Calculate the derivative of a complex function examples. Rules for calculating derivatives

Since you came here, you probably already managed to see this formula in the textbook

and make a face like this:

Friend, don't worry! In fact, everything is simple to disgrace. You will definitely understand everything. Only one request - read the article slowly try to understand every step. I wrote as simply and clearly as possible, but you still need to delve into the idea. And be sure to solve the tasks from the article.

What is a complex function?

Imagine that you are moving to another apartment and therefore you are packing things in big boxes. Let it be necessary to collect some small items, for example, school stationery. If you just throw them in a huge box, they will get lost among other things. To avoid this, you first put them, for example, in a bag, which you then put in a large box, after which you seal it. This "hardest" process is shown in the diagram below:

It would seem, where does the mathematics? And besides, a complex function is formed in EXACTLY THE SAME way! Only we “pack” not notebooks and pens, but \ (x \), while different “packages” and “boxes” serve.

For example, let's take x and "pack" it into a function:

As a result, we get, of course, \(\cos⁡x\). This is our "bag of things". And now we put it in a "box" - we pack it, for example, into a cubic function.

What will happen in the end? Yes, that's right, there will be a "package with things in a box", that is, "cosine of x cubed."

The resulting construction is a complex function. It differs from the simple one in that SEVERAL “impacts” (packages) are applied to one X in a row and it turns out, as it were, “a function from a function” - “a package in a package”.

In the school course, there are very few types of these same “packages”, only four:

Let's now "pack" x first into an exponential function with base 7, and then into a trigonometric function. We get:

\(x → 7^x → tg⁡(7^x)\)

And now let's “pack” x twice into trigonometric functions, first into and then into:

\(x → sin⁡x → ctg⁡ (sin⁡x)\)

Simple, right?

Now write the functions yourself, where x:
- first it is “packed” into a cosine, and then into an exponential function with base \(3\);
- first to the fifth power, and then to the tangent;
- first to the base logarithm \(4\) , then to the power \(-2\).

See the answers to this question at the end of the article.

But can we "pack" x not two, but three times? No problem! And four, and five, and twenty-five times. Here, for example, is a function in which x is "packed" \(4\) times:

\(y=5^(\log_2⁡(\sin⁡(x^4)))\)

But such formulas will not be found in school practice (students are more fortunate - they can be more difficult☺).

"Unpacking" a complex function

Look at the previous function again. Can you figure out the sequence of "packing"? What X was stuffed into first, what then, and so on until the very end. That is, which function is nested in which? Take a piece of paper and write down what you think. You can do this with a chain of arrows, as we wrote above, or in any other way.

Now the correct answer is: first x was “packed” into the \(4\)th power, then the result was packed into the sine, it, in turn, was placed in the logarithm base \(2\), and in the end the whole construction was shoved into the power fives.

That is, it is necessary to unwind the sequence IN THE REVERSE ORDER. And here is a hint how to do it easier: just look at the X - you have to dance from it. Let's look at a few examples.

For example, here is a function: \(y=tg⁡(\log_2⁡x)\). We look at X - what happens to him first? Taken from him. And then? The tangent of the result is taken. And the sequence will be the same:

\(x → \log_2⁡x → tg⁡(\log_2⁡x)\)

Another example: \(y=\cos⁡((x^3))\). We analyze - first x was cubed, and then the cosine was taken from the result. So the sequence will be: \(x → x^3 → \cos⁡((x^3))\). Pay attention, the function seems to be similar to the very first one (where with pictures). But this is a completely different function: here in the cube x (that is, \(\cos⁡((x x x)))\), and there in the cube the cosine \(x\) (that is, \(\cos⁡ x·\cos⁡x·\cos⁡x\)). This difference arises from different "packing" sequences.

The last example (with important information in it): \(y=\sin⁡((2x+5))\). It is clear that here we first performed arithmetic operations with x, then we took the sine from the result: \(x → 2x+5 → \sin⁡((2x+5))\). And this is an important point: despite the fact that arithmetic operations are not functions in themselves, here they also act as a way of “packing”. Let's delve a little deeper into this subtlety.

As I said above, in simple functions x is "packed" once, and in complex functions - two or more. Moreover, any combination of simple functions (that is, their sum, difference, multiplication or division) is also a simple function. For example, \(x^7\) is a simple function, and so is \(ctg x\). Hence, all their combinations are simple functions:

\(x^7+ ctg x\) - simple,
\(x^7 ctg x\) is simple,
\(\frac(x^7)(ctg x)\) is simple, and so on.

However, if one more function is applied to such a combination, it will already be a complex function, since there will be two “packages”. See diagram:

Okay, let's get on with it now. Write the sequence of "wrapping" functions:
\(y=cos(⁡(sin⁡x))\)
\(y=5^(x^7)\)
\(y=arctg⁡(11^x)\)
\(y=log_2⁡(1+x)\)
The answers are again at the end of the article.

Internal and external functions

Why do we need to understand function nesting? What does this give us? The point is that without such an analysis we will not be able to reliably find the derivatives of the functions discussed above.

And in order to move on, we will need two more concepts: internal and external functions. This is a very simple thing, moreover, in fact, we have already analyzed them above: if we recall our analogy at the very beginning, then the inner function is the “package” and the outer one is the “box”. Those. what X is “wrapped” in first is an internal function, and what the internal is “wrapped” in is already external. Well, it’s understandable why - it’s outside, it means external.

Here in this example: \(y=tg⁡(log_2⁡x)\), the function \(\log_2⁡x\) is internal, and
- external.

And in this one: \(y=\cos⁡((x^3+2x+1))\), \(x^3+2x+1\) is internal, and
- external.

Perform the last practice of analyzing complex functions, and finally, let's move on to the point for which everything was started - we will find derivatives of complex functions:

Fill in the gaps in the table:

Derivative of a complex function

Bravo to us, we still got to the "boss" of this topic - in fact, the derivative of a complex function, and specifically, to that very terrible formula from the beginning of the article.☺

\((f(g(x)))"=f"(g(x))\cdot g"(x)\)

This formula reads like this:

The derivative of a complex function is equal to the product of the derivative of the external function with respect to the constant internal function and the derivative of the internal function.

And immediately look at the parsing scheme "by words" to understand what to relate to:

I hope the terms "derivative" and "product" do not cause difficulties. "Complex function" - we have already dismantled. The catch is in the "derivative of the external function with respect to the constant internal." What it is?

Answer: this is the usual derivative of the outer function, in which only the outer function changes, while the inner one remains the same. Still unclear? Okay, let's take an example.

Let's say we have a function \(y=\sin⁡(x^3)\). It is clear that the inner function here is \(x^3\), and the outer
. Let us now find the derivative of the outer with respect to the constant inner.

If we follow the definition, then the derivative of a function at a point is the limit of the increment ratio of the function Δ y to the increment of the argument Δ x:

Everything seems to be clear. But try to calculate by this formula, say, the derivative of the function f(x) = x 2 + (2x+ 3) · e x sin x. If you do everything by definition, then after a couple of pages of calculations you will simply fall asleep. Therefore, there are simpler and more effective ways.

To begin with, we note that the so-called elementary functions can be distinguished from the whole variety of functions. These are relatively simple expressions, the derivatives of which have long been calculated and entered in the table. Such functions are easy enough to remember, along with their derivatives.

Derivatives of elementary functions

Elementary functions are everything listed below. The derivatives of these functions must be known by heart. Moreover, it is not difficult to memorize them - that's why they are elementary.

So, the derivatives of elementary functions:

Name	Function	Derivative
Constant	f(x) = C, C ∈ R	0 (yes, yes, zero!)
Degree with rational exponent	f(x) = x n	n · x n − 1
Sinus	f(x) = sin x	cos x
Cosine	f(x) = cos x	− sin x(minus sine)
Tangent	f(x) = tg x	1/cos 2 x
Cotangent	f(x) = ctg x	− 1/sin2 x
natural logarithm	f(x) = log x	1/x
Arbitrary logarithm	f(x) = log a x	1/(x ln a)
Exponential function	f(x) = e x	e x(nothing changed)

If an elementary function is multiplied by an arbitrary constant, then the derivative of the new function is also easily calculated:

(C · f)’ = C · f ’.

In general, constants can be taken out of the sign of the derivative. For example:

(2x 3)' = 2 ( x 3)' = 2 3 x 2 = 6x 2 .

Obviously, elementary functions can be added to each other, multiplied, divided, and much more. This is how new functions will appear, no longer very elementary, but also differentiable according to certain rules. These rules are discussed below.

Derivative of sum and difference

Let the functions f(x) and g(x), whose derivatives are known to us. For example, you can take the elementary functions discussed above. Then you can find the derivative of the sum and difference of these functions:

1. (f + g)’ = f ’ + g ’
2. (f − g)’ = f ’ − g ’

So, the derivative of the sum (difference) of two functions is equal to the sum (difference) of the derivatives. There may be more terms. For example, ( f + g + h)’ = f ’ + g ’ + h ’.

Strictly speaking, there is no concept of "subtraction" in algebra. There is a concept of "negative element". Therefore, the difference f − g can be rewritten as a sum f+ (−1) g, and then only one formula remains - the derivative of the sum.

f(x) = x 2 + sinx; g(x) = x 4 + 2x 2 − 3.

Function f(x) is the sum of two elementary functions, so:

f ’(x) = (x 2+ sin x)’ = (x 2)' + (sin x)’ = 2x+ cosx;

We argue similarly for the function g(x). Only there are already three terms (from the point of view of algebra):

g ’(x) = (x 4 + 2x 2 − 3)’ = (x 4 + 2x 2 + (−3))’ = (x 4)’ + (2x 2)’ + (−3)’ = 4x 3 + 4x + 0 = 4x · ( x 2 + 1).

Answer:
f ’(x) = 2x+ cosx;
g ’(x) = 4x · ( x 2 + 1).

Derivative of a product

Mathematics is a logical science, so many people believe that if the derivative of the sum is equal to the sum of the derivatives, then the derivative of the product strike"\u003e equal to the product of derivatives. But figs to you! The derivative of the product is calculated using a completely different formula. Namely:

(f · g) ’ = f ’ · g + f · g ’

The formula is simple, but often forgotten. And not only schoolchildren, but also students. The result is incorrectly solved problems.

A task. Find derivatives of functions: f(x) = x 3 cosx; g(x) = (x 2 + 7x− 7) · e x .

Function f(x) is a product of two elementary functions, so everything is simple:

f ’(x) = (x 3 cos x)’ = (x 3)' cos x + x 3 (cos x)’ = 3x 2 cos x + x 3 (−sin x) = x 2 (3cos x − x sin x)

Function g(x) the first multiplier is a little more complicated, but the general scheme does not change from this. Obviously, the first multiplier of the function g(x) is a polynomial, and its derivative is the derivative of the sum. We have:

g ’(x) = ((x 2 + 7x− 7) · e x)’ = (x 2 + 7x− 7)' · e x + (x 2 + 7x− 7) ( e x)’ = (2x+ 7) · e x + (x 2 + 7x− 7) · e x = e x(2 x + 7 + x 2 + 7x −7) = (x 2 + 9x) · e x = x(x+ 9) · e x .

Answer:
f ’(x) = x 2 (3cos x − x sin x);
g ’(x) = x(x+ 9) · e x .

Note that in the last step, the derivative is factorized. Formally, this is not necessary, but most derivatives are not calculated on their own, but to explore the function. This means that further the derivative will be equated to zero, its signs will be found out, and so on. For such a case, it is better to have an expression decomposed into factors.

If there are two functions f(x) and g(x), and g(x) ≠ 0 on the set of interest to us, we can define a new function h(x) = f(x)/g(x). For such a function, you can also find the derivative:

Not weak, right? Where did the minus come from? Why g 2? But like this! This is one of the most complex formulas - you can’t figure it out without a bottle. Therefore, it is better to study it with specific examples.

A task. Find derivatives of functions:

There are elementary functions in the numerator and denominator of each fraction, so all we need is the formula for the derivative of the quotient:

By tradition, we factor the numerator into factors - this will greatly simplify the answer:

A complex function is not necessarily a formula half a kilometer long. For example, it suffices to take the function f(x) = sin x and replace the variable x, say, on x 2+ln x. It turns out f(x) = sin ( x 2+ln x) is a complex function. She also has a derivative, but it will not work to find it according to the rules discussed above.

How to be? In such cases, the replacement of a variable and the formula for the derivative of a complex function help:

f ’(x) = f ’(t) · t', if x is replaced by t(x).

As a rule, the situation with the understanding of this formula is even more sad than with the derivative of the quotient. Therefore, it is also better to explain it with specific examples, with detailed description every step.

A task. Find derivatives of functions: f(x) = e 2x + 3 ; g(x) = sin ( x 2+ln x)

Note that if in the function f(x) instead of expression 2 x+ 3 will be easy x, then we get an elementary function f(x) = e x. Therefore, we make a substitution: let 2 x + 3 = t, f(x) = f(t) = e t. We are looking for the derivative of a complex function by the formula:

f ’(x) = f ’(t) · t ’ = (e t)’ · t ’ = e t · t ’

And now - attention! Performing a reverse substitution: t = 2x+ 3. We get:

f ’(x) = e t · t ’ = e 2x+ 3 (2 x + 3)’ = e 2x+ 3 2 = 2 e 2x + 3

Now let's look at the function g(x). Obviously needs to be replaced. x 2+ln x = t. We have:

g ’(x) = g ’(t) · t' = (sin t)’ · t' = cos t · t ’

Reverse replacement: t = x 2+ln x. Then:

g ’(x) = cos( x 2+ln x) · ( x 2+ln x)' = cos ( x 2+ln x) · (2 x + 1/x).

That's all! As can be seen from the last expression, the whole problem has been reduced to calculating the derivative of the sum.

Answer:
f ’(x) = 2 e 2x + 3 ;
g ’(x) = (2x + 1/x) cos ( x 2+ln x).

Very often in my lessons, instead of the term “derivative”, I use the word “stroke”. For example, the stroke of the sum is equal to the sum of the strokes. Is that clearer? Well, that's good.

Thus, the calculation of the derivative comes down to getting rid of these very strokes according to the rules discussed above. As a final example, let's return to the derivative power with a rational exponent:

(x n)’ = n · x n − 1

Few know that in the role n may well be a fractional number. For example, the root is x 0.5 . But what if there is something tricky under the root? Again, a complex function will turn out - they like to give such constructions in tests and exams.

A task. Find the derivative of a function:

First, let's rewrite the root as a power with a rational exponent:

f(x) = (x 2 + 8x − 7) 0,5 .

Now we make a substitution: let x 2 + 8x − 7 = t. We find the derivative by the formula:

f ’(x) = f ’(t) · t ’ = (t 0.5)' t' = 0.5 t−0.5 t ’.

We make a reverse substitution: t = x 2 + 8x− 7. We have:

f ’(x) = 0.5 ( x 2 + 8x− 7) −0.5 ( x 2 + 8x− 7)' = 0.5 (2 x+ 8) ( x 2 + 8x − 7) −0,5 .

Finally, back to the roots:

The operation of finding a derivative is called differentiation.

As a result of solving problems of finding derivatives of the simplest (and not very simple) functions by defining the derivative as the limit of the ratio of the increment to the increment of the argument, a table of derivatives and precisely defined rules of differentiation appeared. Isaac Newton (1643-1727) and Gottfried Wilhelm Leibniz (1646-1716) were the first to work in the field of finding derivatives.

Therefore, in our time, in order to find the derivative of any function, it is not necessary to calculate the above-mentioned limit of the ratio of the increment of the function to the increment of the argument, but only need to use the table of derivatives and the rules of differentiation. The following algorithm is suitable for finding the derivative.

To find the derivative, you need an expression under the stroke sign break down simple functions and determine what actions (product, sum, quotient) these functions are related. Further, we find the derivatives of elementary functions in the table of derivatives, and the formulas for the derivatives of the product, sum and quotient - in the rules of differentiation. The table of derivatives and differentiation rules are given after the first two examples.

Example 1 Find the derivative of a function

Solution. From the rules of differentiation we find out that the derivative of the sum of functions is the sum of derivatives of functions, i.e.

From the table of derivatives, we find out that the derivative of "X" is equal to one, and the derivative of the sine is cosine. We substitute these values ​​in the sum of derivatives and find the derivative required by the condition of the problem:

Example 2 Find the derivative of a function

Solution. Differentiate as a derivative of the sum, in which the second term with a constant factor, it can be taken out of the sign of the derivative:

If there are still questions about where something comes from, they, as a rule, become clear after reading the table of derivatives and the simplest rules of differentiation. We are going to them right now.

Table of derivatives of simple functions

1. Derivative of a constant (number). Any number (1, 2, 5, 200...) that is in the function expression. Always zero. This is very important to remember, as it is required very often
2. Derivative of the independent variable. Most often "x". Always equal to one. This is also important to remember
3. Derivative of degree. When solving problems, you need to convert non-square roots to a power.
4. Derivative of a variable to the power of -1
5. Derivative of the square root
6. Sine derivative
7. Cosine derivative
8. Tangent derivative
9. Derivative of cotangent
10. Derivative of the arcsine
11. Derivative of arc cosine
12. Derivative of arc tangent
13. Derivative of the inverse tangent
14. Derivative of natural logarithm
15. Derivative of a logarithmic function
16. Derivative of the exponent
17. Derivative of exponential function

Differentiation rules

1. Derivative of the sum or difference
2. Derivative of a product
2a. Derivative of an expression multiplied by a constant factor
3. Derivative of the quotient
4. Derivative of a complex function

Rule 1If functions

are differentiable at some point , then at the same point the functions

and

those. the derivative of the algebraic sum of functions is equal to the algebraic sum of the derivatives of these functions.

Consequence. If two differentiable functions differ by a constant, then their derivatives are, i.e.

Rule 2If functions

are differentiable at some point , then their product is also differentiable at the same point

and

those. the derivative of the product of two functions is equal to the sum of the products of each of these functions and the derivative of the other.

Consequence 1. The constant factor can be taken out of the sign of the derivative:

Consequence 2. The derivative of the product of several differentiable functions is equal to the sum of the products of the derivative of each of the factors and all the others.

For example, for three multipliers:

Rule 3If functions

differentiable at some point and , then at this point their quotient is also differentiable.u/v , and

those. the derivative of a quotient of two functions is equal to a fraction whose numerator is the difference between the products of the denominator and the derivative of the numerator and the numerator and the derivative of the denominator, and the denominator is the square of the former numerator.

Where to look on other pages

When finding the derivative of the product and the quotient in real problems, it is always necessary to apply several differentiation rules at once, so more examples on these derivatives are in the article."The derivative of a product and a quotient".

Comment. You should not confuse a constant (that is, a number) as a term in the sum and as a constant factor! In the case of a term, its derivative is equal to zero, and in the case of a constant factor, it is taken out of the sign of the derivatives. it typical mistake, which occurs at the initial stage of the study of derivatives, but as the solution of several one-two-part examples, the average student no longer makes this mistake.

And if, when differentiating a product or a quotient, you have a term u"v, wherein u- a number, for example, 2 or 5, that is, a constant, then the derivative of this number will be equal to zero and, therefore, the entire term will be equal to zero (such a case is analyzed in example 10).

Other common mistake- mechanical solution of the derivative of a complex function as a derivative of a simple function. That's why derivative of a complex function devoted to a separate article. But first we will learn to find derivatives of simple functions.

Along the way, you can not do without transformations of expressions. To do this, you may need to open in new windows manuals Actions with powers and roots and Actions with fractions .

If you are looking for solutions to derivatives with powers and roots, that is, when the function looks like , then follow the lesson " Derivative of the sum of fractions with powers and roots".

If you have a task like , then you are in the lesson "Derivatives of simple trigonometric functions".

Step by step examples - how to find the derivative

Example 3 Find the derivative of a function

Solution. We determine the parts of the expression of the function: the entire expression represents the product, and its factors are sums, in the second of which one of the terms contains a constant factor. We apply the product differentiation rule: the derivative of the product of two functions is equal to the sum of the products of each of these functions and the derivative of the other:

Next, we apply the rule of differentiation of the sum: the derivative of the algebraic sum of functions is equal to the algebraic sum of the derivatives of these functions. In our case, in each sum, the second term with a minus sign. In each sum, we see both an independent variable, the derivative of which is equal to one, and a constant (number), the derivative of which is equal to zero. So, "x" turns into one, and minus 5 - into zero. In the second expression, "x" is multiplied by 2, so we multiply two by the same unit as the derivative of "x". We get the following values ​​of derivatives:

We substitute the found derivatives into the sum of products and obtain the derivative of the entire function required by the condition of the problem:

And you can check the solution of the problem on the derivative on .

Example 4 Find the derivative of a function

Solution. We are required to find the derivative of the quotient. We apply the formula for differentiating a quotient: the derivative of a quotient of two functions is equal to a fraction whose numerator is the difference between the products of the denominator and the derivative of the numerator and the numerator and the derivative of the denominator, and the denominator is the square of the former numerator. We get:

We have already found the derivative of the factors in the numerator in Example 2. Let's also not forget that the product, which is the second factor in the numerator in the current example, is taken with a minus sign:

If you are looking for solutions to such problems in which you need to find the derivative of a function, where there is a continuous pile of roots and degrees, such as, for example, then welcome to class "The derivative of the sum of fractions with powers and roots" .

If you need to learn more about the derivatives of sines, cosines, tangents and other trigonometric functions, that is, when the function looks like , then you have a lesson "Derivatives of simple trigonometric functions" .

Example 5 Find the derivative of a function

Solution. In this function, we see a product, one of the factors of which is the square root of the independent variable, with the derivative of which we familiarized ourselves in the table of derivatives. According to the product differentiation rule and the tabular value of the derivative of the square root, we get:

You can check the solution of the derivative problem on derivative calculator online .

Example 6 Find the derivative of a function

Solution. In this function, we see the quotient, the dividend of which is the square root of the independent variable. According to the rule of differentiation of the quotient, which we repeated and applied in example 4, and the tabular value of the derivative of the square root, we get:

To get rid of the fraction in the numerator, multiply the numerator and denominator by .

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 this lesson, we will learn how to find derivative of a complex function. The lesson is a logical continuation of the lesson How to find the derivative?, on which we analyzed the simplest derivatives, and also got acquainted with the rules of differentiation and some technical methods 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 final result of applying the formula 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 is the following:

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 would look like a perversion funny. 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, you first need to take the derivative of the external 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 is the following:

Under the dash, we have a tricky function again! But it's already easier. It is easy to see that the inner function is the arcsine and the outer function is the degree. According to the rule of differentiation of a complex function, you first need to take the derivative of the degree.