Since you came here, you probably already saw this formula in the textbook

and make a face like this:

Friend, don't worry! In fact, everything is simply outrageous. You will definitely understand everything. Just one request - read the article slowly, try to understand every step. I wrote as simply and clearly as possible, but you still need to understand 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 packing things into large boxes. Suppose you need to collect some small items, for example, school writing materials. If you just throw them into 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 “complex” process is presented in the diagram below:

It would seem, what does mathematics have to do with it? Yes, despite the fact that complex function formed in the EXACT SAME way! Only we “pack” not notebooks and pens, but \(x\), while the “packages” and “boxes” are different.

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”. Now let’s put it in a “box” - pack it, for example, into a cubic function.

What will happen in the end? Yes, that’s right, there will be a “bag of things in a box,” that is, “cosine of X cubed.”

The resulting design is a complex function. It differs from simple one in that SEVERAL “influences” (packages) are applied to one X in a row and it turns out as if “function from function” - “packaging within packaging”.

IN school course There are very few types of these “packages”, only four:

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

\(x → 7^x → tg⁡(7^x)\)

Now let’s “pack” X twice into trigonometric functions, first in , and then in:

\(x → sin⁡x → cotg⁡ (sin⁡x)\)

Simple, right?

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

Find the answers to this task at the end of the article.

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 luckier - theirs may be more complicated☺).

"Unpacking" a complex function

Look at the previous function again. Can you figure out the “packing” sequence? What X was stuffed into first, what then, and so on until the very end. That is, which function is nested within which? Take a piece of paper and write down what you think. You can do this with a chain with 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 a sine, it, in turn, was placed into the logarithm to the base \(2\), and in the end this whole construction was stuffed into a power fives.

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

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

\(x → \log_2⁡x → tg⁡(\log_2⁡x)\)

Another example: \(y=\cos⁡((x^3))\). Let's analyze - first we cubed X, and then took the cosine of the result. This means the sequence will be: \(x → x^3 → \cos⁡((x^3))\). Pay attention, the function seems to be similar to the very first one (where it has pictures). But this is a completely different function: here in the cube is x (that is, \(\cos⁡((x·x·x)))\), and there in the cube is 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's clear what they did here first arithmetic operations with x, then took the sine of the result: \(x → 2x+5 → \sin⁡((2x+5))\). And this 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 simple function. For example, \(x^7\) is a simple function and so is \(ctg x\). This means that all their combinations are simple functions:

\(x^7+ ctg x\) - simple,
\(x^7· cot x\) – simple,
\(\frac(x^7)(ctg x)\) – simple, etc.

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

Okay, go ahead 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 fact is that without such an analysis we will not be able to reliably find derivatives of the functions discussed above.

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

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

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

Complete the last practice of analyzing complex functions, and let's finally move on to what we were all started for - we will find derivatives of complex functions:

Fill in the blanks in the table:

Derivative of a complex function

Bravo to us, we finally 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 a constant internal function and the derivative of the internal function.

And immediately look at the parsing diagram, according to the words, so that you understand what to do with what:

I hope the terms “derivative” and “product” do not cause any difficulties. “Complex function” - we have already sorted it out. The catch is in the “derivative of an external function with respect to a constant internal function.” What it is?

Answer: this is the usual derivative of an external function, in which only the external function changes, and the internal one remains the same. Still not clear? Okay, let's use an example.

Let us have a function \(y=\sin⁡(x^3)\). It is clear that the internal function here is \(x^3\), and the external
. Let us now find the derivative of the exterior with respect to the constant interior.

Derivative of a complex function. Examples of solutions

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 examined the simplest derivatives, and also became acquainted with the rules of differentiation and some technical methods finding derivatives. Thus, if you are not very good with derivatives of functions or some points in this article are not entirely clear, then first read the above lesson. Please get in a serious mood - the material is not simple, 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 at the table at the rule (No. 5) for differentiating a complex function:

Let's figure it out. First of all, let's pay attention to the entry. Here we have two functions - and , and the function, figuratively speaking, is nested within the function . A function of this type (when one function is nested within another) is called a complex function.

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

! These definitions are not theoretical and should not appear in the final design of assignments. I apply 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 an entire expression, so it will not be possible to find the derivative directly from the table. 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 the sine cannot be “torn into pieces”:

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

First step what you need to do when finding the derivative of a complex function is to understand which function is internal and which is external.

When simple examples It seems clear that a polynomial is embedded under the sine. But what if everything is not obvious? How to accurately determine which function is external and which is internal? To do this, I suggest using the following technique, which can be done mentally or in a draft.

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

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

Secondly will need to be found, so sine – will be an external function:

After we SOLD OUT With internal and external functions, it's time to apply the rule of differentiation of complex functions.

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

At first we find the derivative of the external function (sine), look at the table of derivatives of elementary functions and notice that . All table formulas are also applicable if “x” is replaced with a complex expression, V in this case:

Please note that the inner function hasn't changed, we don't touch it.

Well, it's quite obvious that

The final result of applying the formula looks like this:

Constant multiplier usually placed at the beginning of the expression:

If there is any misunderstanding, write the solution down 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 down:

Let's figure out where we have an external function and where we have an internal one. To do this, we try (mentally or in a draft) to calculate the value of the expression at . What should you do first? First of all, you need to calculate what the base is equal to: therefore, the polynomial is the internal function:

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

According to the formula, you first need to find the derivative of the external function, in this case, the degree. Looking for in the table the required formula: . We repeat again: any tabular formula valid not only for “x”, but also for complex expressions. Thus, the result of applying the rule for differentiating a complex function is as follows:

I emphasize again that when we take the derivative of the external function, our internal function does not change:

Now all that remains is to find a very simple derivative of the internal function and tweak the result a little:

Example 4

Find the derivative of a function

This is an example for independent decision(answer at the end of the lesson).

To consolidate your 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 the external and where the internal function is, why the tasks are solved this way?

Example 5

a) Find the derivative of the 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 power. Thus, first we bring the function into the form appropriate for differentiation:

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

We again represent the degree 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 give the expression in parentheses to common denominator and write everything down as one fraction. It’s beautiful, of course, but when you get cumbersome long derivatives, 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 you to solve on your own (answer at the end of the lesson).

It is interesting to note that sometimes instead of the rule for differentiating a complex function, you can use the rule for differentiating a quotient , but such a solution will look like a funny perversion. Here 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 move the minus out of the derivative sign, and raise the cosine into the numerator:

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

We find the derivative of the internal function and reset the cosine back down:

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

Example 9

Find the derivative of a function

This is an example for you to solve on your own (answer at the end of the lesson).

So far we have looked at 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

Let's understand the attachments of this function. Let's try to calculate the expression using the experimental value. How would we count on a calculator?

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

This arcsine of one should then be squared:

And finally, we raise seven to a power:

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

Let's start deciding

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 exponential function: The only difference is that instead of “X” we have complex expression, which does not negate the validity of this formula. So, the result of applying the rule for differentiating a complex function is as follows:

Under the stroke we have a complex function again! But it’s already simpler. It is easy to verify that the inner function is the arcsine, the outer function is the degree. According to the rule for differentiating a complex function, you first need to take the derivative of the power.

On which we examined the simplest derivatives, and also became acquainted with the rules of differentiation and some technical techniques for finding derivatives. Thus, if you are not very good with derivatives of functions or some points in this article are not entirely clear, then first read the above lesson. Please get in a serious mood - the material is not simple, 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 at the table at the rule (No. 5) for differentiating a complex function:

Let's figure it out. First of all, let's pay attention to the entry. Here we have two functions - and , and the function, figuratively speaking, is nested within the function . A function of this type (when one function is nested within another) is called a complex function.

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

! These definitions are not theoretical and should not appear in the final design of assignments. I use 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 an entire expression, so it will not be possible to find the derivative directly from the table. 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 the sine cannot be “torn into pieces”:

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

First step what you need to do 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 embedded under the sine. But what if everything is not obvious? How to accurately determine which function is external and which is internal? To do this, I suggest using the following technique, which can be done mentally or in a draft.

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

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

Secondly will need to be found, so sine – will be an external function:

After we SOLD OUT with internal and external functions, it’s time to apply the rule of differentiation of complex functions .

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

At first we find the derivative of the external function (sine), look at the table of derivatives of elementary functions and notice that . All table formulas are also applicable if “x” is replaced with a complex expression, in this case:

Please note that the inner function hasn't changed, we don't touch it.

Well, it's quite obvious that

The result of applying the formula in its final form it looks like this:

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

If there is any misunderstanding, write the solution down 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 down:

Let's figure out where we have an external function and where we have an internal one. To do this, we try (mentally or in a draft) to calculate the value of the expression at . What should you do first? First of all, you need to calculate what the base is equal to: therefore, the polynomial is the internal function:

And only then is the exponentiation 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 look for the required 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 for differentiating a complex function next:

I emphasize again that when we take the derivative of the external function, our internal function does not change:

Now all that remains is to find a very simple derivative of the internal function and tweak the result a little:

Example 4

Find the derivative of a function

This is an example for you to solve on your own (answer at the end of the lesson).

To consolidate your 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 the external and where the internal function is, why the tasks are solved this way?

Example 5

a) Find the derivative of the 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 power. Thus, first we bring the function into the form appropriate for differentiation:

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

We again represent the degree 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 reduce the expression to a common denominator in brackets and write everything down as one fraction. It’s beautiful, of course, but when you get cumbersome long derivatives, 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 you to solve on your own (answer at the end of the lesson).

It is interesting to note that sometimes instead of the rule for differentiating a complex function, you can use the rule for differentiating a quotient , but such a solution will look like an unusual perversion. 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 move the minus out of the derivative sign, and raise the cosine into the numerator:

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

We find the derivative of the internal function and reset the cosine back down:

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

Example 9

Find the derivative of a function

This is an example for you to solve on your own (answer at the end of the lesson).

So far we have looked at 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

Let's understand the attachments of this function. Let's try to calculate the expression using the experimental value. How would we count on a calculator?

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

This arcsine of one should then be squared:

And finally, we raise seven to a power:

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

Let's start deciding

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 for differentiating a complex function next.

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

Everything seems to be clear. But try using this formula to calculate, 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 from the entire variety of functions we can distinguish the so-called elementary functions. It's relative simple expressions, the derivatives of which have long been calculated and listed in the table. Such functions are quite easy to remember - along with their derivatives.

Derivatives of elementary functions

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

So, derivatives of elementary functions:

Name	Function	Derivative
Constant	f(x) = C, C ∈ R	0 (yes, zero!)
Power 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/sin 2 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 particularly elementary, but also differentiable with respect to certain rules. These rules are discussed below.

Derivative of sum and difference

Let the functions be given f(x) And g(x), the derivatives of which 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 " 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 + sin x; g(x) = x 4 + 2x 2 − 3.

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

f ’(x) = (x 2 + sin x)’ = (x 2)’ + (sin x)’ = 2x+ cos x;

We reason 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+ cos x;
g ’(x) = 4x · ( x 2 + 1).

Derivative of the product

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

(f · g) ’ = f ’ · g + f · g ’

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

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

Function f(x) is the 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 factor is a little more complicated, but general scheme this doesn't change. Obviously, the first factor 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 .

Please note that in the last step the derivative is factorized. Formally, this does not need to be done, but most derivatives are not calculated on their own, but to examine the function. This means that further the derivative will be equated to zero, its signs will be determined, and so on. For such a case, it is better to have an expression factorized.

If there are two functions f(x) And g(x), and g(x) ≠ 0 on the set we are interested in, we can define new feature 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? And 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 on specific examples.

Task. Find derivatives of functions:

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

According to tradition, let's factorize the numerator - this will greatly simplify the answer:

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

What should I do? In such cases, replacing a variable and formula for the derivative of a complex function helps:

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

As a rule, the situation with understanding 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.

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 it will work out elementary function f(x) = e x. Therefore, we make a replacement: let 2 x + 3 = t, f(x) = f(t) = e t. We look for the derivative of a complex function using the formula:

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

And now - attention! We perform the reverse replacement: 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 it 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 last expression, the whole problem was reduced to calculating the derivative 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 “prime.” For example, a prime from the amount equal to the sum strokes. Is that clearer? Well, that's good.

Thus, calculating the derivative comes down to getting rid of these same strokes according to the rules discussed above. As last example Let's return to the derivative power with a rational exponent:

(x n)’ = n · x n − 1

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

Task. Find the derivative of the 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 replacement: let x 2 + 8x − 7 = t. We find the derivative using the formula:

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

Let's do the reverse replacement: 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: