions per minute. Two revolutions
per minute is indirectly telling us how this angle theta is changing. Because if the light
beam is making two revolutions per minute, then since they're two pi radians in a revolution,
that amounts to a change of four pi radians per minute for the angle theta. Therefore
I'd really like to write down the equation that has to do with theta and x. And from
trig, I know that tangent of theta is opposite over adjacent. So I can write down tangent
theta equals x divided by one half. Or in other words, tangent theta is 2x. This is
the equation that I need that relates x and theta. Now I'm going to derive both sides
with respect to time t. And I get second squared theta, d theta dt equals two times dx dt.
Next, I can plug in numbers and solve for my quantity of interest, which is dx dt when
x equals one. We already figured out from the two revolutions per minute, that d theta
dt is four pi. Now secant theta is one over cosine theta. And since cosine theta is adjacent
ever had partners, it's reciprocal is high partners over adjacent. So in our picture,
that gives us h over one half. Well, when x equals one, H, is going to be the square
root of one squared plus a half squared by the Pythagorean Theorem. And we'll divide
that by one half. And simplifying, we get the square root of five fourths divided by
one half, which ends up as the square root of five. So let's plug these values into our
equation involving derivatives. And we get the square root of five squared per second
squared times four pi, four d theta dt
equals two times dx dt. Solving for dx dt, we get dx dt is five times four pi divided
by two, or 10 pi. So what are the units here on dx dt, since our distance has been in miles,
and our time is in minutes, this is 10 pi miles per minute. If I want to convert this
to more standard units of miles per hour, I can just multiply my 10 pi miles per minute
by 60 minutes per hour to get 600 pi miles per hour. That works out to about 1885 miles
per hour, which is pretty darn fast. In this related rates problem, we related rotations
per minute, to a change in angle per minute. And we use the trig equation to relate angle
and side length. solving a right triangle means finding the length of all the sides
and the measures of the angles given partial information. In this example, we're given
the length of one side and the measure of one angle, plus we know the measure of this
right angle is 90 degrees, we need to find the measure of the third angle labeled capital
A, and the length of the two sides labeled lowercase b and lowercase C. To find the measure
of angle A, let's use the fact that the measures of the three angles of a triangle add up to
180 degrees. So that means that 49 degrees plus 90 degrees plus a is equal to 180 degrees.
So A is equal to 180 degrees minus 90 degrees minus 49 degrees, which works out to 41 degrees.
To find the length of the side D we have a couple of possible options. We could use the
fact that tan of 49 degrees, which is opposite over adjacent is B over 23. So b is 23 times
tan 49 degrees, which works out to 26.46 units. Alternatively, we could use the fact that
tan of 41 degrees is 23 over b since now if we're looking at the angle here 23 is our
opposite and B is an adjacent That's a little bit harder to solve algebraically. But we
can write B tan 41 degrees equals 23, which means that B is 23 divided by tan 41 degrees.
With a calculator that works out again to 26.46. The reason we want to use 10 in this
problem and not say sine or cosine is because 10 of say 49 degrees relates, and the unknown
side that we're looking for be to the side that we know the measure of, if we had use
sine instead, would be saying that sine of 49 is B over C, and we'd have two unknowns,
which would make it difficult to solve. Next, to find the side length C, we can have a few
options, we could use a trig function again, for example, we could use the cosine of 49
degrees, that's adjacent over hypotenuse, which is 23 oversee. Solving for C, we get
that C is 23 over cosine 49, which works out to 35.06 units. Another option would be to
use the Pythagorean Theorem to find C. Since we know 23 squared plus b squared equals c
squared. In other words, that's 23 squared plus 26.46 squared equals c squared, which
means that C is the square root of that song, which works out again, to 35.06. To review,
the ideas we used were, the sum of the angles is equal to 180 degrees. We used facts like
tangent of an angle being opposite over adjacent and similar facts about sine and cosine. And
we use the Pythagorean Theorem. This allowed us to find all the angles and side lengths
of the triangle, knowing just the side length of one side and the angle of one of the non
right angles to begin with. In this next example, we don't know any of the angles except for
the right angle, but we know to have the side lengths.
To find the unknown angle theta, we can use the fact that cosine theta is adjacent overhype
hotness, so that's 10 over 15. Cosine is a good trig function to use here, because this
equation relates our unknown angle to our two known sides. So we just have one unknown
in our equation to solve for. To solve for theta, we just take the cosine inverse of
10/15, which is 0.8411 radians, or 48.19 degrees. To find the measure of angle fee, we could
use the fact that sine of fee is 10 over 15 and take sine inverse of 10/15. But probably
a little easier, let's just use the fact that these three angles do 180 degrees. That tells
us that fee plus 90 plus 48.19 is equal to 180. Which means that fee is 41.81. Finally,
we can find x either using a trig function, or by using the Pythagorean Theorem. To find
it using a trig function, we could write down something like tan of 48.19 degrees is x over
10. To find that using the Tyrion theorem, we'd write down 10 squared plus x squared
equals 15 squared. I'll use a Pythagorean theorem and find the x by doing the square
root of 15 squared minus 10 squared. That gives me an answer of 11.18. Notice that we
use many of the same ideas as in the previous problem. For example, the fact that the sum
of the angles is 180. The Pythagorean theorem and the trig functions like tan, sine and
cosine, we also use the inverse trig functions to get from an equation like this one to the
angle. This video showed how it's possible to find the length of all the sides of a right
triangle, and the measures of all the angles given partial information. For example, the
measure of one angle and one side or from two sides. This video gives some definitions
and facts related to maximum and minimum values. functions. function f of x has an absolute
maximum at the x value of C. If f of c is greater than or equal to f of x for all x
in the domain of f. The point with x&y coordinates of C FRC is called an absolute maximum point.
And the y value f of c is called the absolute maximum value. Now, if I draw a graph of f,
the y value f of c is the highest value that that function ever achieves. And an absolute
maximum point is just a point where it achieves that maximum value. Now, it's possible for
a function to have more than one absolute maximum point, if there happens to be a tie
for the highest value. But a function has at most one absolute maximum value. A function
f of x has an absolute minimum that x equals C, if f of c is less than or equal to f of
x, for all x in the domain of f. In this case, the point c f of c is called an absolute minimum
point. And the y value f of c is called the absolute minimum value. In the graph of f
of x, f of c is now the lowest point that the function achieves anywhere on its domain,
and C SOC, are the coordinates of a point where the function achieves that minimum value.
For example, this function has an absolute minimum value of about negative eight has
an absolute minimum point with coordinates three, negative eight. If this function stops
here, and just has a domain from zero to four, then the function has an absolute maximum
value of 10 at the absolute maximum point with coordinates for 10. If however, the function
keeps going in this direction, it will not have an absolute maximum value at all. absolute
maximum and minimum values can also be called global maximum and minimum values.
In addition to absolute maximum mins, we can talk about local maximums. So a function f
of x has a local maximum at x equals C. If f of c is greater than or equal to f of x,
for all x, near C. By near C, we mean there's some open interval around C for which this
was true. For our graph of f, we have a local maximum right here. Even though it's not the
highest point anywhere around since there's a higher point up here, this is the highest
point in an open interval around see the point C FFC is called a local maximum point. And
the y value f FC is called a local maximum value. Similarly, a function f of x has a
local minimum at x equals C, if f of c is less than or equal to f of x for all x, near
C. And the point c f of c is called a local minimum point. And the y value f of c is called
a local minimum value. A function might have many local minimum values. In this example,
assuming that the domain is zero to four, we have a local minimum point right here.
Because it's the lowest point anywhere nearby. It also happens to be an absolute minimum
point. Now turning our attention to local maximums, we have a local maximum point right
here with coordinates about one two. Since f of one is as high or higher than f of x
for any x value in an open interval around one. In this example, the absolute maximum
point of 410 does not count as a local maximum point. Simply because we can't take an open
interval on both sides of for the function doesn't exist on the right side. And so for
that sort of technical reason, we end up with an absolute maximum point That's not a local
maximum point here. local maximum and minimum values can also be called relative maximum
and minimum values. Please take a look at this graph and pause the video for a moment
to mark all local maximum minimum points, as well as all global, that is absolute maximum
points. See if you can find the absolute maximum value and the absolute minimum value for the
function. I'm going to mark the local maximum points in green, and the absolute maximum
points in red. The function definitely has a local men
here. Since this is the lowest point anywhere nearby and then open interval, and there's
a local max point here. There's also a local min point here, where the function also hits
a low point and open interval. But that local man is also an absolute man. So I'll mark
it half green and half red. There's also a local min point here at the point three, two,
since this point is the as low or lower than any point in an open interval. And the function
is defined in an open interval around three, even though it's discontinuous there. In fact,
this point is tied for local minimum, with all the points on this interval here, between
two and three, there are as low or lower than all points in an open interval around them.
The point 04 doesn't count as a local max, because the function is not defined on the
other side of zero. So there's no open interval to to consider. This point is also not an
absolute maximum because the function gets higher over here. In fact, as long as this
trend continues, the function f of x has no absolute maximum value at all, because its
values just keep getting higher and higher as x goes off to infinity. There's one more
point that I want to consider. And that's this point here, at three, three and a half.
Well, it's tempting to say that f has a local maximum here, it looks like it's the highest
point in the ground. But in fact, there is no point here at three 3.5. Right, the functions
value at three is actually down here, too. So there's no point here to be a local maximum
point. And if you start looking at points really close to that point, those aren't local
maximums either, because you can always find a point just a little bit higher as you get
closer and closer, but don't quite reach this missing point of three 3.5. So we have all
the absolute and local maximum points marked. And now to find the absolute maximum value.
Well, we just said that there is none. But the absolute minimum value is the y value
of this absolute minimum point here. So I'd say that's about 0.5. Here I've drawn the
graph of a function. What do you notice about the derivative of this function at its local
maximum and minimum points, please pause the video and think about it. Well, the local
maximum minimum points are here, here and here. And at two of those points, the derivative
f prime of c equals zero. And that the third point, f prime of c does not exist, because
the function has a corner. A number c is called a critical number for a function f if f prime
of c does not exist, or f prime of c exists and equals zero. So in other words, all of
these local maximum minimum points for this example, they're all critical points. And
this is true in general, if f has a local max or min at C, then C must be a critical
number for F. We also say that the point c f of c is a critical point for F. It's important
not to read too much into this statement. The statement says that if f has a local max
or man at C then C must be a critical number. But the converse doesn't hold. In other words,
if c is a critical number, then f may or may not have a local max or men at sea. One example
to keep in mind is the function f of x equals x cubed at a value of C of zero. Since f prime
of x is 3x squared, we have that f prime of zero equals zero. So zero is a critical number.
But notice that F does not have a local maximum man at x equals zero.
In this video, we defined absolute and local, maximum and minimums. We also defined critical
numbers, which are numbers c, where f prime of c equals zero, or f prime of c does not
exist. We noted that if f has a local max or min at C, then C is a critical number.
But not necessarily advice. In this video, we'll see how the first derivative and the
second derivative can help us find local maximums and local minimums for a function. Recall
that f of x has a local maximum at x equals C. If f of c is greater than or equal to f
of x, for all x, in an open interval around C, F of X has a local minimum at x equals
C. If f of c is less than or equal to f of x, for all x, in an open interval around C.
In this example, the function f has a local maximum at x equals six at x equals 11. And
a local minimum at x equals about 10. We've seen before that if f has a local max or local
min at x equals C, then f prime of c is equal to zero or it does not exist. Number C at
which f prime of c is zero or does not exist, are called critical numbers. But you have
to be careful, because it is possible for F to have a critical number at C. That is
a place where if privacy is equal to zero, or does not exist, but not have a local max,
or min at x equals C. In fact, this happens in the graph above, at x equals two, since
f prime of two is zero, but there's no local max or min there. Please pause the video for
a moment and try to figure out what's different about the derivative of f in the vicinity
of x equals two, where there's no local max or min, and in the vicinity of x equals 610
and 11 where there are local maxes and mins. Near the critical point at x equals two, the
derivative is positive on the left and positive again on the right. But near the local maximums,
the derivative is positive on the left and negative on the right. And near the local
minimum, the derivative is negative on the left and positive on the right. These observations
help motivate the first derivative test for finding local maximums and minimums. The first
derivative test says that if f is a continuous function near x equals C, and if c is a critical
number, then we can decide if f has a local maximum or minimum at x equals c by looking
at the first derivative near x equals C. More specifically, if we know that f prime of x
is positive for x less than c, and negative for x greater than c, then our function looks
something like this. Or maybe like this, your x equals C, and so we have a local max at
x equals C. If on the other hand, f prime of x is negative for x less than c, and positive
for x greater than c, then our function looks something like this. Or maybe like this, your
x equals C and so we have local men at x equals see. If our first derivative is positive on
both sides of C or negative on both sides of C, then we do not have a look All extreme
point at all at x equals C. Instead, our graph might look something like this, or maybe like
this. The first derivative test is great, because it lets us locate local extreme points
just by looking at the first derivative. The second derivative test gives us an alternative
for finding local maximum points by using the second derivative. Specifically, the second
derivative test tells us that if f is continuous near x equals C, then if f prime of c is equal
to zero, and f double prime of c is greater than zero,
then f has a local min at x equals C. If on the other hand, f prime of c equals zero,
and f double prime of c is less than zero, then f has a local max at x equals C. Note
that if f double prime of c is equal to zero, or does not exist, then the second derivative
test is inconclusive. We might have a local max or a local man at x equals C, or we might
not. So we'd have to use a different method like the first derivative test to find out.
In this video, we introduced the first derivative test and the second derivative test, which
allow us to determine if a function has a local minimum or a local maximum at a certain
value of x. In this video, I'll work through two examples of finding extreme values, that
is, maximum values and minimum values of functions. In the first example, we're asked to find
the absolute maximum and minimum values for this rational function g of x on the interval
from zero to four, these maximum and minimum values could occur at critical numbers in
the interior of the interval, or they could occur at the endpoints of the interval. So
we'll need to check the critical numbers, and check the endpoints and compare our values.
To find the critical numbers, those are the numbers where g prime of x is equal to zero
or does not exist. So let's take the derivative g prime of x using the quotient rule. So we
get x squared plus x plus two squared on the denominator, and then we have low times d
high, the derivative of the numerator is one minus high times the derivative of the denominator,
that's 2x plus one. Before we figure out where that zero or doesn't exist, let's simplify
it a little bit. So we can multiply out the numerator. I'll distribute the negative sign.
And I'll add together like terms in the numerator, I'm just leaving the denominator alone on
all these steps. So our simplified numerator is going to be minus x squared plus 2x plus
three. Now that I've simplified the derivative, I can figure out where it's equal to zero
and where it doesn't exist. Let me clear a little space. Now the only way that g prime
of x could not exist, is if the denominator is zero. But on our interval, where x is between
zero and 4x squared plus x plus two is always greater than or equal to two. So the denominator
is never zero on this interval. In fact, it turns out that x squared plus x plus two is
never zero, even if we look outside this interval. And you can check that if you want to use
the quadratic formula. But in any case, we don't have to worry about the places where
g prime of x does not exist. So we only have to worry where g prime of x is equal to zero.
To find where g prime of x is equal to zero, we just have to check where the numerator
is equal to zero. So I'll set negative x squared plus 2x plus three equal to zero and multiply
both sides there by negative one and a factor I get that x equals three or x equals minus
one. So these are my critical numbers. But notice that one of these critical numbers
negative one doesn't even lie within my interval, so I don't have to worry about it. All I have
to worry about is x equals 3x equals three is one place where my function g could have
an absolute maximum or minimum. So let's figure out G's value there by plugging in three for
x that evaluates to to 14th, or one set So we've checked the critical numbers. Now let's
go ahead and check the endpoints. Those are the point the x values of zero, and four,
since our interval is from zero to four. Plugging in, we get that g of zero is negative one
half, and g of four is 320 seconds. I sometimes like to make a table of all these candidate
values.
The candidate x values are 03, and four and the corresponding g of x values we found were
negative a half 1/7, and 320 seconds. Now to find the absolute maximum and minimum values,
all I have to do is figure out which one of these y values is the biggest and which is
the smallest. Well, clearly negative one half is a smallest, so that's the absolute minimum
value. And we just need to compare 1/7 and 320 seconds to see which is bigger. Now, 1/7
is the same as 320 firsts, which is going to be bigger than 320 seconds. So one half
1/7 is our absolute max value. We can confirm this by looking at a graph of our function
g. Remember, we're just interested in the interval from zero to four. So we're just
interested in this section of the graph. And it does look like the minimum value is here
at when x equals zero minimum value of negative one half like we found, and the maximum value.
Well, I'm not sure exactly where it is from this graph, but it does look like it's somewhere
around three. And that is a value of something around 1/7. So the graph does confirm what
we found as a more precise answer using calculus. For the next example, let's find the absolute
extreme values for the function f of x, which is the absolute value of x minus x squared,
on the interval from negative two to two. As before, we can find absolute extreme values
by checking first the critical numbers, and then also the endpoints of the interval negative
two and two. To find the critical numbers, we need to take the derivative of our function.
But because our function involves the absolute value, it's a little tricky to take the derivative.
Instead, let's first rewrite f using piecewise notation. Recall that, if we're looking at
the absolute value of x, when x is bigger than or equal to zero, absolute value of x
is just x. So f of x will be x minus x squared. On the other hand, when x is less than zero,
the absolute value of x is negative x. So f of x will be negative x minus x squared.
Now to take the derivative, we can take the derivative of each piece. So when x is bigger
than zero, I don't want to take the derivative when x equals zero, because there might be
funny things happening, you know, a cost per corner, so I'm just gonna worry about when
x is bigger than zero, and when x is less than zero, for now, when x is bigger than
zero, I can just use the power rule I get one minus 2x is the derivative when x is less
than zero, I get negative one minus 2x. And now to find where I have critical numbers,
I need to find where f prime of x is equal to zero, or f prime of x does not exist. Well,
f prime of x equals zero, where one minus 2x equals zero for x bigger than zero. And
where one, negative one minus 2x is equal to zero for x less than zero. So that corresponds
to x equal one half for x bigger than zero, and x equals negative one half for x less
than zero. So those are my first two critical numbers. And that is they do lie within my
interval that I'm interested in. But I also have to worry about where f prime of x does
not exist. And the candidate x value for that is where x equals 01 way to convince ourselves
the derivative does not exist when x equals zero is to look at the fact that the derivative
is very close to one for x values, very close to zero from the right side and very close
to negative one for x values from the left side. So the graph of the function is going
to have to be sloping down with a slope near negative one for x less than zero and up with
a slope near one for x greater than zero, and so it'll end up having a cost per corner
there.
Also notice that even if I weren't 100% sure that the derivative didn't exist at x equals
zero, it's not going to hurt to consider this x equals zero as a possible additional candidate
for the absolute max or min value. So I've got my three critical numbers, and my end
points are just going to be x equals negative two and x equals two. So let me make a chart
of values my x values to consider are negative two, negative one half 01, half and positive
two, and my corresponding f of x values are going to be, let's say absolute value of negative
two minus negative two squared works out to two minus four, which is negative two, I plug
in negative one half, I get one half minus 1/4, which is 1/4. plug in zero, I get zero.
and plugging in one half, I get 1/4. And plugging in two, I get negative two again. So now my
biggest value is going to be 1/4. So that's my absolute max value, and my smallest value
is going to be negative two. So that's my absolute min value, I can confirm what I found
looking at the graph. So here I've graphed my function, y equals absolute value of x
minus x squared on the interval from negative two to two. And I can see indeed, that my
absolute min is going to be a value of negative two, it occurs at two absolute minimum points,
and my absolute maximum is going to be a value of about 1/4. And that occurs at two absolute
maximum points. And that concludes this video on finding extreme values. the mean value
theorem relates the average rate of change of a function on an interval to its instantaneous
rate of change, or derivative. Let's assume that f is a function defined on a closed interval
a, b, and maybe defined in some other places to let's assume that f is continuous on the
whole closed interval. And that is differentiable on the interior of the interval, then the
mean value theorem says that there must be some number c in the interval a, b, such that
the average rate of change of f on a B is equal to the derivative of f at C. In symbols,
we can write the average rate of change as f of b minus F of A over B minus A. And that
has to equal f prime at C for some number C. On the graph, the average rate of change
of f is the slope of the secant line. And so the mean value theorem says that there's
some number c, somewhere in between a and b, so that the slope of the secant line is
exactly the same as the slope of the tangent line at that x value of C. The number c is
not necessarily unique. So I encourage you to pause the video and see if you can draw
a graph of a function where there's more than one c value that works. So you might have
drawn something maybe like this. Now, if we draw our secant line, there's two values of
c, where the slope of the tangent line is equal to the slope of the secant line. In
this example, we're asked to verify the mean value theorem for a particular function on
a particular interval. Verify means that we need to check the hypotheses of the theorem
hold. And also the the conclusion holds. The hypotheses are that f is continuous on the
closed interval, one three, and that is differentiable on that interior of that interval. Both of
these facts are true, because f is a polynomial. Now we need to verify that the conclusion
of the mean value theorem holds. In other words, we need to find a number c in the interval
one, three, such that the derivative of f at C is equal to the average rate of change
of f on the interval from one to three.
Now f prime of x is 6x squared minus eight. So f prime at any number c is just six c squared
minus eight. We can also compute f of three just by plugging in and get 31 and f of one
is negative five. Plugging in these values into our equation, we get that six c squared
minus eight has to equal 31 minus negative five over To, in other words, six c squared
minus eight had better equal 18, which means that six c squared needs to equal 24. So C
squared has to equal four, which means that C has to equal plus or minus two. Since negative
two is not in the interval from one to three, we're left with a c value of positive two.
So C equals two is the number we're looking for. And at C equals to f prime is equal to
18, which is the average rate of change of f on the interval. we've verified the mean
value theorem. In this example, we're told that f of one is seven, and that the derivative
of f is bounded between negative three and negative two. On the interval one six, we're
asked to find the biggest and smallest values that f of six could possibly be. Well, the
mean value theorem gives us one way of relating the derivative of the function to its values
on the endpoints of the interval. More specifically, the mean value theorem tells us that the average
rate of change F of six minus f of one over six minus one is equal to the derivative f
prime of c, for some C, in the interval, one, six. Since the derivative is bounded between
negative three and negative two, we know that the average rate of change is bounded between
negative three and negative two. We know that f of one is seven. And now we can solve this
inequality for f of six. Multiply the inequality by five and add seven. And now we can see
that negative eight is the smallest possible value for F six and negative three is the
largest possible roles, there is an important special case of the mean value theorem. If
f is a function defined on the closed interval a b, and f of x is continuous on that whole
closed interval, differentiable on the interior of the interval. And if f of a is equal to
f
of b,
then there's a number c in the interval a, b, such that f prime of c is zero. If we look
at a graph of such a function that has equal values at a, and at B, we can see where its
derivative has to be zero at a maximum, or a minimum in between A and B. To see why the
rolls there is a special case of the mean value theorem. Think about what the mean value
theorem would say about this function, it would say there is a C, such that f prime
of c is equal to the average rate of change of the function. But since f of b and F of
A are the same, by our assumption, this average rate of change is just zero. And so the mean
value theorem, its conclusion is that there's a C, such that f prime of c equals zero, which
is exactly the conclusion of rules theorem. In this video, we saw that for a function
that's continuous on a closed interval, and differentiable on the interior of that interval,
the average rate of change of the function is equal to the instantaneous rate of change
of the function, f prime of c for some C in the interval. This video gives two proofs
of the mean value theorem for integrals. the mean value theorem for integrals says that
for continuous function f of x, defined on interval from a to b, there's some number
c between A and B, such that f of c is equal to the average value of f. The first proof
that I'm going to give uses the intermediate value theorem. Recall that the intermediate
value theorem says that if we have a continuous function f defined on an interval, which I'll
call x 1x, two, if we have some number l in between, f of x one and f of x two, then f
has to achieve the value l somewhere between x one and x two. Keeping in mind the intermediate
value theorem, let's turn our attention back to the mean value theorem for integrals. Now,
it's possible that our function f of x might be constant on the interval from a to b. But
if that's true, then our mean value theorem for integrals holds easily, because f AV is
just equal to that constant, which is equal to f of c for any c between A and B. So let's
assume that f is not constant, will it continue continuous function on a closed interval has
to have a minimum value and a maximum value, which I'll call little m, and big M. Now,
we know that F's average value on the interval has to be between its maximum value and its
minimum value. If you don't believe this, consider the fact that all of F's values on
the interval have to lie between big M and little m. And if we integrate this inequality,
we get little m times b minus a is less than or equal to the integral of f is less than
or equal to big M times b minus a. Notice that the first and the last integrals, were
just integrating a constant. Now if I divide all three sides by b minus a, I can see that
little m is less than or equal to the average value of f is less than or equal to big M
as I wanted. Now, I just need to apply the intermediate value theorem with F's average
as my number L and little m and big M as my values of f of x one and f of x two. The intermediate
value theorem says that F average is achieved by f of c for some C in between my x one and
x two. And therefore, for some C in my interval a b. And that proves the mean value theorem
for integrals. Now I'm going to give a second proof for the mean value theorem for integrals.
And this time, it's going to be as a corollary to the regular mean value theorem for functions.
Recall that the mean value theorem for functions, says that if g of x is continuous on a closed
interval,
and differentiable on the interior of that interval, then there's some number c in the
interval, such that the derivative of g at C is equal to the average rate of change of
G, across the whole interval from a to b. Let's keep the mean value theorem for functions
in mind, and turn our attention back to the mean value theorem for integrals. I'm going
to define a function g of x to be the integral from a to x of f of t dt, where F is the function
given to us in the statement of the mean value theorem for integrals. Notice that g of A
is just the integral from a to a, which is zero, while g of B is the integral from a
to b of our function. Now, by the fundamental theorem of calculus, our function g of x is
continuous and differentiable on the interval a, b, and g prime of x is equal to f of x.
And by the mean value theorem for functions, we know that g prime of c has to equal g of
b minus g of a over b minus a, for some numbers, C and the interval a b, if we substitute in
the three facts above, into our equation below, we get f of c is equal to the integral from
a to b of f of t dt minus zero over b minus a, which is exactly the conclusion that we
wanted to reach. This shows that the mean value theorem for integrals really is the
mean value theorem for functions where our function is an integral. And this completes
the second proof of the main value theorem for integrals. So now I've proved the mean
value theorem for integrals in two different ways. And I've used a lot of the great theorems
of calculus along the way in this video, We'll solve inequalities involving polynomials like
this one, and inequalities involving rational expressions like this one. Let's start with
a simple example, maybe a deceptively simple example, if you see the inequality, x squared
is less than four, you might be very tempted to take the square root of both sides and
get something like x is less than two as your answer. But in fact, that doesn't work. To
see why it's not correct, consider the x value of negative 10. Negative 10 satisfies the
inequality, x is less than two, since negative 10 is less than two. But it doesn't satisfy
the inequality x squared is less than four, since negative 10 squared is 100, which is
not less than four. So these two inequalities are not the same. And it doesn't work to solve
a quadratic inequality just to take the square root of both sides, you might be thinking
part of why this reasoning is wrong, as we've ignored the negative two option, right? If
we had the equation, x squared equals four, then x equals two would just be one option,
x equals negative two would be another solution. So somehow, our solution to this inequality
should take this into account. In fact, a good way to solve an inequality involving
x squares or higher power terms, is to solve the associated equation first. But before
we even do that, I like to pull everything over to one side, so that my inequality has
zero on the other side. So for our equation, I'll subtract four from both sides to get
x squared minus four is less than zero. Now, I'm going to actually solve the associated
equation, x squared minus four is equal to zero, I can do this by factoring 2x minus
two times x plus two is equal to zero. And I'll set my factors equal to zero, and I get
x equals two and x equals minus two. Now, I'm going to plot the solutions to my equation
on the number line. So I write down negative two and two, those are the places where my
expression x squared minus four is equal to zero.
Since I want to find where x squared minus four is less than zero, I want to know whether
this expression x squared minus four is positive or negative, a good way to find that out is
to plug in test values. So first, I plug in a test value in this area, the number line,
something less than negative two, say x equals negative three. If I plug in negative three
into x squared minus four, I get negative three squared minus four, which is nine minus
four, which is five, that's a positive number. So at negative three, the expression, x squared
minus four is positive. And in fact, everywhere on this region of the number line, my expression
is going to be positive, because it can jump from positive to negative, without going through
a place where it's zero, I can figure out whether x squared minus four is positive or
negative on this region, and on this region of the number line by plugging in test value
similar way, evaluate the plug in between negative two and two, a nice value is x equals
00 squared minus four, that's negative four and negative number. So I know that my expression
x squared minus four is negative on this whole interval. Finally, I can plug in something
like x equals 10, something bigger than two, and I get 10 squared minus four. Without even
computing that I can tell that that's going to be a positive number. And that's all that's
important. Again, since I want x squared minus four to be less than zero, I'm looking for
the places on this number line where I'm getting negatives. So I will share that in on my number
line. It's in here, not including the endpoints, because the endpoints are where my expression
x squared minus four is equal to zero and I want it strictly less than zero, I can write
my answer. As an inequality, negative two is less than x is less than two, or an interval
notation as soft bracket negative two, two soft bracket. Our next example, we can solve
similarly, first, we'll move everything to one side so that our inequality is x cubed
minus 5x squared minus 6x is greater than or equal to zero. Next, we'll solve the associated
equation by factoring. So first, I'll write down the equation. Now I'll factor out an
x. And now I'll factor the quadratic. So the solutions to my equation are x equals 0x equals
six and x e equals negative one, I'll write the solutions to the equation on the number
line. So that's negative one, zero, and six. That's where my expression x times x minus
six times x plus one is equal to zero. But I want to find where it's greater than or
equal to zero. So again, I can use test values, I can plug in, for example, x equals negative
two, either to this version of expression, or to this factored version. Since I only
care whether my answer is positive or negative, it's sometimes easier to use the factored
version. For example, when x is negative two, this factor is negative. But this factor,
x minus six is also negative when I plug in negative two for x. Finally, x plus one, when
I plug in negative two for x, that's negative one, that's also negative. And a negative
times a negative times a negative gives me a negative number. If I plug in something
between negative one and zero, say x equals negative one half, then I'm going to get a
negative for this factor, a negative for this factor, but a positive for this third factor.
Negative times negative times positive gives me a positive for a test value between zero
and six, let's try x equals one. Now I'll get a positive for this factor a negative
for this factor, and a positive for this factor. positive times a negative times a positive
gives me a negative. Finally, for a test value bigger than six, we could use a x equals 100,
that's going to give me positive positive positive. So my product will be positive.
Since I want values where my expression is greater than or equal to zero, I want the
places where n equals zero. And the places where it's positive. So my final answer will
be close bracket negative one to zero, close bracket union, close bracket six to infinity.
As our final example, let's consider the rational inequality, x squared plus 6x plus nine divided
by x minus one is less than or equal to zero. Although it might be tempting to clear the
denominator and multiply both sides by x minus one, it's dangerous to do that, because x
minus one could be a positive number. But it could also be a negative number. And when
you multiply both sides by a negative number, you have to reverse the inequality. Although
it's possible to solve the inequality this way, by thinking of cases where x minus one
is less than zero or bigger than zero, I think it's much easier just to solve the same way
as we did before. So we'll start by rewriting so that we move all terms to the left and
have zero on the right, well, that's already true. So the next step would be to solve the
associated equation. That is x squared plus 6x plus nine over x minus one is equal to
zero. That would be where the numerators 0x squared plus 6x plus nine is equal to zero,
so we're x plus three squared is zero, or x equals negative three, there's one extra
step we have to do for rational expressions. And that's we need to find where the expression
does not exist. That is, let's find where the denominator is zero. And that said, x
equals one. I'll put all those numbers on the number line, the places where my rational
expression is equal to zero, and the place where my rational expression doesn't exist,
then I can start in with test values. For example, x equals minus four, zero and two
work. If I plug those values into this expression here, I get a negative answer, a negative
answer and a positive answer. The reason I need to conclude the values on my number line
where my denominator is zero is because I can my expression can switch from negative
to positive by passing through a place where my rational expression doesn't exist, as well
as passing by passing flew to a place where my rational expression is equal to zero. Now
I'm looking for where my original expression was less than or equal to zero. So that means
I want the places on the number line where my expression is equal to zero, and also the
places where it's negative. So My final answer is x is less than one, or an interval notation,
negative infinity to one. In this video, we solved polynomial and rational inequalities
by making a number line. And using test values to make a sign chart. The first and second
derivative of a function can tell us a lot about the shape of the graph of the function.
In this video, we'll see what f prime and f double prime can tell us about where the
function is increasing and decreasing, is concave up and concave down and has inflection
points. We say that a function is increasing. If f of x one is less than f of x two, whenever
x one is less than x two. In other words, the graph of the function goes up. As x increases
from left to right, we say the function f is decreasing. If f of x one is greater than
f of f two, whenever x one is less than x two. In other words, the height of the function
goes down as we move from left to right. In this graph, it's a little hard to say what's
happening when x is near two, is it completely horizontal, or is the graph slightly increasing?
If we assume it's slightly increasing, then, in this example, f of x is increasing as x
ranges from zero to six, and again as x ranges from 10 to 11. The graph is decreasing for
x values between six and 10. And for x values between 11 and 12, the first derivative of
f can tell us where the function is increasing and decreasing. In particular, if f prime
of x is greater than zero for all x on an interval, then f is increasing on this interval.
This makes sense, because f prime being greater than zero means the tangent line has positive
slope. Similarly, if f prime of x is less than zero for all excellent interval, then
f is decreasing on this interval. That's because a negative derivative means the tangent line
has a negative slope.
A precise proof of these facts can be found in the textbook, or in another video, we say
that a function is concave up on an interval from a to b. If informally, it looks like
a bowl that could hold water on that interval. More formally, the function is concave up
on that interval. If all the tangent lines for the function on that interval, lie below
the graph of the function. The function is concave down on the interval from a to b.
If informally, it looks like an upside down ball that would spill water on that interval.
Or more formally, the function is concave down. If all the tangent lines lie above the
graph of the function on an interval. In this example, f is concave up around here. And
again around here. On the left piece, it looks like part of a ball that could hold water.
So we can say that f is concave up on the intervals from two to four, and the interval
from eight to 11. f is concave down on this piece, and this piece and this piece, so we
can say that f is concave down on the interval from zero to two, from four to eight, and
from 11 to 12. The concavity of a function is related to its second derivative. Here,
where the function is concave up, its derivative is going from essentially zero to larger positive
values. So the first road was increasing, which means the second derivative is positive.
On this section of the graph, which is also concave up, the driven is going from negative
values to zero. That's an increase in the first derivative. So that means the second
derivative here must be positive. And in this piece, where the first term is going from
zero to positive values, the first derivative is also increasing, so the second derivative
is also positive. On the parts of the function that are concave down, we can see that in
this example, the second derivative is negative. Here, the first derivative is going from positive
towards zero, that's a decrease in the first derivative or a negative second derivative.
Here the first derivative is going from positive to zero to negative, that's also a decreasing
first derivative or a negative second derivative. And the same thing happens on this section
here. In general, we can use the second derivative to predict the concavity of a function. The
concavity test says that if the second derivative is positive for all x on an interval, then
the function f is concave up on that interval. Similarly, if the second derivative is negative
for all x on the interval, then the function f is concave down on that interval. One way
to remember the concavity test is that a positive second derivative gives us a happy face. So
the smile is supposed to be a concave up function. And a negative second derivative gives us
a sad face where the smile or the frown, I guess, is a concave down function. Next, let's
talk about inflection points. A function has an inflection point at x equals C, if it's
continuous at C, and it changes concavity at C. In other words, f has an inflection
point at x equals C. If f changes from concave up to concave down at x equals C, or it changes
from concave down to concave up. In this graph, if we draw the concavity regions again, we
see that F has an inflection point at x equals two, where the function changes from concave
down to concave up at x equals four, where the function changes from concave up to concave
down at x equals eight, and again, at x equals 11. Since concavity, has to do with the second
derivative being positive or negative, inflection points happen where the second derivative
changes sign from positive to negative, or from negative to positive. And that's exactly
what the inflection point test says. If f double prime of x changes sign
at x equals C, then f has an inflection point at x equals C. Now in order to change from
positive to negative or negative positive, f double prime has to go through zero, or
go through a point where it doesn't exist. But you have to be careful, just because f
double prime is zero, it doesn't exist, does not guarantee that you necessarily have an
inflection point, because it could be zero and still be positive on both sides or negative
on both sides. For example, if f of x is x to the fourth, then f prime of x is for x
cubed, and f double prime of x is 12x squared. So f double prime at zero is certainly zero.
But there is no inflection point at x equals zero. In fact, the graph of f of x equals
x to the fourth looks kinda like a flattened quadratic, and so there's no change in calm
cavity, f is concave up on both sides of x equals zero. In this video, without the first
derivative can tell us where the function is increasing and decreasing, while the second
derivative can tell us where the function is concave up and concave down. And the second
derivative, changing sign from positive to negative or negative or positive can tell
us where we have inflection points. Since lines are much easier to work with, and more
complicated functions, it can be extremely useful to approximate a function near a particular
value with its tangent line. That's the central idea of this video. Let's start with an example.
Suppose that F of T is the temperature in degrees Fahrenheit at time t measured in hours,
where t equals zero represents midnight. Suppose that f of six is 60 degrees, and the derivative
f prime of six is three degrees per hour. What's your best estimate for the temperature
at 7am and at 8am? Please pause the video for a moment to make your estimate. The temperature
at 6am is 60 degrees. So the temperature at 7am, which we're calling F of seven is approximately
60 degrees. But we can do better than this. At 6am, the temperature is rising, in fact,
it's rising at a rate of three degrees per hour. If this rate of change continues, then
by 7am, the temperature will have risen three degrees and reached 63 degrees. And by 8am,
the temperature will have had two hours to rise from 60 degrees by a rate at a rate of
three degrees per hour. So f of eight should be about 60 degrees plus the three degrees
per hour times two hours or 66 degrees. These estimates use all the information, we're given
both the value of the temperature at six, and its rate of change. Let's see what these
estimates mean graphically, in terms of the tangent line. I'll draw a rough graph of temperature
over time. And I'll also draw in the tangent line at time six. At time six, the height
of the function and the tangent line is equal to 60 degrees. The tangent line has slope
three degrees per hour. So that's a rise of a run of three, which means that seven o'clock,
which is one hour, after six o'clock, the tangent line has risen by three degrees. And
at eight o'clock, the tangent line has risen by another three degrees. So at seven o'clock,
our tangent line has had 63 degrees, and at eight o'clock, our tangent line has height
66 degrees. When making these estimates, here, we were actually using the tangent line to
approximate our function. Our actual temperature function may be rising more steeply than the
tangent line, or it possibly could be rising less steeply, like in this picture. But either
way, the tangent line is a good approximation for our function. When time is near six o'clock.
The idea of approximating a function with its tangent line is a very important idea
that works for any differentiable function. Let f of x be any differentiable function
and let A be an arbitrary x value. Let's suppose we know the value of f at a, we'll call it
F of A. And let's say we want to find the value of f, add an x value near a, let's call
it a plus delta x where delta x means a small number. If we can't compute f of a plus delta
x directly, we can try to approximate it using the tangent line.
We know that the tangent line has a slope given by f prime of a. And so when we go over
by a run of delta x, the tangent line goes up by a rise of f prime of A times delta x.
So the height of the tangent line is going to be f of a plus f prime of A times delta
x. The linear approximation principle says that we can approximate our function with
our tangent line. In other words, f of a plus delta x is approximately equal to f of a plus
f prime of a delta x. Remember that delta x is supposed to be a small number, because
if you get too far away from a, your tangent lines no longer going to be a good approximation
of your function. But how small is small enough is sort of a judgment call. Sometimes the
approximation principle is written with different symbols, if we let x equal a plus delta x,
so x is a number close to a, then delta x is x minus a. And we can rewrite the approximation
principle, as f of x is approximately f of a plus f prime of A times x minus a. The quantity
on the right side here is sometimes referred to as l of x, and called the linearization
of f at A. That is the linearization of f of a is l of x, which is equal to f of a plus
f prime of A times x minus a. So the approximation principle can also be written as f of x is
approximately equal to l of x. Let's look a little more closely at this linearization
equation and what it has to do with the tangent line. Suppose we were going to try to write
down the equation of the tangent line at x equals a, well, the equation for any line
can be given in point slope form as y minus y naught equals the slope times x minus x
naught. Since we're looking for the tangent line that goes through the point A f of a,
we can set x naught equal to a and y naught equal to f of a. Also, the slope of the tangent
line is just f prime of a. So we can rewrite this as y minus F of A equals f prime of A
times x minus a solving for y, we get y equals f of a plus f prime of A times x minus a.
So this equation for the tangent line is really just the equation that we have for the linearization.
But linearization is really just a fancy word for the tangent line. There's a lot of notation
and definitions on this page. But there's only one important principle that you need
to remember. And that's the idea that you can approximate a function with its tangent
line. If you can keep that idea and this picture in mind, then it's easy to come up with this
approximation principle. And its alternative forms. Let's use the approximation principle
in an example, the approximation principle tells us that f of a plus delta x is approximately
equal to f of a plus f prime of A times delta x, we need to figure out what f should be
what a should be, and what delta x should be. Since we're trying to figure out the square
root of 59, it makes sense to make our function the square root function. For a, we'd like
to pick something that is easy to compute f of a, well, what's the number close to 59,
that is easy to compute the square root of 64 springs to mind. So let's set a equal to
64. Since we're trying to compute the square root of 59, we want a plus delta x to be 59.
In other words, 64 plus delta x is 59. And so delta x should be negative five, it's fine
to have a negative number for delta x. Now plugging into our approximation formula, we
have f of 59 is approximately equal to f of 64, plus f prime of 64 times negative five.
Since f of x is the square root of x, or in other words, x to the one half power, f prime
of x is going to be one half x to the minus one half power, or one over two times the
square root of x. So f prime of 64 is one over two times the square root of 64, which
is 1/16. I can rewrite my red equation to say the square root of 59 is approximately
the square root of 64 plus 1/16 times negative five, which is eight minus five sixteenths,
or 7.6875. using a calculator, I can get a more exact value of the square to 59. my calculator
says 7.68114575, up to eight decimal places. Let me draw the picture that goes along with
this approximation. We have the square root function, and at x equals 64, we're looking
at the tangent line. Our delta x here is a negative five, and gets us down to 59. So
we're using the value of our tangent line right here to approximate our actual square
root function right here. As you can see from the picture, it looks like the tangent line
value should be slightly bigger than the actual value. And in fact, that's what we get. Then
next example is very similar. We call it the linearization of a function is just the equation
for its tangent line. Namely, the linearization at a is f of a plus f prime of A times x minus
a. And the approximation principle says that f of x, the function is approximately equal
to its linearization. Its tangent line, at least when x is near a. This is basically
the same formula that we use in the last problem, we're just calling our value x this time instead
of a plus delta x. Since we're trying to estimate sine of A value, it makes sense to let our
function be sine of x. For a, we want to pick a number that's close to 33 degrees, for which
it's easy to calculate sine of that number. Well, sine of 30 degrees is easy to calculate.
So let's make a equal to 30 degrees, but let's put it in radians and call it pi over six.
in calculus, we pretty much always want to use radians for sine and cosine especially
when taking derivatives. Since the derivative formula, D sine x dx equals cosine of x only
works when x is in radians, our x needs to be 33 degrees, since that's the value, we
want to estimate the sine of, we need to multiply by pi over 180 degrees to convert it to radians.
So that becomes 11 pi over 60 radians. Let's plug in for F and a first to get the linearization.
And then we'll plug in for x next. So the linearization of our function is going to
be sine of pi over six, plus the derivative of sine at pi over six times x minus pi over
six. That is l of x is one half, since sine of pi over six is one half, plus cosine of
pi over six times x minus pi over six, cosine of pi over six is the square root of three
over two. So this is our equation for the tangent line, or the linearization of sine
of x at pi over six. Now we know that sine of x is approximately equal to as linearisation,
as long as x is near pi over six. So in particular, sine of our 33 degrees in radians, which is
11 pi over 60 is approximately equal to one half plus a square root of three over two
times 11 pi over 60 minus pi over six.
That simplifies to one half plus the square root of three over two times pi over 60. And
now I'm going to cheat a little bit and use my calculator to get a decimal value for this
of about 0.5453. Now if I use my calculator to find sine of 11 pi over 60, directly, remember,
that's the same thing as sine of 33 degrees. my calculator tells me it is 0.5446, approximately.
So you can see our approximation using the linearization is very close to the calculators,
more accurate value. Notice that in this example, the approximate value based on the linearization
is slightly higher than the actual value. And you can see why from a graph of sine.
The tangent line at pi over six lies slightly above the graph of sine x. Therefore, the
approximate value based on the linearization will be slightly bigger than the actual value
of sine of 33 degrees. In this video, we use several formulas to express one key idea.
The main formulas were the approximation principle, the linear approximation, and the linearization.
The key idea is that a differentiable function can be approximated near a value x equals
A by the tangent line at x equals a. The differential is a new vocabulary word wrapped around the
familiar concept of approximating a function with its tangent line. This figure should
look familiar from the previous video on linear approximation is the same picture. Suppose
we have a differentiable function, f of x, and we know the value of f at some x value
a. That is, we know the value of f evey, but we don't know the value of f at some nearby
x value a plus delta x. That is we don't know f of a plus delta x. So we draw the tangent
line to f of x at x equals a. And we use the tangent line at a plus delta x as an approximation
for the function at a plus delta x. Since the tangent line has slope of f prime of a,
the rise of a run is f prime of a. So if this run here is delta x, this rise has to be f
prime of A times delta x. So the height of the tangent line here at a plus delta x is
going to be f of a plus f prime of A times delta x. That's just the height here plus
the extra height here. And since we're using that height to approximate In our function,
we say that f of a plus delta x is approximately equal to f of a plus f prime of A times delta
x equivalently. If I subtract F of A from both sides, I get f of a plus delta x minus
f of a is approximately equal to f prime of A times delta x. This equation is just the
approximation principle that we've seen before. And this is a very slight alteration of it.
So there's nothing new yet. But now I'm going to wrap some new notation around this familiar
concept. The differential dx is another way of writing delta x, you can think of it as
a small change in the value of x. The differential df is defined as f prime of x dx, or equivalently
f prime of x delta x. Sometimes this is written as d y instead, but d y just means the same
thing here as df. Sometimes it's handy to specify the differential add a particular
value of x, like a value of x equals a, and this is written df equals f prime of a dx,
or f prime of a delta x. Notice that the value of a is not apparent when you just write down
d f, or d y.
Finally, the change in f, which is written delta f, is defined as f of x plus delta x
minus f of x for some value of x, for example, f of a plus delta x minus f of a, this can
also be written as the change in y. Using these new definitions, we can now rewrite
our approximation principle to say, delta f is approximately equal to d f, the change
in the function is approximately equal to the differential course this could also be
written as the change in Y is approximately equal to d y. In the picture, we can now write
d x for the run, and D, F, for the rise of the tangent line. Pause the video and take
a moment to find delta f in this picture, though delta f is f of a plus delta x minus
f of a, so that's this height here. I'll write that as delta f, or delta y. So the approximation
principle, written in differential notation, is just saying that the rise of the function,
delta f is approximated by the rise in the tangent line df. Let's use the differential
and an example. For the function f of x equal to x times ln x. Let's first find the differential
df. We know that df is equal to f prime of x dx. And f prime of x by the product rule
is equal to x times the derivative of ln x, which is one over x plus the derivative of
x, which is one times ln x. So in other words, one plus ln x. Therefore, df is equal to one
plus ln x times dx. When x equals two, and delta x equals negative point three, well
delta x is the same thing as dx, we can just plug in those values and get df is one plus
ln of two times negative 0.3. as a decimal, that's approximately negative 0.5079. Now
delta f is defined as f of x plus delta x minus f of x. So for our function, that's
x plus delta x times ln of x plus delta x minus x ln x. Plugging in the given values
for x and delta x, we get delta f is two minus 0.3 times ln of two minus 0.3 minus two ln
two, which according to my calculator is negative zero. Point 4842. And we see that the change
in the function between two and two minus point three is closely approximated by the
change in the tangent line. As expected. The differential is often used to estimate error,
as in this example, suppose that the radius of a sphere is measured as eight centimeters
with a possible error of point five centimeters. So the sphere that we measure looks something
like this, but the actual sphere might be slightly bigger, or slightly smaller, we want
to use the differential to estimate the resulting error in computing the volume of the sphere.
Well, the volume of a sphere is given by four thirds pi r cubed, where r is the radius.
If our radius changes by point five centimeters, our volume will change by substantially more.
And that change in volume is the error the resulting error in measuring volume. But instead
of computing delta V directly, we're asked to approximate it using the differential.
So we're going to use the fact that delta V is approximately equal to dv, which is easier
to compute.
By definition, dv is equal to the derivative of our function, I'll just call that v prime
as a function of r times Dr. Now, v prime of r is equal to four pi r squared, just by
taking the derivative. And here, we're interested in an R value of eight, and a value of Dr.
Same as delta r of 0.5 centimeters. So dv is going to be four pi r squared Dr. And when
I plug in R and D R, I get four pi times eight squared times 0.5, which is 128 pi, or as
a decimal 402.1 centimeters. That's our error estimate, which seems quite a bit bigger than
our original error of point five in measuring the radius. Now, the relative error of a function
is its error over the original value of the function. So in our case, it's the change
in volume over the actual volume. Since we're using the differential instead, we'll compute
the relative error as dv over V. Now, the volume when r is eight centimeters, is four
thirds times pi times eight cubed. And dv, we already saw was four pi times eight squared
times 0.5. So dv divided by V is given by this ratio, which simplifies to 0.1875. So
an 18.75% relative there. To me, the relative error gives a better sense for the error than
the absolute error estimate above. This video introduced the idea of the differential, we
said that we could think of dx as just being another way of writing delta x, but df represents
the rise in the tangent line, and is equal to f prime of x times dx. Whereas delta f
is the rise in the actual function F. And that's f of x plus delta x minus f of x. On
the picture, dx is the run, df is the rise in the tangent line. And delta f is the rise
in the actual function. In the language of differentials, we can restate the approximation
principle to say that the change in f can be approximated by the differential. In the
past, we've encountered limits, like the limit as x goes to two of x minus two over x squared
minus four. We can't evaluate this limit just by plugging in two for x, because x minus
two goes to zero, and x squared minus four goes to zero as x goes to two. This is known
as a zero over zero indeterminate form. It's called indeterminate because we can't tell
what the limit is going to be just by the fact that the numerator goes to zero and the
denominator goes to zero. It depends on how fast the numerator and the denominator are
going to zero compared to each other. And the final limit of the quotient could be any
number at all, or it could be infinity or it could not even exist. In the past, we've
been able to evaluate some limits in zero over zero indeterminate form by using algebraic
tricks to rewrite the quotients. In this video, we'll introduce lopi talls rule, which is
a very powerful technique for evaluating limits and indeterminate forms. A limited of the
form the limit as x goes to a of f of x over g of x is called a zero over zero indeterminate
form, if the limit as x goes to a of f of x is equal to zero, and the limit as x goes
to a of g of x is equal to zero. A limit and this form is called an infinity over infinity
and determinant form. If the limit as x goes to a of f of x is equal to infinity or minus
infinity. And the limit as x goes to a of g of x is equal to infinity or minus infinity.
We saw an example of a zero over zero indeterminate form in the introductory slide.
One example of a infinity over infinity and determinant form is the limit as x goes to
infinity of 3x squared minus 2x plus seven divided by negative 2x squared plus 16. Notice
that as x goes to infinity, the numerator goes to infinity while the denominator goes
to negative infinity. In these definitions of indeterminate form, it's possible for a
to be negative infinity or infinity, like it is in this example, but it doesn't have
to be loopy. talls rule can be applied when f and g are differentiable functions. And
the derivative of g is nonzero in some open interval around a except possibly in a under
these conditions, if the limit as x goes to a of f of x over g of x is zero over zero
or infinity over infinity indeterminant form than the limit as x goes to a of f of x over
g of x is the same thing as the limit as x goes to a of f prime of x over g prime of
x, provided that the second limit exists, or as plus or minus infinity. Let's look at
loopy tiles rule in action. In this example, as x goes to infinity, the numerator x goes
to infinity and the denominator three to the x also goes to infinity. So we have an infinity
over infinity indeterminate form. So let's try applying lopi tiles rule, our original
limit should equal the limit as x goes to infinity of the derivative of the numerator,
which is one divided by the derivative of the denominator, which is ln of three times
three to the x, provided that the second limit exists or as infinity or negative infinity.
In the second limit, the numerators just fixed at one. And the nominator goes to infinity
as x goes to infinity. Therefore, the second limit is just zero. And so the original limit
evaluates to zero as well. In this example, we have a zero over zero indeterminate form,
because as x goes to zero, sine of x and x, both go to zero in the numerator, and sine
of x cubed goes to zero in the denominator. So using low Patel's rule, I'll try to evaluate
instead, the limit I get by taking the derivative of the numerator and the derivative of the
denominator, the derivative of sine x minus x is cosine of x minus one, and the derivative
of sine x cubed is three times sine x squared times cosine x using the chain rule. Now let
me try to evaluate the limit again, as x goes to zero, cosine of x goes to one. So the numerator
here goes to zero. As x goes to zero, sine of x goes to zero and cosine of x goes to
one, so the denominator also goes to zero. So I still have a zero over zero indeterminate
form. And I might as well try applying loopy toss rule again. But before I do, I want to
point out that cosine of x is going to one. So the cosine of x here really isn't affecting
my limit. And in fact, I could rewrite my limit of a product as a product of limits
where the second limit is just one and can be ignored from here on out. Now apply lopatok
rule on this first limit, which is a little bit easier to take the derivatives and so
the derivative of the top is minus sine x. And the derivative of the bottom is six times
sine x times cosine x. Now let's try to evaluate again, as x goes to zero, our numerator is
going to zero, and our denominator is also going to zero. But hang on, we don't have
to apply lobby towels rule again, because we can actually just simplify our expression,
the sine x on the top cancels with the sine x on the bottom. And we can just rewrite our
limit as the limit of negative one over six cosine of x, which evaluates easily to negative
one, six.
In this example, I want to emphasize that it's a good idea to simplify after each application
of lopi talls rule. If you don't simplify, like we did here, then you might be tempted
to apply loopy towels rule and additional time when you don't need to, which might make
the problem more complicated. Instead of simpler to solve this video, we were able to evaluate
zero over zero and infinity over infinity indeterminate forms by replacing the limit
of f of x over g of x with the limit of f prime of x over g prime of x, provided that
second limit exists. This trick is known as lopi tels rule. We've seen that lopatok rule
can be used to evaluate limits of the form zero over zero, or infinity over infinity.
In this video, we'll continue to use lopi towels rule to evaluate additional indeterminate
forms, like zero times infinity, infinity to the 00 to the zero, and one to the infinity.
In this example, we want to evaluate the limit of a product. Notice that as x goes to zero
from the positive side, sine x goes to zero, and ln x goes to negative infinity. Remember
the graph of y equals ln x. So this is actually a zero times infinity indeterminate form.
Even though the second factor is going to negative infinity, we still call it a zero
times infinity and indeterminate form, you can think of the Infinity here as standing
for either positive or negative infinity. It it's indeterminant. Because as x goes to
zero, the sine x factor is pulling the product towards zero, while the ln x factor is pulling
the product towards large negative numbers. And it's hard to predict what the limit of
the product will actually be. But the great thing is, I can actually rewrite this product
to look like an infinity over infinity and determinant form, or a zero over zero and
determinant form. Instead of sine x times ln x, I can rewrite the limit as ln x divided
by one over sine x. Now as x goes to zero, my numerator is going to negative infinity.
And since sine x is going to zero through positive numbers, my denominator one over
sine x is going to positive infinity. So I have an infinity over infinity indeterminate
form. Now, I could instead choose to leave the sine x in the numerator, and instead,
put a one over ln x in the denominator. If I do this, then as x goes to zero through
positive numbers, sine x goes to zero. And since ln x goes to negative infinity, one
over ln x goes to zero. And so I have a zero over zero indeterminate form. Sometimes it
can be difficult to decide which of these two ways to rewrite a product as a quotient.
One rule of thumb is to take the version that makes it easier to take the derivative of
the numerator and denominator. Another trick is just to try one of the ways and if you
get stuck, go back and try the other. I'm going to use the first method of rewriting
it because I recognize that one over sine x can be written as cosecant of x. And I know
how to take the derivative of cosecant x. Using low Beatles rule on this infinity over
infinity and determinant form, I can rewrite my limit as the limit of what I get when I
take the derivative of the numerator, that's whenever x divided by the derivative of the
denominator, that's negative cosecant x cotangent x. As always, I want to simplify my expression
before going any further. I can rewrite my trig functions in the denominator in terms
of sine and cosine. cosecant x is one over sine x cotangent x is cosine of x over sine
of x. Now flipping and multiplying, I get the limit as x goes to zero plus of one over
x times sine squared of x over negative cosine of x. In other words, the limit of negative
sine squared x over x cosine x,
we know that cosine of x goes to one as x goes to zero. So I can rewrite this as the
limit of negative sine squared x over x times the limit of something that goes to one. So
I once again have a zero over zero indeterminate form. And I can apply lopatok rule yet again,
taking the derivative of the top, I get negative two, sine x, cosine of x. And the job of the
bottom is just one. Now I'm in a good position just to evaluate the limit by plugging in
zero for x in the numerator, I get negative two times zero times one, the denominator
is just one, so my final limit is zero. In this limit, we have a battle of forces. As
x is going to infinity, one over x is going to zero. So one plus one over x is going to
one, but the exponent x is going to infinity, it's hard to tell what's going to happen here.
If we had one, to any finite number, that would be one. But anything slightly bigger
than one, as we raise it to a bigger and bigger powers, we would expect to get infinity. So
our limit has an independent permanent form, it's hard to tell whether the answer is going
to be one infinity, or maybe something in between. Whenever I see an expression with
a variable in the base, and a variable in the exponent, I'm tempted to use logarithms.
If we set y equal to one plus one over x to the x, then if I take the natural log of both
sides, I can use my log roles to rewrite that by multiplying by x in the front. Now, if
I wanted to take the limit as x goes to infinity of ln y, that would be the limit of this product,
x times ln one plus one over x. As x goes to infinity, the first factor x goes to infinity.
One plus one over x goes to just one and ln one is going to zero. So we have a infinity
times zero indeterminate form, which we can try to rewrite as an infinity over infinity,
or a zero over zero indeterminate form. Let's rewrite this as the limit of ln one plus one
over x divided by one over x. This is indeed a zero over zero in determinant form. So we
can use lobi tiles rule and take the derivative of the top and the bottom, the derivative
of the top is one over one plus one over x times the derivative of the inside, which
would be negative one over x squared. And the derivative on the bottom, the derivative
of one over x is negative one over x squared, we can actually cancel out these two factors,
and rewrite our limit as the limit as x goes to infinity of one over one plus one over
x, which is just equal to one, since one over x is going to zero. So we found that the limit
of ln y is equal to one, but we're really interested in finding the limit of y, which
we can think of as e to the ln y. Since ln y is going to one, e to the ln y must be going
to e to the one. In other words, E. So we found that our original limit is equal to
E. And in fact, you may recognize that this limit is one of the ways of defining IE.
In the previous example, we had a one to the internet Today in determinant form, and we
took logs and use log roles to write that as an infinity times zero and determinant
form. Well, the same thing can be done if we have an infinity to the zero indeterminate
form,
or
a zero to the zero indeterminate form. So one to the infinity, infinity to the zero,
and zero to the zero, are all indeterminate forms that can be handled using lobi toss
rule. In this video, we saw that a zero times infinity indeterminate form could be converted
to a zero over zero, or infinity over infinity indeterminate form by rewriting f of x times
g of x as f of x divided by one over g of x, or as g of x divided by one over f of x.
We also saw how to use lopi talls rule on these three sorts of indeterminate forms by
taking the ln of y, where y is our f of x to the g of x that we want to take the limit
of. This video is about Newton's method for finding the zeros of a function, f of x. In
other words, the values of x that make f of x equal to zero. The zeros of a function can
also be thought of as the x intercepts of its graph. Suppose we want to find a solution
to the equation either the x equals 4x. This equation cannot be solved using standard algebraic
methods. For example, taking the ln of both sides doesn't really help because we still
get x equals ln of 4x, which is just as hard to solve. Instead, we can look for approximate
solutions. Looking at the graph of y equals e to the x and y equals 4x, we see there should
be two solutions, one at approximately x equals little more than two and the other around
x equals maybe point three or point four. Newton's method will allow us to make much
more accurate approximations to the solution of this equation, then we can do by just glancing
at the graph. To use Newton's method, instead of looking at the equation, e to the x equals
4x. We'll look at the equation E to the X minus 4x equals zero. And in fact, we'll define
the function f of x to be e to the x minus 4x. And look for zeros of that function. After
all, finding a zero of this function is the same as finding a solution to our original
equation. So now we're trying to solve the equivalent problem of finding the zero of
the function f of x equals e to the x minus 4x. That's the function that's drawn below.
I'm going to focus on this zero, the one near to, and I'm just going to make an initial
guess anything reasonably close to the actual zero should do. So I'll just put an initial
guess right here, and I'll call it x one. Now x one is not actually zero of my function,
and I'll write the point on the graph above it as x one, f of x one. To get a better estimate
for the zero of my function, I'm going to make use of the tangent line to my function
that goes through this point. So the second step will be to find this tangent line. Since
the tangent line is a reasonably good approximation to the function, the point where the tangent
line crosses the x axis should be closer to the point where the function itself crosses
the x axis, which is the point I'm looking for. So the third step will be to find the
x intercept for the tangent line. I'll call this x intercept, x two. Now I'm just going
to repeat this process. I'll use x two as my next guess. I'll follow it up to the function
where I have the point x to f of x two, and then I'll draw a new tangent line and get
a new intercept. I can repeat this process as often as I need to, to get a sufficiently
accurate approximation to my actual zero of my function. Now that I've described the process
graphically, let's find some equations that go along with this picture. If I start with
the initial guess, of x one, then the tangent line through x one, f of x one is given by
the x Bayesian y equals f of x one plus f prime of x one times x minus x one.
You might remember this equation from a section on linearization. And it's really just comes
from the formula y minus y one equals m x minus x one that holds for any line, where
m here is the derivative at x one, and y one is f of x one. plugging into that equation,
we have y minus f of x one equals f prime of x one times x minus x one, which simplifies
to y equals f of x one plus f prime of x one times x minus x one. So that's where this
linearization equation comes from. It's just the equation of the tangent line. Now, if
we want to find the x intercept of the tangent line,
we just said
the tangent line equation equal to zeros, we have zero equals f of x one plus f prime
of x one times x minus x one, and we solve for x. so this can subtract f of x one from
each side, divided by f prime of x one and solve for x. We're calling this new x intercept
x two. So x two is x one minus f of x one over f prime of x one. Now we have our second
guess, x two, and we can again find the tangent line through x two, f of x two, that tangent
line will be given by the same sort of equation. And if we then find the x intercept, the same
algebraic steps, get us to the analogous equation x three equals x two minus f of x two over
f prime of x two. And more generally, as we repeat this process over and over again, our
n plus one guess is going to be given by x n plus one equals x n minus f of x n over
f prime of x n. That's the J equation at the core of Newton's method. Now that we've got
the theory down, let's grind through the problem at hand with some numbers. Our function has
the equation f of x equals e to the x minus 4x. So f prime of x is e to the x minus four.
So from Newton's methods equation, we have in general, X sub n plus one is X sub n minus
e to the x sub n minus four times X sub n over e to the x sub n minus four. Let's start
with for example, x sub one equals three, then x sub two is going to be three minus
e cubed minus four times three over e cubed minus four. plugging this into a calculator,
I get x sub two equals 2.4973, and so on. Now to cube compute x sub three, I have to
take this whole number and plug that in to my formula. I've written it out as just 2.49.
But for accuracy, when I actually computed my calculator, I'll use the entire number.
my calculator gives me this answer for x of three and continue in this process, I can
get x of 4x 5x sub six. If I compute one more, x of seven, I noticed that I have no change
to my value in the number of digits that the calculator spits out. So at this point, my
Newton's methods iterations have converged. And I have an answer that's accurate to about
eight decimal places. I found one zero for my function. And if I wanted to find the second
zero, the one over here, I would just need to start with an initial value that's close
to this x coordinate, perhaps an initial value of zero might be good. In this video, we developed
an algorithm for getting increasingly accurate approximations to the zero of a function.
The central equation that we used was this one which tells us how to get from one approximation,
X sub n to the next 1x sub n plus one. When we go from a function, say 3x plus sine x
to its derivative, in this case, three plus cosine x, that's called differentiating, or
finding a derivative. Anti differentiating, or finding an antiderivative, takes us the
other direction, from a derivative to a function that has that as its derivative.
For example, if g prime of x is 3x squared, that's the derivative. What could g of x the
original function be? Well, g of x could be x cubed, since the derivative of x cubed is
3x squared. Or it could also be g of x equals x cubed plus seven, for example, or g of x
equals x cubed plus any constant, where I write a general constant with a capital C.
That's because the derivative of a constant zero, so the derivative of x cubed plus a
constant is just going to be 3x squared, no matter what the constant is. A function capital
F of X is called an antiderivative of lowercase F of X on an interval a, b, if the derivative,
capital F prime of X is equal to lowercase F of X on that interval a b. In other words,
we can think of little f as being the derivative of the function capital F. In the above example,
x cubed is an antiderivative of 3x squared. And in fact, x cubed plus C for any constant
C is also an antiderivative of 3x squared. When we add on a general constancy, that's
sometimes referred to as a general antiderivative, we found a general family of anti derivatives
for the function 3x squared. But could there be other anti derivatives, other functions
whose derivative is 3x squared. In fact, there are no others. And one way to think about
this intuitively, is if you have two functions with the same derivative, it's like having
two runners in a race that always speed up and slow down at exactly the same times. If
one of those runners starts ahead of the other, then the distance between them will always
stay exactly the same. That's the vertical distance drawn here on the graph. And that's
the constant C, that separates one antiderivative y equals x cubed from another y equals x cubed
plus C. And in general, if capital F of X is an antiderivative for a little f of x,
then all other anti derivatives can be written in the form capital F of x plus C for some
constancy. A more rigorous justification of this fact, can be proved using the mean value
theorem, as I'll do in a separate video. If you know the derivatives for some standard
functions, then it's pretty easy to get some anti derivatives. For example, the antiderivative
of one is x. Since the derivative of x is one, if we want to make that a general antiderivative,
we can add a constant C, the antiderivative of x is x squared over two because when I
take the derivative of x squared over two, the two that I pulled down and multiply cancels
with the two in the denominator, leaving me x. Again, I can make this a more general antiderivative
by adding a constant C. More generally, the antiderivative of x to the n for any and that's
not equal to negative one is given by x to the n plus one divided by n plus one plus
a constant C. I can check this by taking the derivative of x to the n plus one over n plus
one. The n here is just a constant. So using the power rule, I get n plus one times x to
the n divided by n plus one that yields x to the n, which is what I want it. We can
think of this rule as the power rule for anti differentiating since it's closely related
to the power rule for differentiating. Now, this rule doesn't apply when n equals negative
one. Notice that we'd be dividing by zero if n were negative one but we can handle the
case when n equals negative one separately, since x to the negative one is one of our
x, we recognize that the antiderivative of one of our x is just ln of the absolute value
of x plus C. Since the derivative of ln of the absolute value of x is one over x, please
pause the video and see how many more anti derivatives you can fill in in this table.
You should get all of these formulas based on the analogous formulas for differentiating,
notice that the antiderivative of sine x is negative cosine x not cosine x, because the
derivative of cosine x is negative sine x. If I have a constant times x to the n, I'm
going to call the constant a instead of C, since I've already got some C's floating around.
If I want the antiderivative of A times x to the n, that's just going to be a times
the antiderivative of x to the n, which is x to the n plus one over n plus one plus a
constant say that's because when I take the derivative of a constant times a function,
I can just pull the constant out. More generally, the antiderivative of a constant A times any
function, little f of x is just going to equal A times the antiderivative of little f of
x, which I'll denote with capital F of X, plus a constant C. The antiderivative of f
of x plus g of x is capital F of X, plus capital G of x plus c, where capital F and capital
G are the antiderivative of lowercase F and lowercase J. This is because the derivative
of a sum is equal to the sum of the derivatives. Let's use this information to compute the
antiderivative for f of x equals five over one plus x squared minus one over two times
the square root of x. First, I'm going to rewrite f of x as five times one over one
plus x squared minus one half times x to the minus one half. I know that the antiderivative
of one over one plus x squared is arctangent of x. And by the power rule for anti differentiating
the antiderivative of x to the minus one half, I get by raising the exponent by one, negative
one half plus one is positive one half, and then dividing by the new exponent by my constant
multiplication rules, I can just multiply by my constants. And that's my antiderivative
capital F of X, I've have to remember the plus C for the general antiderivative, I can
simplify a little bit by canceling these one halves. And I get a final answer of five times
arc tan of x minus a squared of x plus C. In this video, we introduced anti derivatives
and build a table of anti derivatives based on our knowledge of derivatives. In this video,
we'll solve problems where we're given an equation for the derivative of the function.
And we're given an initial condition, something like f of one equals seven. And we have to
find the function f of x. In this first example, suppose g prime of x is e to the x minus three
times sine x. and g of two pi is five, we need to find g of x. Well, g of x is an antiderivative
of e to the x minus three sine x. So g of x is of the form e to the x plus three cosine
of x plus a constant C. That's because the derivative of e to the x is e to the x, the
derivative of cosine x is minus sine x, and the derivative of a constant is just zero.
Now I need to find the value of the constant C that makes this initial condition come true.
If I plug in two pi for x, I get e to the two pi plus three times cosine of two pi plus
C, and that needs to equal five. Since cosine of two pi is one, I have that e to the two
pi plus three plus c equals five. And so C is equal to two minus e to the two pi. So
my function g of x is equal to either the x plus three cosine x Plus two minus e to
the two pi. In this example, we're given the second derivative of f. And we're given two
initial conditions, f of one is zero, and f of zero is two.
To start, I'm going to rewrite f double prime of x in a more manageable form. by distributing,
I get the square root of x times x minus the square root of x over x. and rewriting with
exponents, we get x to the three halves minus x to the minus one half. Next, I'm going to
find f prime of x, which is the antiderivative of f double prime of x. So f prime of x, using
the power rule for anti derivatives, I raised the exponent of three halves by one to get
five halves, and then divide by the new exponent five halves. Similarly, I raised negative
one half by one to get one half and divide by the new exponent one half. And I'll add
on a constant see, let me simplify a little here, instead of dividing by five halves,
multiply by two fifths, and instead of dividing by one half, multiply by two. Now I've got
an expression for f prime of x, but I need an expression for f of x, which is the antiderivative
of f prime, and so all anti differentiate again. So now I have two fifths times x to
the seven halves over seven halves minus two times x to the three halves over three halves,
the antiderivative of a constant C is C times x, and then I'll add on a new constant D.
After simplifying a bit, I'm ready to use my initial conditions, in order to solve for
my constant C and D. When I plug in zero for x, all my x terms vanish, I'm just left with
D, so d has to equal two. So I can rewrite my function, setting D equal to two. And now
my second condition says that f of one equals zero. So plugging in one for x, I get 430
fifths minus four thirds plus c plus two, and that has to equal zero, which means that
c is negative two, minus 430 fifths, plus four thirds, which simplifies to minus 80
to 100, and fifths. If we plug that in for C, we get a final answer for f of x. And that
finishes the problem. In this final example, we're not given any equations, so we have
to make them up ourself, we're told that we're standing at the edge of a cliff, at height
30 meters, with throw a tomato up in the air at an initial velocity of 20 meters per second.
The tomato then falls down to the ground due to gravity. And we want to find how long that
takes and what its velocity is at impact. We know that the acceleration due to gravity
is negative 9.8 meters per second squared. If we're working in metric units. The similar
figure if we're working in units of feet, is negative 32 feet per second squared. The
negative sign is because gravity is pulling objects down towards the ground in the negative
direction. We're also given the initial condition, that velocity at time zero is 20 meters per
second. That's a positive velocity because we're throwing the tomato up. And we're told
that the initial position s of zero is 30 meters. So let's start with the equation that
acceleration is negative 9.8. In other words, S double prime of t is negative 9.8. Therefore,
S prime of t is negative 9.8 t plus a constant c one. And from my initial condition about
velocity, I know that s prime of zero is 20. So in other words, S prime of zero is negative
9.8 times zero plus c one that has to equal 20. Which means that c one is equal to 20.
Substituting in 20 for C one, I can rewrite s prime of t Now I can find f of t, the antiderivative
of S prime. And that's going to be negative 9.8 times t squared over two plus 20 t plus
a second constant C two.
Using my second initial condition, f of zero equals 30. I can plug in zero for t, and get
an expression that equals 30. Since all the terms drop out, besides the C two, that tells
me that C two is 30. And so I can find my equation position by substituting 34 c two.
Now I want to find out how long it takes the tomato to hit the ground. So that's going
to be the time when s of t equals zero. Setting zero equal to my expression for S of t, I
can use the quadratic formula to solve for t, and I get t equals approximately negative
1.17, or 5.25. The negative time doesn't make sense for the problem. So I'm left with a
time of impact of 5.25 seconds. Now to find the velocity of the impact, I need to plug
that time into my velocity equation. In other words, my equation for S prime. So s prime
of 5.25, is 9.8 times 5.25, plus 20, which simplifies to negative 31.45 meters per second,
probably enough to squash the tomato. And that's all for this video on finding anti
derivatives using initial conditions. In this video, I'll use the mean value theorem to
show that the antiderivative zero has to be a constant. And then any two anti derivatives
of the same function have to differ by a constant. In a previous video, I stated the fact that
if f of x is one antiderivative of a function, little f of x, than any other antiderivative
of that same function can be written in the form capital F of x plus C for some constant
C. In other words, any two antiderivative of the same function have to differ by a constant.
To prove this fact, let's first note that if the derivative of a function, g prime of
x is equal to zero on an interval than the original function, g of x must equal C for
some constant C. This statement follows from the mean value theorem, because the mean value
theorem tells us that for any x one and x two in our interval, the average rate of change
between x one and x two is equal to the derivative at some number x three in between x one and
x two. But by assumption, g prime is zero everywhere on the interval, so g prime of
x three must be equal to zero. This means that our numerator, g of x want to minus g
of x one has to equal zero. In other words, g of x two is equal to g of x one, but that's
true for any x one and x two. So all the values of G are equal, and G must be a constant.
The second observation that I want to make is that if G one and G two are two functions,
which have the same derivative, then g one of x must equal g two of x plus C for some
constant C. This statement follows from the previous statement, because if G one prime
of x is equal to g two prime of x, then g one prime of x minus g two prime of x must
equal zero. But that means if I look at the function g one of x minus g two of x, and
take its derivative, that has to equal zero, since the derivative The difference is the
difference of the derivatives. Now our previous statement tells us that if the derivative
of a function is zero, the function must be a constant, and therefore, g one of x minus
g two of x equals C for some constant C, which means that g one of x is equal to g two of
x plus C, which is what we wanted to prove. So we've proved that any two functions with
the same derivative have to differ by a constant or in other words, If capital F of X is one
antiderivative of a function than any other antiderivative must be of the form capital
F of x plus C. This concludes the proof that any two anti derivatives of a particular function
must differ by a constant. This video will review summation notation, that is, the sigma
notation used to write a song.
In this expression, using the greek capital letter sigma, the letter I is called the index.
The number one is called the lower limit of summation, or the starting index. And the
number five is called the upper limit of summation, or the ending index, we evaluate this expression
by summing up to the I, for all values of i starting from one and ending at five and
stepping through the integers. In other words, we start with i equals one and take two to
the one. And then we add to it two to the two, two to the Three, two to the four, and
two to the five. If we do the arithmetic, this comes out to 62. In the second expression,
our index is J, and we start with J equals three and go up to J equals seven, once again,
stepping through integer values. So we have to add up 1/3 plus 1/4, plus 1/5, plus one,
six, and then our last term is one seven. This sum is equal to 153, over 140. When we're
given a psalm like this one, it can be handy to write it in sigma notation, because it's
more compact. But to do so, we have to look for the pattern between the terms. In this
case, the terms all differ by three. So I can think of nine as being six plus three,
and 12, as being six plus twice, three, and 15 as being six plus three times three, and
so on. In fact, we can even think of six itself as being six plus zero times three to fit
this pattern. Now we can write the sum as sigma of six plus i times three, where I ranges
from zero to four. Here, we're thinking of six plus three as six plus one times three.
Now, there are other ways to write this sum and sigma notation. For example, we could
notice that each of the terms is a multiple of three. And in fact, six is three times
two, nine is three times three, and so on. And so we could write our sum as sigma of
three times n, where n ranges from two to six. The choice of the letter we use for the
index doesn't matter at all. For example, we could also write this as sigma from K equals
two to six of three times k. Here, k and n play the same role. Please pause the video
for a moment and try to write this next example in sigma notation. Since the denominators
are powers of two, we could write the denominators as two to the i, where i ranges from two to
five. The numerators are one less than the denominators, so we can write the numerators
as two to the i minus one was we're adding these terms up, we write sigma, and we I go
from two to five. In this video, we reviewed summation notation, or sigma notation for
writing sums. In this video, we'll approximate the area under a curve using tall skinny rectangles,
this will introduce the idea of an integral. Let's start by approximating the area under
this curve y equals x squared in between x equals zero, and x equals three by approximating
it with six rectangles.
There's several ways to do this. For example, I could draw rectangles for the right side
of each rectangle is as tall as the curve. We'll call this using write endpoints. Alternatively,
we could line up our rectangles, so the left side of each rectangle is as high as the curve.
We'll call this use Left endpoints. Notice that the leftmost rectangle in this picture
is degenerate and has height zero. If we're using the picture with the right endpoints,
then the base of each rectangle has size one half. And the height of each rectangle is
given by the value of our function, y equals x squared on the right side of the rectangle.
So for example, the area of the first rectangle is its base times its height, which is point
five times 0.5 squared. The 0.5 squared comes from me evaluating the function at this point
0.5 to get the height. Similarly, the area of the second rectangle is going to be also
base times height basis, still point five, and now the height is going to be one squared,
or one. If we continue like this, and add up all our areas, we get the area of all six
rectangles is given by this expression. Notice that there are six terms here, one for each
rectangle, each rectangle has base of point five, and has height given by the right endpoint,
we can write this in sigma notation as the sum of 0.5 times 0.5 times i squared, where
I ranges from one to six. This works because the numbers in parentheses here are all multiples
of point five. The first one is point five times one, and the last one, three is point
five times six. Now if we compute the sum, we get 91 eighths, which is 11.375. Notice
from the picture, that the sum of areas of rectangles is an over estimate for the area
under the curve, we can do the same sort of computation for this green picture using left
endpoints, and we'll get an under estimate for the area under the curve, I invite you
to try it for yourself before going on with the video. For the green rectangles, the first
rectangle has area zero, the second rectangle has area given by its base of 0.5 times its
height of 0.5 squared. And if we compute all six areas and add them up, we get a similar
expression to the previous one, only this time, we end with a 2.5 squared, which is
the height of our last rectangle. One way to write this in sigma notation, is still
starting with i equals one for the first rectangle to six for the last one, we use the base.
And then for the height, we use 0.5 times I minus one,
squared.
This works because when I is one, i minus one is zero, so we start with a height of
zero like we should. And when I is six, we get point five times six minus one, which
is point five times five, or 2.5, just like we want it to be, if we add up the sum, we
get an answer of 55 eighths, which is equal to 6.875, which is an underestimate for the
true area under the curve. Now there's a big gap between 6.875 and 11.375. So it'd be nice
to get tighter bounds on the area. One way to do this is by using more rectangles, for
example, 12 rectangles instead of six. Again, we could choose to use right endpoints, which
gives us an over estimate of area in this case, or we could use left endpoints, which
gives us an under estimate. The area of the eyes rectangle is given by the base times
the height, and the base is going to be in this case 0.25. While the height is given
by the functions value on the right endpoint, the right endpoint of the eyes rectangle is
going to be 0.25 times I and the function is the squaring function. So that height will
be given by point two, five i squared. The area of all the rectangles can then be given
by the sum from i equals one to 12 for the 12 rectangles of 0.25 times 0.25 i
squared.
If we work out that sum, it comes to 10.156. Again, we can do the same thing with left
endpoints. Now the area of the eye, the rectangle, is given by base times height, which is point
two, five. And now the height is given by the value of the function on the left endpoint.
So that's going to be 0.25 times I minus one. And we need to square that, since our function
y equals x squared is giving the height of my rectangles. So the area of all the rectangles
together is going to be the sum from the first rectangle to the 12th rectangle of 0.25 times
0.25, times I minus one
squared.
That works out to 7.906. So we're honing in on the actual area under the curve. Now it's
somewhere between about eight and about 10, we can keep getting better and better estimates
of area by using more and more rectangles. For example, if we want to use 100 rectangles,
then our area of all rectangles using right endpoints is going to be given by the sum
from i equals one to 100 of the basis times the heights. Now the base of each rectangle
will be 100 of the length here from zero to three. In other words, the base will be three
over 100. The eyes right endpoint, which I'll call X sub i, is going to be just three over
100 times I, since you get to the right eye threat endpoint by taking I copies of a rectangle
of width three one hundredths. Therefore, the eyes height is going to be given by this
is right endpoint squared, or three one hundredths times i squared. So we can write our sum of
areas as sigma from i equals one to 100 of three one hundredths times three one hundredths
times i squared. The formula using left endpoints is similar. The if left endpoint is going
to be three one hundredths, times I minus one, since if we're using left endpoints,
we only have to travel through i minus one rectangles to get to the left endpoint of
the rectangle. So our area for the left endpoints becomes the sum from i equals one to 100 of
three over 100 times three over 100 times I minus one squared. These two psalms work
out to be 9.1435 and 8.8654. At this point, you might be willing to wager a guess that
our exact area under the curve is probably going to be nine. But to determine the exact
area for sure, let's do this process of dividing into rectangles one more time. And this time,
we'll just use an rectangles where n is some big number. Since we're dividing an interval
of length three into n little pieces, the width of each sub interval, in other words,
the base of each rectangle, we'll be given by three over n. I'll call this delta x as
a little tiny bit of x. Now the right endpoint, x by is given by three over n times i. Since
we have to travel through I rectangles, each of width three over n on order to get to that
right endpoint. So our height, H sub i is given by the functions value on that right
endpoint. We can work out similar expressions for the picture using left endpoints here.
Our estimate of area using right endpoints is then the sum from i equals one to n of
three over n times three I over n squared. And our estimate using left endpoints is a
sum from i equals one to n of three over n times three, i minus one over n squared, the
more rectangles we use, in other words, the bigger the value of n, the closer our estimated
area will be to our exact area under the curve. And therefore, the exact area is given by
the limit the limit as n goes to infinity of this song, which is known as a Riemann
sum,
there are really two possible limits, we could use right endpoints or left endpoints. But
as the picture suggests, these two limits should turn out to be the same thing. In fact,
there are other options between sides using right endpoints and left endpoints, we could,
for example, use the midpoints of our intervals to compute our areas of rectangles. And that
limit should also end up as the same thing. So we have an expression for the area under
the curve y equals x squared
between the
values of x equals one and x equals three. And that's given by the limit of this psalm
called a Riemann sum. I'll stick with the right endpoint version for now, to compute
the exact area, we have to evaluate this limit, which is tricky. I'm going to start by rewriting.
Since three and n don't involve the index I, I can pull them outside of this summation
sign. I'll clean this up a little bit. Now will you need to use the fact that the sum
of the first n squares of the integers is equal to n times n plus one times two n plus
one
over six,
we can check that formula for a few values of n. For example, if n equals two, we're
summing out one squared plus two squared, which is five. And we're plugging in two times
two plus one times four plus one over six from the formula, which also equals five.
If we use this formula, in our limit calculation, we get this expression, which simplifies to
nine halves, by dividing 27 by six, we can cancel a copy of n and get n plus one times
two n plus one over n squared. I'm going to pull out the nine halves. And now I observe
I have the limit of a rational expression, where the highest power term on the numerator
is going to be two n squared, the highest power and there's nominators just the n squared,
so that's going to be a limit of two, multiply that by by nine halves, and I get a limit
of nine, just like I expected from the previous work. So that was a big production. But we
did successfully find the area under the curve, and it was nine. In this video, we approximated
the area under a curve by taking the limit as the number of rectangles goes to infinity
of the area of the rectangles, which is given by a psalm called a Riemann sum of the basis
times the heights of the rectangles. The basis are often written as delta x, and the heights
are given by the function value on the left endpoint, or the right endpoint, or some other
point in the interval. For our purposes, f was always x squared. But this sort of expression,
called a Riemann sum, can be used more generally, to evaluate the area under any continuous
function. In previous sections, we thought of the definite integral as representing area,
and we've computed it as a number. In this section, we'll think of the integral itself
as a function of the bounds of integration. And we'll describe the first part of the fundamental
theorem relating the derivative and the integral. Suppose f of x has the graph shown here, and
let g of x be the integral from one to x of f of t dt. I'm using t as my variable inside
my integrand Here, just to distinguish it from the variable x that I'm using in my bounds
of integration, this expression just means the net area between one and some value x
on the x axis. I'll call geovax, the accumulated area function, because as x increases, g of
x, measures how much net area has accumulated. Let's calculate and plot some values of g
of x. g of one is the integral from one to one of f of t, dt, that's just zero. Since
the bounds of integration here the same, g of two is the integral from one to two of
f of t, dt. That's the net area from one to two, which is to square units. g of three
is the integral from one to three. Now, we've added on an additional two units here, and
an additional one unit up here from this triangle, for a total of five, g of four is g of three
with some additional area tacked on the additional
area measures three units. So g of four is eight. Please pause the video and fill in
the next few values of J.
When we go from g of four to g of five, we add on an extra unit of area. So g of five
is nine. As we go from g of five to g of six, we start accumulating negative area, because
f is now below the x axis. So here I've accumulated one unit of negative area, which means that
g of six is one less than g of five. In other words, gf six is eight, g of seven is five.
Since we accumulate three more units of negative area, to find g of zero, the integral from
one to zero of f of t dt, I'm going to rewrite this integral as negative the integral from
zero to one of f of t dt. Since there are two units of area between zero and one, g
of zero is negative to apply all these values of G on these coordinate axes, and connect
the dots to get an idea of what g of x looks like. Now let's think about the derivative
g prime of x. We know that g prime of x is positive, where g of x is increasing, but
g of x is increasing, wherever we're adding on positive area, that is when f of x is positive.
So we have that g prime of x is positive, where f is x is positive.
Also,
g prime of x is negative, where g of x is decreasing. That happens when we're adding
on negative area because f of x is negative. So we can see that g prime of x is negative,
where f of x is negative. Also, g prime is zero at this local maximum, where f is zero.
At that instant, we're not adding on any positive or negative
area.
If we look a little closer, we can see the rate at which g of x is increasing depends
on the height of f of x. When f of x is tall, or high, we're adding on area very quickly.
While when f of x is low or small, we're adding on area more slowly. So the rate of change
of G. In other words, g prime of x is behaving very much like the function f of x itself.
And in fact, it turns out that g prime of x is equal to f of x. This is the first part
of the fundamental theorem of calculus.
The Fundamental Theorem of Calculus Part One says that of f of x is a continuous function
on the closed interval from a to b, then for any x in this interval, the function g of
x, the integral from a to x of f of t dt is continuous on the interval a b and differentiable
on the inside of this interval, and for Furthermore, g prime of x is equal to f of x, as we saw
in the previous example. The proof of this fact relies on a limit definition of derivative,
and can be found in a later video. For now, let's do some examples based on this fact.
First, let's find the derivative with respect to x of the integral from five to x of the
square root of t squared plus three dt. The Fundamental Theorem of Calculus tells us that
this expression here thought of as a function of x is differentiable, and its derivative
is just the integrand function evaluated on x. This is great, we don't have to do any
work here at all. To evaluate the derivative, we just plug in x, where we see the T here,
the derivative and the second expression is also the square root of x squared plus three,
it might seem odd that these two expressions have the same derivative. But remember, in
both cases, we're taking the derivative of the accumulated area function. And the rate
at which area accumulates doesn't depend on the lower bound of the integral. That is,
it doesn't depend on where we start counting, it just depends on the height of the function
at x.
For this third example, remember that the integral from x to four is the same thing
as the negative of the integral from four to x. So we get the negative of the derivative
from four to x. and applying the fundamental theorem of calculus, this lesson is negative
the square root of x squared plus three, it makes sense that we should get a negative
answer for this example. When we're integrating from X to four, then as x increases, our area
actually decreases. So our accumulated area function should have a negative derivative.
This last example is more complicated, because instead of just having x as our upper bound,
we have a function of x sine of x, we can think of sine of x as being the inside function,
and the accumulated area function as being an outside function and apply the chain rule.
In general, the chain rule says that we have the derivative with respect to x of a function
of u of x, then that's the same thing as the derivative with respect to u of that function
at U times the derivative of the inside function u of x with respect to x, applying the chain
rule to our accumulated area function, where you have x is sine x, we have that the derivative
respect to x of the integral from four to sine x of the integral of t squared plus three
DT can be written as the derivative by two you, I've accumulated area function from four
to you have the integrand times the derivative with respect to x of our u of x, which is
just sine x, we can apply the fundamental theorem of calculus to calculate the first
derivative. By just plugging in you for T, we get you squared plus three. And then the
derivative of sine x, of course, is just cosine of x. Since we want our final answer to be
entirely in terms of x, we're going to rewrite this as the square root of sine x squared
plus three times cosine x, or just the square root of sine squared x plus three times cosine
of x. We could have gotten this answer more quickly by just plugging in this entire expression
sine of x in where we saw the T here in the integrand, and then multiplying the answer
by the derivative of sine x due to the chain rule.
This video introduced the fundamental theorem of calculus part one that says that the derivative
of the integral of a function is just the original function in some sense, taking the
derivative and does the process of taking the integral. derivatives and integrals are
closely related. inverse operations. This video introduces the second part of the fundamental
theorem of calculus. Another way of relating derivatives and integrals. Part Two of the
fundamental theorem of calculus says that if f is a continuous function on the closed
interval a b, then the integral from a to b of f of x dx is equal to capital F of b
minus capital F of A, where capital F is any antiderivative for lowercase F. That is, capital
F is a function whose derivative is lowercase F. The proof of part two of the fundamental
theorem of calculus follows directly from part one. And I'll give that proof in another
video. But here, I just want to make a few comments about what this theorem means. If
we think of f of x as the derivative of capital F of x, then this is saying that the integral
of the derivative is equal to the original function evaluated on the endpoints. I also
want to comment on the phrasing any antiderivative. Suppose capital G of X is a different antiderivative
for lowercase F. We know that any two anti derivatives differ by a constant. So we know
that g of x has to equal capital F of X plus some constant. So if we take g of b minus
g of a, that's going to be the same thing as f of b plus c minus f of a plus C. And
since this constant sees subtract out to cancel here, this is just f of b minus f of a. So
this difference is the same value, no matter which antiderivative of lowercase f we use.
And that's why we can say that capital F can be any antiderivative. Part Two of the fundamental
theorem of calculus is super useful, because it allows us to compute integrals simply by
finding anti derivatives and evaluating them. Finding anti derivatives tends to be really
easy. Computing integrals, using the Riemann sum definition is really hard. And so because
of the fundamental theorem of calculus, we don't have to go through all those lengthy
and tedious computations, we've involving limits of areas of rectangles, all we have
to do to evaluate an integral is find an antiderivative, and evaluate it. Let's see how this works
in some examples. In this first example, the antiderivative of 3x squared is x cubed. And
the antiderivative of negative four over x is minus four times ln absolute value of x.
We could add a plus C to make it a general antiderivative, but we don't really need it.
The fundamental theorem says that we can use any antiderivative, so we might as well use
the simplest one, where c equals zero. Now we need to evaluate this antiderivative on
the endpoints of negative one and negative five. And we usually write this as a vertical
line with a negative one at the bottom and a negative five at the top to mean evaluation.
In other words, the notation capital F
of X
between A and B means capital F of b minus capital F of A, which is what we need to compute
for our antiderivative here. So now we just plug in negative five for x, and then we subtract
what we get when we plug in negative one for x. In this example, you can see why it's important
to write the antiderivative of one over x as ln absolute value of x not just ln of x,
because ln of the absolute value of five, which is ln a five actually has an answer,
whereas ln of negative five would not exist. I can simplify this expression a little bit,
I get negative 125 minus four ln five minus negative one, plus four ln of one. Since ln
of one is zero, this becomes negative 124 minus four ln five. That's about negative
130 point 438. In this next example, we need to find the antiderivative for this expression
y squared minus y plus one over the square root Why. Now we can't take the antiderivative
separately of the numerator and the denominator, because that's just not how the quotient rule
works for differentiation. So it doesn't work that way for anti differentiation either.
Instead, let's try to simplify this expression to make it look more like something we can
take the antiderivative of. So I'm going to rewrite the denominator as y to the one half
power. And dividing by y to the one half is the same thing as multiplying by y to the
negative one half distributed, distributing and adding exponents, I get y to the three
halves minus y to the one half plus y to the negative one half. Now that's something I
can take the antiderivative of just using the power rule and reverse y to the three
halves, becomes y to the five halves by adding one to the exponent, I divide by the new exponent.
Now here, I get y to the three halves divided by three halves, and here are negative one
half plus one is positive one half, I need to evaluate this between four and one. let
me simplify a little bit. And now I'll substitute in values. Now four to the five halves is
the same thing as four to the one half raised to the fifth power. So that's two to the fifth
power, or 32. Similarly, for the three halves is four to the one half cubed, so that's two
cubed, or eight, and four to the one half is just two. And one to any power is just
one. And after some arithmetic, I get an answer of 146 15th.
The Fundamental Theorem of Calculus, part two can be stated this way, for a function
capital F with continuous derivative, the integral of the derivative is equal to the
original function evaluated on the bounds of integration. In this video, I'll prove
both parts of the fundamental theorem of calculus. The first part of the fundamental theorem
of calculus says that if f of x is a continuous function, then the function g of x defined
as the integral from a constant A to the variable x of f of t dt is differentiable, and has
derivative equal to the original function,
f of x.
To prove this theorem, let's start with the limit definition of derivative. The derivative,
g prime of x, by definition, is the limit as h goes to zero of g of x plus h minus g
of x over h. Now g of x is defined as an integral from a to x. So g of x plus h is going to
be the integral from a to x plus h, just plugging in x plus h for x of f of t dt. g of x is
the integral from a to x
of f of t dt.
By properties of integrals. The integral from a to x plus h minus the integral from a to
x is just the integral from x to x plus h. Now informally, the integral from x to x plus
h can be closely approximated by a skinny rectangle with height, f of x and width, H.
And so this limit is approximately the limit as h goes to zero of f of x times h over h,
which is just f of x. But let's make this argument a little more precise. Let's let
capital M be the maximum value that f of x achieves on this little sub interval, and
lowercase m be the minimum value achieved. In this picture, they occur on the endpoints
of the interval from x to x plus h, but they could also occur somewhere in the interior.
But we know that f of x does have to have a minimum value and a maximum value, since
it's a continuous function by assumption on a closed interval. Now we know that the integral
of f of t dt from x to x plus h has to be less than or equal to capital M times h and
bigger than a to lowercase m times h. This is one of the properties of integrals, and
can be verified visually by comparing this shaded red area to the small blue rectangle,
which has area lowercase m times h, and comparing it to the area of the big rectangle, which
has area capital M times h equivalently, the integral from x to x plus H of F of t dt divided
by h has to be less than or equal to capital M and greater than or equal to little m. But
the intermediate value theorem, which holds for all continuous functions, says that this
intermediate value that lies between the minimum and maximum value of f has to be achieved
as f of c for some C in the interval. Therefore, I cannot replace this integral in the limit
expression above by just simply the value f of c for some c between x and x plus h.
The value of C here depends on x and h. But as h goes to zero, C has to get closer and
closer to x. And since f is continuous, this means that this limit is equal to
f of x.
We've now proved the first part of the fundamental theorem of calculus, that the derivative of
g exists and equals f of x. The second part of the fundamental theorem of calculus says
that if f is continuous, then the integral from a to b of f of x dx is equal to the antiderivative
of lowercase F, which I'll denote by capital F, evaluated at B minus that antiderivative
evaluated today. Part Two of the fundamental theorem follows directly from part one. Let's
let g of x be defined as the integral from a to x of f of t dt. Then part one of the
fundamental theorem of calculus tells us that g prime of x exists and equals lowercase f
of x. In other words, capital G is an antiderivative for a lowercase F. Now g of the minus g of
A is by definition, the integral from a to b of f of t dt minus the integral of a from
a to a of f of t dt. The second integral is zero, since the bounds of integration are
identical. So part two of the fundamental theorem of calculus is true if I use the antiderivative
capital
G.
But the theorem is supposed to be true for any antiderivative. So let's let capital F
be any antiderivative of lowercase F, we know that capital F of x has to equal capital G
of X plus some constant since any two antiderivative for the same function differ by a constant,
and therefore, capital F of b minus capital F of A is going to equal capital G of B plus
C minus capital G of A plus C. The constant C cancels out, and we just get capital G of
b minus capital G of A, which we already saw was equal to the integral from a to b of lowercase
f of t dt. So the left side of this equation is equal to the right side. And the fundamental
theorem of calculus Part two is proved for any antiderivative. This completes the proof
of the fundamental theorem of calculus. This video is about the substitution method for
evaluating integrals, also known as u substitution. As the first example, let's try to integrate
to x sine of x squared dx. Now sine of x squared is the composition of the function sine and
the function x squared. And notice that the function x squared has derivative to x, which
is sitting right here and the integrand. I'm going to make the substitution u equals x
squared, and then I'll write d u is equal to 2x dx. That's differential notation. To
find d u, I take the derivative have X squared and then multiply by the differential dx,
I can then rewrite the integrand as sine of u. And the 2x. dx becomes do after making
this substitution, I can integrate, because the antiderivative of sine of u is negative
cosine of u. And I'll add on the constant of integration. I'm not finished yet, my original
problem was in terms of x, and now I've got a function in terms of u. So let's substitute
back in since u is equal to x squared, I can replace that, and I have my final answer.
To verify that this final answer is correct, that it really is the antiderivative of what
we started with, let's take the derivative of our answer and make sure we get back the
function to x sine of x
squared.
If we take the derivative of negative cosine of x squared plus C, then we get the derivative
of a constant is zero, so we have the derivative of negative cosine, that's equal to sine of
the inside function x squared times the derivative of the inside function using the chain rule.
And we do in fact, get back to the integrand that we started with. Notice that we use the
chain rule when taking the derivative to check our answer. Let's try some more examples of
use substitution. When looking for what to substitute as you, it's good to look for a
chunk that's in the integrand, whose derivative is also in the integrand. It's also good enough
to just have a constant multiple of the derivative in the integrand. So in the first example,
we might use the chunk one plus 3x squared, as are you. The derivative of that expression
is six times x. And even though six times x isn't completely in the integrand, we do
have a factor of x in the numerator, that's just a constant multiple away from the derivative
of 6x. So let's write out d u, that's going to be 6x dx. And I'm going to go ahead and
rewrite this as x dx is equal to one six d U. writing it this way, it makes it easy to
substitute one six d u for x dx. And then in my denominator, my one plus 3x squared
becomes u. I can rewrite this as one six times the integral of one over u d u you and I recognize
that the antiderivative of one over u is ln absolute value of u. Substituting back in
for you, I get a final answer of one six ln absolute value of one plus 3x squared, plus
say, the absolute value signs are not really necessary in this example, since one plus
3x squared is always positive. As our next example, let's look at the integral of e to
the 7x dx. one chunk with us here is u equals 7x. If we do that, then d u is just seven
dx.
And so we have dx is equal to 1/7 do substituting in we have the integral of e to the u times
1/7 d u, I can pull the 1/7 out and integrate e to the u to the app, just either the U and
substituting in for back for 7x, I get e to the 7x plus C. I encourage you to pause the
video to check that these two answers are correct. By taking derivatives. You'll notice
that you use the chain rule each time. Next, let's do an example with a definite integral,
the integral from E to E squared of ln x over x dx. If we set u equal to ln x, then d u
is the derivative of ln x, that's one of our x times dx. This is a much better choice of
you than say setting u equal to x from the denominator, because then d u would just be
dx. And when we did the substitution, nothing would really change. For definite integrals,
we need to deal with the bounds of integration here e and d squared. We really have two options,
worry about them now or worry about them later. I'll show you the worry about them. Now method
first. Our bounds of integration E and E squared are values of Bax as we convert everything
in our integral from x to you, we need to convert the bounds of integration from values
of x to values of u also. Now, when x is equal to e, u is equal to ln IV, which is one, just
using this equation. Similarly, when x is equal to E squared, u is equal to ln t squared,
which is two. So as I rewrite my integral, I'm going to replace the bounds with one and
two. And now my ln x becomes my u and my dx divided by x becomes my do, they're going
to grow of UD u is equal to use squared over two, and I evaluate this between the bounds
of u equals two and u equals one to get two squared over two minus one squared over two,
which is one half. Notice that when we did the problem this way, we never actually had
to get back to our variable x, we stayed in the variable u and evaluated. The second way
of dealing with the bounds of integration is to worry about them later. Let's go back
to the beginning of the problem. We're just about to substitute u equals ln x and d u
equals one over x dx. Instead of substituting in for the bounds of integration, I'm going
to temporarily ignore them and just evaluate the indefinite integral ln x over x dx, which
I can substitute in as you times do, we can evaluate that to get you squared over two.
Normally we'd have a plus c constant. But since we're ultimately going to be doing a
definite integral, we don't really need the constant here. Now, just like when we're doing
indefinite integrals, I'm going to get back to the variable x by substituting back in
for you U is ln of x. So I square that and divided by two, and then I can go back to
my original bounds of integration, those bounds are the x values of E squared, and E. Plugging
in those bounds, I get ln of E squared quantity squared over two minus ln of E squared over
two, which evaluates to two squared over two minus one over two, which is again, one half.
This video gave some examples of use substitution to evaluate integrals. This method works great
in examples like this one, where there's a chunk that you can call you whose derivative
or at least a constant multiple of its derivative is also in the integrand.
You've already seen how u substitution works in practice. In this video, I'll try to explain
why it works. u substitution is based on the chain rule. Recall the chain rule says if
we take the derivative of a function, capital F of lowercase g of x with respect to x, we
get the derivative of capital F evaluated on the inside function g of x times the derivative
of g of x.
If we just write that equation in the opposite order, we have that an expression of the form
f prime of g of x times g prime of x can all be wrapped up as the derivative of a composite
function, f of g of x. Now if I take the integral of both sides of this equation with respect
to x, on the right side, I'm taking the integral
of a derivative.
Well, the integral or antiderivative of a derivative is just the original function,
capital f of g of x plus C. Now when we do use substitution, we're really just writing
this equation down. We are seeing an expression of the form f prime of g of x times g prime
of x dx, we're recognizing you as g of x and d u as g prime of x dx. So we're rewriting
this expression as the integral of capital F prime of u d u do and that integrates to
just capital F of u plus C. And then we're substituting back in for you To get capital
f of g of x plus C, the beginning and end of this process are exactly the same as the
left side and right side of our chain rule expression above. So when you're doing u substitution
to integrate, you can thank this chain rule that's behind it all. This video introduces
the idea of an average value of a function. To take the average of a finite list of numbers,
we just add the numbers up and divide by n, the number of numbers. In summation notation,
we write the sum from i equals one to n of Q i all divided by n. But defining the average
value of a continuous function is a little different. Because a function can take on
infinitely many values on an interval from a to b, we could estimate the average value
of the function by sampling it at a finite Li many evenly spaced x values. I'll call
them x one through x n. And let's assume that they're spaced a distance of delta x apart,
then the average value of f at these sample points is just the sum of the values of f
divided by n, the number of values are in summation notation, the sum from i equals
one to n of f of x i all divided by n. This is an approximate average value of f, since
we're just using n sample points. But the approximation gets better as the number of
sample points n gets bigger and bigger. So we could define the average as the limit as
n goes to infinity of the sample average. I'd like to make this look more like a Riemann
sum. So I need to get delta x in there. So I'm just going to multiply the top and the
bottom by delta x. And notice that n times delta x is just the length of the interval
b minus a. Now as the number of sample points goes to infinity, delta x, the distance between
them goes to zero. So I can rewrite my limit as the limit as delta x goes to zero, of the
sum of FX II times delta x divided by b minus a. Now the limit of this Riemann sum in the
numerator is just the integral from a to b of f of x dx. And so the average value of
the function is given by the integral on the interval from a to b divided by the length
of the interval. Notice the similarity between the formula for the average value of a function
and the formula for the average value of a list of numbers, the integral for the function
corresponds to the summation sign for the list of numbers. And the length of the interval
B minus A for the function corresponds to n, the number of numbers in the list of numbers.
Now let's work an example. For the function g of x equals one over one minus 5x. On the
interval from two to five, we know that the average value of G is given by the integral
from two to five of one over one minus 5x dx divided by the length of that interval,
I'm going to use use of the tuition to integrate, so I'm going to set u equal to one minus 5x.
So d u is negative five dx. In other words, dx is negative 1/5 times do. Looking at my
bounds of integration, when x is equal to two, u is equal to
one minus five times two, which is negative nine. And when x is equal to five, u is equal
to negative 24. substituting into my integral, I get the integral from negative nine to negative
24 of one over u times negative 1/5. Do and that's divided by three. Now dividing by three
is same as multiplying by 1/3. And as I integrate, I'm going to pull the negative 1/5 out and
then take the integral of one over u, that's ln of the absolute value of u evaluated in
between negative 24 and negative nine. The absolute value signs are important here because
they prevent me from trying to take the natural log of negative numbers to evaluate
get negative 1/15 times ln of 24 minus ln of nine, I can use my log rolls to simplify
and get negative 1/15 ln of 24 over nine, that's negative 1/15 ln of eight thirds, and
as a decimal, that's approximately negative 0.0654. So I found the average value of G.
Now my next question is, does g ever achieve that average value, in other words, is there
a number c in the interval from two to five for which GFC equals its average value? Well,
one way to find out is just to set GFC equal to G's average value. In other words, set
one over one minus five c equal to negative 1/15 ln of eight thirds, and try to solve
for C. There are lots of ways to solve this equation. But I'm going to take the reciprocal
of both sides, subtract one from both sides and divide by negative five. This simplifies
to three over ln of eight thirds, plus 1/5, which is approximately 3.25. And that x value
does lie inside the interval from two to five. So we've demonstrated that g does achieve
its average value over the interval. But in fact, we could have predicted this to be true.
Gs average value has to lie somewhere between GS minimum value and maximum value on this
interval. And since G is continuous on the interval from two to five, it has to achieve
every value that lies in between as minimum and maximum, including its average value.
The same argument shows that for any continuous function, the function must achieve its average
value on an interval. And this is known as the mean value theorem for integrals. Namely,
for any continuous function f of x on an interval from a to b, there has to be at least one
number c, between A and B, such that f of c equals its average value, or, in symbols,
f of c equals the integral from A b of f of x dx divided by b minus a. This video gave
the definition of an average value of a function, and stated the mean value theorem for integrals.
If we rewrite the formula for average value a little, then we can see a geometric interpretation
for average value, the area of the box with height the average value is the same as the
area under the curve. This video gives two proofs of the mean value theorem for integrals.
the mean value theorem for integrals says the for continuous function f of x, defined
on an interval from a to b, there's some number c between A and B, such that f of c is equal
to the average value of f. The first proof that I'm going to give us is the intermediate
value theorem. Recall that the intermediate value theorem says that if we have a continuous
function f, defined on an interval, which I'll call x 1x, two, if we have some number
l in between f of x one and f of x two, then f has to achieve the value out somewhere between
x one and x two. Keeping in mind the intermediate value theorem, let's turn our attention back
to the mean value theorem for integrals. Now it's possible that our function f of x might
be constant on the interval from a to b. But if that's true, then our mean value theorem
for integrals holds easily, because f AV is just equal to that constant, which is equal
to f OC for any c between A and B. So let's assume that f is not constant. Well, like
continuous function on a closed interval has to have a minimum value and a maximum value,
which I'll call little m, and big M. Now we know that F's average value on the interval
has to be between its maximum value and its minimum value. If you don't believe this,
consider the fact that all of us values on the interval have to lie between big M and
little m. And if we integrate this inequality
We get little m times b minus a is less than or equal to the integral of f is less than
or equal to big M times b minus a. Notice that the first and the last integrals, we're
just integrating a constant. Now if I divide all three sides by b minus a, I can see that
little m is less than or equal to the average value of f is less than or equal to big M
as I wanted. Now, I just need to apply the intermediate value theorem, with F average
as my number L and little m and big M as my values of f of x one and f of x two. The intermediate
value theorem says that F average is achieved by f of c for some C in between my x one and
x two. And therefore, for some C in my interval a b. And that proves the mean value theorem
for integrals. Now I'm going to give a second proof for the mean value theorem for integrals.
And this time, it's going to be as a corollary to the regular mean value theorem for
functions. Recall that the mean value theorem for functions, says that if g of x is continuous
on a closed interval,
and differentiable on the interior of that interval, then there's some number c in the
interval, such that the derivative of g at C is equal to the average rate of change of
G, across the whole interval from a to b. Let's keep the mean value theorem for functions
in mind, and turn our attention back to the mean value theorem for integrals. I'm going
to define a function g of x to be the integral from a to x of f of t dt, where F is the function
given to us in the statement of the mean value theorem for integrals. Notice that g of A
is just the integral from a to a, which is zero, while g of B is the integral from a
to b of our function. Now, by the fundamental theorem of calculus, our function g of x is
continuous and differentiable on the interval a, b, and g prime of x is equal to f of x.
And by the mean value theorem for functions, we know that g prime of c has to equal g of
b minus g of a over b minus a, for some numbers, C and the interval a b, if we substitute in
the three facts above, into our equation below, we get f of c is equal to the integral from
a to b of f of t dt minus zero over b minus a, which is exactly the conclusion that we
wanted to reach. This shows that the mean value theorem for integrals really is the
mean value theorem for functions where our function is an integral. And this completes
the second proof of the main value theorem for integrals. So now I've proved the mean
value theorem for integrals in two different ways. And I've used a lot of the great theorems
of calculus along the way.