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10 класс, 40 урок, Определение производной

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hello the topic of this video lesson is

determining the derivative let's

look at the graph of some function

y equal to f from x

here we have the x axis y axis let's

take the value of this function at

some point x the

corresponding value of the function is f

from x let's give the argument x an increment we get the

point x plus delta x and the value of the

corresponding function will be f from x

plus delta x

now if we gradually

decrease this increment that is. x

+ 2x we will shift here all the time

then the correspondence why the function

will also

tend to f from x consider

the increment of the function delta y

when rotating the function is the difference between the

values of the function at the point x plus delta x

and f from x

now consider this limit

we will tend delta x to zero that is,

gradually decrease it and

find the ratio of the increment of the function the increment of the argument

argument

so according to the definition of the increment of the function the ratio of the

increment of the function to the

increment of the argument when this

increment tends to zero is called the

derivative of the function and is denoted as y prime

prime

let's consider an example let us

have a function y from x is equal to and x the graphs

as we know this is just a straight line let's

burn the argument x some

transformation x plus delta x and find the

value of the function at the point x plus delta x

that is, here instead of x we must

put x plus delta x this will be x

plus delta x from here we can find the increment the

increment the

increment of the function is the difference between this

value and this we will get

delta x and now we will find the limit

delta x tends to zero the

ratio of the increment of the function to

the increment of the argument as we said by

definition this is the derivative this

will be equal to the limit

delta x tends to zero delta y we

calculated is equal to delta x divide by

delta x dx divided by delta x

cancels out we get 1 this means this

limit will be equal to one this means the

derivative of this function is equal to 1

since we have y this is x then we can write the

first formula the derivative of x is

one let's now consider this

function let's say we have a straight line y is equal to

and to x plus b this is our y from x we find y

from x plus delta x that is we find the

value of the function of the point at the point x plus z

and x we get to instead of x we put x

plus delta x plus b now we find the transformations the

transformations the

transformation is different the corresponding

from this value and this if we

write this out we get that kx will be destroyed

ska xd will go away would remain to delta x

then the limit

with delta x tending to zero from the

ratio of the increment of the function to the

increment of the argument delta x will be

equal to and this means that the

derivative of this function will be

equal to in other words the derivative

of kx + b

will be equal to although consider an example

let's try to find the derivatives of the function and

advertising x square for this y prime

we have is equal to the limit delta x tends

to zero but the ratio of the increment of the

function to the increment of the argument the increment of the

function this will be the value of the

function at point x plus delta x squared minus the value of the function at point x x squared

divided by delta x

we get that x squared

minus x squared will leave

2 x delta x plus

delta x squared divided by

delta x we can reduce by delta x then here we have

dtx here goes squared here will be

delta x will remain 2 x plus bx if

delta x tends to zero it will remain

2 x thus we got that the

derivative of x squared is equal to 2 x

let's look at another example let's

find the derivative of the function y is equal to

one x also find the limit delta x

tends to zero

y from x plus delta x so instead of x

I'll leave x plus x minus

1 by x divided by delta x we

here I put on a common denominator and

send this denominator down the common

denominator will be x by x plus delta x

multiplied by delta x of this then delta x from above

above

we get x minus x minus delta x now x minus x we have mutually

cancel delta x is reduced with

delta x then from above I will be left with just

-13 delta x tending to 0 we have this will be

0 will remain minus 1

divided by x multiplied by x that is -1

by x squared thus we got

that one divided by x the derivative

is -1 by x-squared the

derivative has several important

applications the first of them is the following

let you have some body that

moves according to the law y is equal to and from

their that is some function then

the speed of this body will be be as a

derivative of this function, that is, the

speed of the body is only a stroke from x, so

knowing how the body moves, we can

find its speed using the derivative, the

second meaning of the derivative is geometric,

geometric,

let's say we have some function y

equal to f from x and

let's say we

want to draw a tangent line to some point of this function, this tangent

line is inclined to the x axis at

some angle alpha, so the

derivative of this function at the point of contact,

let's say this. let's say x 0 that is the

derivative of the function f from x at point x0 and that

is the tangent of the angle of inclination

corresponding to the tangent

this is the geometric meaning of the

derivative on this this video lesson and wants

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10 класс, 40 урок, Определение производной