The rule of differentiation of a complex function examples. Function derivative

It is absolutely impossible to solve physical problems or examples in mathematics without knowledge about the derivative and methods for calculating it. The derivative is one of the most important concepts of mathematical analysis. We decided to devote today's article to this fundamental topic. What is a derivative, what is its physical and geometric meaning, how to calculate the derivative of a function? All these questions can be combined into one: how to understand the derivative?

Geometric and physical meaning of the derivative

Let there be a function f(x) , given in some interval (a,b) . The points x and x0 belong to this interval. When x changes, the function itself changes. Argument change - difference of its values x-x0 . This difference is written as delta x and is called argument increment. The change or increment of a function is the difference between the values of the function at two points. Derivative definition:

The derivative of a function at a point is the limit of the ratio of the increment of the function at a given point to the increment of the argument when the latter tends to zero.

Otherwise it can be written like this:

What is the point in finding such a limit? But which one:

the derivative of a function at a point is equal to the tangent of the angle between the OX axis and the tangent to the graph of the function at a given point.

The physical meaning of the derivative: the time derivative of the path is equal to the speed of the rectilinear motion.

Indeed, since school days, everyone knows that speed is a private path. x=f(t) and time t . Average speed over a certain period of time:

To find out the speed of movement at a time t0 you need to calculate the limit:

Rule one: take out the constant

The constant can be taken out of the sign of the derivative. Moreover, it must be done. When solving examples in mathematics, take as a rule - if you can simplify the expression, be sure to simplify .

Example. Let's calculate the derivative:

Rule two: derivative of the sum of functions

The derivative of the sum of two functions is equal to the sum of the derivatives of these functions. The same is true for the derivative of the difference of functions.

We will not give a proof of this theorem, but rather consider a practical example.

Find the derivative of a function:

Rule three: the derivative of the product of functions

The derivative of the product of two differentiable functions is calculated by the formula:

Example: find the derivative of a function:

Solution:

Here it is important to say about the calculation of derivatives of complex functions. The derivative of a complex function is equal to the product of the derivative of this function with respect to the intermediate argument by the derivative of the intermediate argument with respect to the independent variable.

In the above example, we encounter the expression:

In this case, the intermediate argument is 8x to the fifth power. In order to calculate the derivative of such an expression, we first consider the derivative of the external function with respect to the intermediate argument, and then multiply by the derivative of the intermediate argument itself with respect to the independent variable.

Rule Four: The derivative of the quotient of two functions

Formula for determining the derivative of a quotient of two functions:

We tried to talk about derivatives for dummies from scratch. This topic is not as simple as it sounds, so be warned: there are often pitfalls in the examples, so be careful when calculating derivatives.

With any question on this and other topics, you can contact the student service. In a short time, we will help you solve the most difficult control and deal with tasks, even if you have never dealt with the calculation of derivatives before.

In the "old" textbooks, it is also called the "chain" rule. So if y \u003d f (u), and u \u003d φ (x), that is

y \u003d f (φ (x))

complex - compound function (composition of functions) then

where , after calculation is considered at u = φ(x).

Note that here we took "different" compositions from the same functions, and the result of differentiation naturally turned out to be dependent on the order of "mixing".

The chain rule naturally extends to the composition of three or more functions. In this case, there will be three or more “links” in the “chain” that makes up the derivative, respectively. Here is an analogy with multiplication: “we have” - a table of derivatives; "there" - multiplication table; “with us” is a chain rule and “there” is a multiplication rule with a “column”. When calculating such “complex” derivatives, of course, no auxiliary arguments (u¸v, etc.) are introduced, but, having noted for themselves the number and sequence of functions participating in the composition, they “string” the corresponding links in the indicated order.

. Here, five operations are performed with “x” to obtain the value of “y”, that is, a composition of five functions takes place: “external” (the last of them) - exponential - e ; then in reverse order is a power law. (♦) 2 ; trigonometric sin (); power. () 3 and finally the logarithmic ln.(). That's why

The following examples will “kill pairs of birds with one stone”: we will practice differentiating complex functions and supplement the table of derivatives of elementary functions. So:

4. For a power function - y \u003d x α - rewriting it using the well-known "basic logarithmic identity" - b \u003d e ln b - in the form x α \u003d x α ln x we get

5. For an arbitrary exponential function, using the same technique, we will have

6. For an arbitrary logarithmic function, using the well-known formula for the transition to a new base, we successively obtain

7. To differentiate the tangent (cotangent), we use the rule for differentiating the quotient:

To obtain derivatives of inverse trigonometric functions, we use the relation which is satisfied by the derivatives of two mutually inverse functions, that is, the functions φ (x) and f (x) connected by the relations:

Here is the ratio

It is from this formula for mutually inverse functions

and
,

In the end, we summarize these and some other, just as easily obtained derivatives, in the following table.

Complex functions do not always fit the definition of a complex function. If there is a function of the form y \u003d sin x - (2 - 3) a r c t g x x 5 7 x 10 - 17 x 3 + x - 11, then it cannot be considered complex, unlike y \u003d sin 2 x.

This article will show the concept of a complex function and its identification. Let's work with formulas for finding the derivative with examples of solutions in the conclusion. The use of the table of derivatives and the rules of differentiation significantly reduce the time to find the derivative.

Basic definitions

Definition 1

A complex function is a function whose argument is also a function.

It is denoted this way: f (g (x)) . We have that the function g (x) is considered an argument f (g (x)) .

Definition 2

If there is a function f and is a cotangent function, then g(x) = ln x is the natural logarithm function. We get that the complex function f (g (x)) will be written as arctg (lnx). Or a function f, which is a function raised to the 4th power, where g (x) \u003d x 2 + 2 x - 3 is considered an entire rational function, we get that f (g (x)) \u003d (x 2 + 2 x - 3) 4 .

Obviously g(x) can be tricky. From the example y \u003d sin 2 x + 1 x 3 - 5, it can be seen that the value of g has a cube root with a fraction. This expression can be denoted as y = f (f 1 (f 2 (x))) . Whence we have that f is a sine function, and f 1 is a function located under the square root, f 2 (x) \u003d 2 x + 1 x 3 - 5 is a fractional rational function.

Definition 3

The degree of nesting is defined by any natural number and is written as y = f (f 1 (f 2 (f 3 (. . . (f n (x)))))) .

Definition 4

The concept of function composition refers to the number of nested functions according to the problem statement. For the solution, the formula for finding the derivative of a complex function of the form

(f(g(x)))"=f"(g(x)) g"(x)

Examples

Example 1

Find the derivative of a complex function of the form y = (2 x + 1) 2 .

Solution

By convention, f is a squaring function, and g(x) = 2 x + 1 is considered a linear function.

We apply the derivative formula for a complex function and write:

f "(g (x)) = ((g (x)) 2) " = 2 (g (x)) 2 - 1 = 2 g (x) = 2 (2 x + 1) ; g "(x) = (2x + 1)" = (2x)" + 1" = 2 x" + 0 = 2 1 x 1 - 1 = 2 ⇒ (f(g(x))) "=f" (g(x)) g"(x) = 2 (2x + 1) 2 = 8x + 4

It is necessary to find a derivative with a simplified initial form of the function. We get:

y = (2x + 1) 2 = 4x2 + 4x + 1

Hence we have that

y"=(4x2+4x+1)"=(4x2)"+(4x)"+1"=4(x2)"+4(x)"+0==4 2 x 2 - 1 + 4 1 x 1 - 1 = 8 x + 4

The results matched.

When solving problems of this kind, it is important to understand where the function of the form f and g (x) will be located.

Example 2

You should find the derivatives of complex functions of the form y \u003d sin 2 x and y \u003d sin x 2.

Solution

The first entry of the function says that f is the squaring function and g(x) is the sine function. Then we get that

y "= (sin 2 x)" = 2 sin 2 - 1 x (sin x)" = 2 sin x cos x

The second entry shows that f is a sine function, and g (x) = x 2 denote the power function. It follows that the product of a complex function can be written as

y " \u003d (sin x 2) " \u003d cos (x 2) (x 2) " \u003d cos (x 2) 2 x 2 - 1 \u003d 2 x cos (x 2)

The formula for the derivative y \u003d f (f 1 (f 2 (f 3 (. . . (f n (x)))))) will be written as y "= f" (f 1 (f 2 (f 3 (. . . ( f n (x)))))) f 1 "(f 2 (f 3 (. . . (f n (x))))) f 2 " (f 3 (. . . (f n (x)) )) . . . f n "(x)

Example 3

Find the derivative of the function y = sin (ln 3 a r c t g (2 x)) .

Solution

This example shows the complexity of writing and determining the location of functions. Then y \u003d f (f 1 (f 2 (f 3 (f 4 (x))))) denote, where f , f 1 , f 2 , f 3 , f 4 (x) is the sine function, the function of raising to 3 degree, a function with a logarithm and base e, a function of the arc tangent and a linear one.

From the formula for the definition of a complex function, we have that

y "= f" (f 1 (f 2 (f 3 (f 4 (x))))) f 1 "(f 2 (f 3 (f 4 (x)))) f 2 "(f 3 (f 4 (x))) f 3 "(f 4 (x)) f 4" (x)

Getting what to find

f "(f 1 (f 2 (f 3 (f 4 (x))))) as the derivative of the sine in the table of derivatives, then f "(f 1 (f 2 (f 3 (f 4 (x))))) ) = cos (ln 3 a r c t g (2 x)) .
f 1 "(f 2 (f 3 (f 4 (x)))) as a derivative of a power function, then f 1 "(f 2 (f 3 (f 4 (x)))) = 3 ln 3 - 1 a r c t g (2 x) = 3 ln 2 a r c t g (2 x) .
f 2 "(f 3 (f 4 (x))) as a logarithmic derivative, then f 2 "(f 3 (f 4 (x))) = 1 a r c t g (2 x) .
f 3 "(f 4 (x)) as a derivative of the arc tangent, then f 3 "(f 4 (x)) = 1 1 + (2 x) 2 = 1 1 + 4 x 2.
When finding the derivative f 4 (x) \u003d 2 x, take 2 out of the sign of the derivative using the formula for the derivative of the power function with an exponent that is 1, then f 4 "(x) \u003d (2 x)" \u003d 2 x "\u003d 2 · 1 · x 1 - 1 = 2 .

We combine the intermediate results and get that

y "= f" (f 1 (f 2 (f 3 (f 4 (x))))) f 1 "(f 2 (f 3 (f 4 (x)))) f 2 "(f 3 (f 4 (x))) f 3 "(f 4 (x)) f 4" (x) = = cos (ln 3 a r c t g (2 x)) 3 ln 2 a r c t g (2 x) 1 a r c t g (2 x) 1 1 + 4 x 2 2 = = 6 cos (ln 3 a r c t g (2 x)) ln 2 a r c t g (2 x) a r c t g (2 x) (1 + 4 x 2)

The analysis of such functions resembles nesting dolls. Differentiation rules cannot always be applied explicitly using a derivative table. Often you need to apply the formula for finding derivatives of complex functions.

There are some differences between a complex view and a complex function. With a clear ability to distinguish this, finding derivatives will be especially easy.

Example 4

It is necessary to consider on bringing such an example. If there is a function of the form y = t g 2 x + 3 t g x + 1 , then it can be considered as a complex function of the form g (x) = t g x , f (g) = g 2 + 3 g + 1 . Obviously, it is necessary to apply the formula for the complex derivative:

f "(g (x)) \u003d (g 2 (x) + 3 g (x) + 1) " \u003d (g 2 (x)) " + (3 g (x)) " + 1 " == 2 g 2 - 1 (x) + 3 g "(x) + 0 \u003d 2 g (x) + 3 1 g 1 - 1 (x) \u003d \u003d 2 g (x) + 3 \u003d 2 t g x + 3; g " (x) = (t g x) " = 1 cos 2 x ⇒ y " = (f (g (x))) " = f " (g (x)) g " (x) = (2 t g x + 3 ) 1 cos 2 x = 2 t g x + 3 cos 2 x

A function of the form y = t g x 2 + 3 t g x + 1 is not considered complex, since it has the sum t g x 2 , 3 t g x and 1 . However, t g x 2 is considered a complex function, then we get a power function of the form g (x) \u003d x 2 and f, which is a function of the tangent. To do this, you need to differentiate by the amount. We get that

y " = (t g x 2 + 3 t g x + 1) " = (t g x 2) " + (3 t g x) " + 1 " = = (t g x 2) " + 3 (t g x) " + 0 = (t g x 2) " + 3 cos 2 x

Let's move on to finding the derivative of a complex function (t g x 2) ":

f "(g (x)) = (t g (g (x)))" = 1 cos 2 g (x) = 1 cos 2 (x 2) g "(x) = (x 2)" = 2 x 2 - 1 \u003d 2 x ⇒ (t g x 2) " \u003d f " (g (x)) g " (x) \u003d 2 x cos 2 (x 2)

We get that y "= (t g x 2 + 3 t g x + 1)" = (t g x 2) " + 3 cos 2 x = 2 x cos 2 (x 2) + 3 cos 2 x

Complex functions can be included in complex functions, and the complex functions themselves can be complex functions of the complex form.

Example 5

For example, consider a complex function of the form y = log 3 x 2 + 3 cos 3 (2 x + 1) + 7 e x 2 + 3 3 + ln 2 x (x 2 + 1)

This function can be represented as y = f (g (x)) , where the value of f is a function of the base 3 logarithm, and g (x) is considered the sum of two functions of the form h (x) = x 2 + 3 cos 3 (2 x + 1) + 7 e x 2 + 3 3 and k (x) = ln 2 x (x 2 + 1) . Obviously, y = f (h (x) + k (x)) .

Consider the function h(x) . This is the ratio of l (x) = x 2 + 3 cos 3 (2 x + 1) + 7 to m (x) = e x 2 + 3 3

We have that l (x) = x 2 + 3 cos 2 (2 x + 1) + 7 = n (x) + p (x) is the sum of two functions n (x) = x 2 + 7 and p (x) \u003d 3 cos 3 (2 x + 1) , where p (x) \u003d 3 p 1 (p 2 (p 3 (x))) is a complex function with a numerical coefficient of 3, and p 1 is a cube function, p 2 cosine function, p 3 (x) = 2 x + 1 - linear function.

We found that m (x) = e x 2 + 3 3 = q (x) + r (x) is the sum of two functions q (x) = e x 2 and r (x) = 3 3 , where q (x) = q 1 (q 2 (x)) is a complex function, q 1 is a function with an exponent, q 2 (x) = x 2 is a power function.

This shows that h (x) = l (x) m (x) = n (x) + p (x) q (x) + r (x) = n (x) + 3 p 1 (p 2 ( p 3 (x))) q 1 (q 2 (x)) + r (x)

When passing to an expression of the form k (x) \u003d ln 2 x (x 2 + 1) \u003d s (x) t (x), it is clear that the function is represented as a complex s (x) \u003d ln 2 x \u003d s 1 ( s 2 (x)) with integer rational t (x) = x 2 + 1, where s 1 is the squaring function, and s 2 (x) = ln x is logarithmic with base e.

It follows that the expression will take the form k (x) = s (x) t (x) = s 1 (s 2 (x)) t (x) .

Then we get that

y = log 3 x 2 + 3 cos 3 (2 x + 1) + 7 e x 2 + 3 3 + ln 2 x (x 2 + 1) = = f n (x) + 3 p 1 (p 2 (p 3 (x))) q 1 (q 2 (x)) = r (x) + s 1 (s 2 (x)) t (x)

According to the structures of the function, it became clear how and what formulas must be applied to simplify the expression when it is differentiated. To familiarize yourself with such problems and to understand their solution, it is necessary to refer to the point of differentiating a function, that is, finding its derivative.

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The proof of the formula for the derivative of a complex function is given. Cases where a complex function depends on one or two variables are considered in detail. A generalization is made to the case of an arbitrary number of variables.

Content

See also: Examples of applying the formula for the derivative of a complex function

Basic formulas

Here we present the derivation of the following formulas for the derivative of a complex function.
If , then
.
If , then
.
If , then
.

Derivative of a complex function of one variable

Let a function of a variable x be represented as a complex function in the following form:
,
where and there are some functions. The function is differentiable for some value of the variable x . The function is differentiable for the value of the variable .
Then the complex (composite) function is differentiable at the point x and its derivative is determined by the formula:
(1) .

Formula (1) can also be written as follows:
;
.

Proof

Let us introduce the following notation.
;
.
Here there is a function of variables and , there is a function of variables and . But we will omit the arguments of these functions so as not to clutter up the calculations.

Since the functions and are differentiable at the points x and , respectively, then at these points there are derivatives of these functions, which are the following limits:
;
.

Consider the following function:
.
For a fixed value of the variable u , is a function of . It's obvious that
.
Then
.

Since the function is a differentiable function at the point , then it is continuous at that point. That's why
.
Then
.

Now we find the derivative.

.

The formula has been proven.

Consequence

If a function of variable x can be represented as a complex function of a complex function
,
then its derivative is determined by the formula
.
Here , and there are some differentiable functions.

To prove this formula, we sequentially calculate the derivative according to the rule of differentiation of a complex function.
Consider a complex function
.
Its derivative
.
Consider the original function
.
Its derivative
.

Derivative of a complex function in two variables

Now let a complex function depend on several variables. First consider case of a complex function of two variables.

Let the function depending on the variable x be represented as a complex function of two variables in the following form:
,
where
and there are differentiable functions for some value of the variable x ;
is a function of two variables, differentiable at the point , . Then the complex function is defined in some neighborhood of the point and has a derivative, which is determined by the formula:
(2) .

Proof

Since the functions and are differentiable at the point , they are defined in some neighborhood of this point, are continuous at the point, and their derivatives at the point exist, which are the following limits:
;
.
Here
;
.
Due to the continuity of these functions at a point, we have:
;
.

Since the function is differentiable at the point , it is defined in some neighborhood of this point, is continuous at this point, and its increment can be written as follows:
(3) .
Here

- function increment when its arguments are incremented by the values and ;
;

- partial derivatives of the function with respect to the variables and .
For fixed values of and , and there are functions of the variables and . They tend to zero as and :
;
.
Since and , then
;
.

Function increment :

. :
.
Substitute (3):

.

The formula has been proven.

Derivative of a complex function of several variables

The above derivation is easily generalized to the case when the number of variables of a complex function is greater than two.

For example, if f is function of three variables, then
,
where
, and there are differentiable functions for some value of the variable x ;
is a differentiable function, in three variables, at the point , , .
Then, from the definition of differentiability of the function , we have:
(4)
.
Since, due to continuity,
; ; ,
then
;
;
.

Dividing (4) by and passing to the limit , we obtain:
.

And finally, consider the most general case.
Let a function of a variable x be represented as a complex function of n variables in the following form:
,
where
there are differentiable functions for some value of the variable x ;
- differentiable function of n variables at a point
, , ... , .
Then
.

See also:

On which we analyzed the simplest derivatives, and also got acquainted with the rules of differentiation and some techniques for finding derivatives. Thus, if you are not very good with derivatives of functions or some points of this article are not entirely clear, then first read the above lesson. Please tune in to a serious mood - the material is not easy, but I will still try to present it simply and clearly.

In practice, you have to deal with the derivative of a complex function very often, I would even say almost always, when you are given tasks to find derivatives.

We look in the table at the rule (No. 5) for differentiating a complex function:

We understand. First of all, let's take a look at the notation. Here we have two functions - and , and the function, figuratively speaking, is nested in the function . A function of this kind (when one function is nested within another) is called a complex function.

I will call the function external function, and the function – inner (or nested) function.

! These definitions are not theoretical and should not appear in the final design of assignments. I use the informal expressions "external function", "internal" function only to make it easier for you to understand the material.

To clarify the situation, consider:

Example 1

Find the derivative of a function

Under the sine, we have not just the letter "x", but the whole expression, so finding the derivative immediately from the table will not work. We also notice that it is impossible to apply the first four rules here, there seems to be a difference, but the fact is that it is impossible to “tear apart” the sine:

In this example, already from my explanations, it is intuitively clear that the function is a complex function, and the polynomial is an internal function (embedding), and an external function.

First step, which must be performed when finding the derivative of a complex function is to understand which function is internal and which is external.

In the case of simple examples, it seems clear that a polynomial is nested under the sine. But what if it's not obvious? How to determine exactly which function is external and which is internal? To do this, I propose to use the following technique, which can be carried out mentally or on a draft.

Let's imagine that we need to calculate the value of the expression with a calculator (instead of one, there can be any number).

What do we calculate first? First of all you will need to perform the following action: , so the polynomial will be an internal function:

Secondly you will need to find, so the sine - will be an external function:

After we UNDERSTAND with inner and outer functions, it's time to apply the compound function differentiation rule .

We start to decide. From the lesson How to find the derivative? we remember that the design of the solution of any derivative always begins like this - we enclose the expression in brackets and put a stroke at the top right:

First we find the derivative of the external function (sine), look at the table of derivatives of elementary functions and notice that . All tabular formulas are applicable even if "x" is replaced by a complex expression, in this case:

Note that the inner function has not changed, we do not touch it.

Well, it is quite obvious that

The result of applying the formula clean looks like this:

The constant factor is usually placed at the beginning of the expression:

If there is any misunderstanding, write down the decision on paper and read the explanations again.

Example 2

Find the derivative of a function

Example 3

Find the derivative of a function

As always, we write:

We figure out where we have an external function, and where is an internal one. To do this, we try (mentally or on a draft) to calculate the value of the expression for . What needs to be done first? First of all, you need to calculate what the base is equal to:, which means that the polynomial is the internal function:

And, only then exponentiation is performed, therefore, the power function is an external function:

According to the formula , first you need to find the derivative of the external function, in this case, the degree. We are looking for the desired formula in the table:. We repeat again: any tabular formula is valid not only for "x", but also for a complex expression. Thus, the result of applying the rule of differentiation of a complex function next:

I emphasize again that when we take the derivative of the outer function, the inner function does not change:

Now it remains to find a very simple derivative of the inner function and “comb” the result a little:

Example 4

Find the derivative of a function

This is an example for self-solving (answer at the end of the lesson).

To consolidate the understanding of the derivative of a complex function, I will give an example without comments, try to figure it out on your own, reason, where is the external and where is the internal function, why are the tasks solved that way?

Example 5

a) Find the derivative of a function

b) Find the derivative of the function

Example 6

Find the derivative of a function

Here we have a root, and in order to differentiate the root, it must be represented as a degree. Thus, we first bring the function into the proper form for differentiation:

Analyzing the function, we come to the conclusion that the sum of three terms is an internal function, and exponentiation is an external function. We apply the rule of differentiation of a complex function :

The degree is again represented as a radical (root), and for the derivative of the internal function, we apply a simple rule for differentiating the sum:

Ready. You can also bring the expression to a common denominator in brackets and write everything as one fraction. It’s beautiful, of course, but when cumbersome long derivatives are obtained, it’s better not to do this (it’s easy to get confused, make an unnecessary mistake, and it will be inconvenient for the teacher to check).

Example 7

Find the derivative of a function

This is an example for self-solving (answer at the end of the lesson).

It is interesting to note that sometimes, instead of the rule for differentiating a complex function, one can use the rule for differentiating a quotient , but such a solution will look like a perversion unusual. Here is a typical example:

Example 8

Find the derivative of a function

Here you can use the rule of differentiation of the quotient , but it is much more profitable to find the derivative through the rule of differentiation of a complex function:

We prepare the function for differentiation - we take out the minus sign of the derivative, and raise the cosine to the numerator:

Cosine is an internal function, exponentiation is an external function.
Let's use our rule :

We find the derivative of the inner function, reset the cosine back down:

Ready. In the considered example, it is important not to get confused in the signs. By the way, try to solve it with the rule , the answers must match.

Example 9

Find the derivative of a function

This is an example for self-solving (answer at the end of the lesson).

So far, we have considered cases where we had only one nesting in a complex function. In practical tasks, you can often find derivatives, where, like nesting dolls, one inside the other, 3 or even 4-5 functions are nested at once.

Example 10

Find the derivative of a function

We understand the attachments of this function. We try to evaluate the expression using the experimental value . How would we count on a calculator?

First you need to find, which means that the arcsine is the deepest nesting:

This arcsine of unity should then be squared:

And finally, we raise the seven to the power:

That is, in this example we have three different functions and two nestings, while the innermost function is the arcsine, and the outermost function is the exponential function.

We start to decide

According to the rule first you need to take the derivative of the outer function. We look at the table of derivatives and find the derivative of the exponential function: The only difference is that instead of "x" we have a complex expression, which does not negate the validity of this formula. So, the result of applying the rule of differentiation of a complex function next.