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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 “most complex” process is presented in the diagram below:

It would seem, what does mathematics have to do with it? Yes, despite the fact that a complex function is formed in EXACTLY THE 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 “impacts” (packages) are applied to one X in a row and it turns out to be “function from function” - “packaging within packaging”.

In the 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 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 is clear that here they first did arithmetic operations with x, then took the sine of 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 \(ctg x\). 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 a very simple thing, moreover, in fact, we have already analyzed them above: if we remember 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 started all about - 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 “word by word” parsing diagram to understand what is 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.

The operation of finding the 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. The first to work in the field of finding derivatives were Isaac Newton (1643-1727) and Gottfried Wilhelm Leibniz (1646-1716).

Therefore, in our time, to find the derivative of any function, you do not need to calculate the above-mentioned limit of the ratio of the increment of the function to the increment of the argument, but you 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 prime sign break down simple functions into components and determine what actions (product, sum, quotient) these functions are related. Next, 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 derivative table 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 a 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 sine is equal to cosine. We substitute these values ​​into the sum of derivatives and find the derivative required by the condition of the problem:

Example 2. Find the derivative of a function

Solution. We differentiate as a derivative of a sum in which the second term has a constant factor; it can be taken out of the sign of the derivative:

If questions still arise about where something comes from, they are usually cleared up after familiarizing yourself with the table of derivatives and the simplest rules of differentiation. We are moving on 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 equal to 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 for a long time
3. Derivative of degree. When solving problems, you need to convert non-square roots into powers.
4. Derivative of a variable to the power -1
5. Derivative of square root
6. Derivative of sine
7. Derivative of cosine
8. Derivative of tangent
9. Derivative of cotangent
10. Derivative of arcsine
11. Derivative of arccosine
12. Derivative of arctangent
13. Derivative of arc cotangent
14. Derivative of the natural logarithm
15. Derivative of a logarithmic function
16. Derivative of the exponent
17. Derivative of an exponential function

Rules of differentiation

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

Rule 1.If the functions

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

and

those. the derivative of an 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 term, then their derivatives are equal, i.e.

Rule 2.If the functions

are differentiable at some point, then their product is 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.

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

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

For example, for three multipliers:

Rule 3.If the functions

differentiable at some point And , then at this point their quotient is also differentiableu/v , and

those. the derivative of the quotient of two functions is equal to a fraction, the numerator of which 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 for things on other pages

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

Comment. You should not confuse a constant (that is, a number) as a term in a 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. This is a typical mistake that occurs at the initial stage of studying derivatives, but as the average student solves several one- and two-part examples, he no longer makes this mistake.

And if, when differentiating a product or quotient, you have a term u"v, in which 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 (this case is discussed in example 10).

Another common mistake is mechanically solving the derivative of a complex function as the derivative of a simple function. That's why derivative of a complex function a separate article is devoted. But first we will learn to find derivatives of simple functions.

Along the way, you can’t do without transforming expressions. To do this, you may need to open the manual in new windows. Actions with powers and roots And Operations with fractions .

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

If you have a task like , then you will take 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 define the parts of the function expression: the entire expression represents a 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 by 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 has 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 turns 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 obtain the following derivative values:

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 to the derivative problem 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 the quotient: the derivative of the quotient of two functions is equal to a fraction, the numerator of which 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 us 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 problems in which you need to find the derivative of a function, where there is a continuous pile of roots and powers, such as, for example, , then welcome to class "Derivative of sums 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 a lesson for you "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, the derivative of which we familiarized ourselves with in the table of derivatives. Using the rule for differentiating the product and the tabular value of the derivative of the square root, we obtain:

You can check the solution to the derivative problem at online derivatives calculator .

Example 6. Find the derivative of a function

Solution. In this function we see a quotient whose dividend is the square root of the independent variable. Using the rule of differentiation of quotients, which we repeated and applied in example 4, and the tabulated value of the derivative of the square root, we obtain:

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

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?, in which we analyzed 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 if 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 finding the derivative right away 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 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 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, 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:

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, you first 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 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 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 a funny 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, 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 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.

If g(x) And f(u) – differentiable functions of their arguments, respectively, at points x And u= g(x), then the complex function is also differentiable at the point x and is found by the formula

A typical mistake when solving derivative problems is mechanically transferring the rules for differentiating simple functions to complex functions. Let's learn to avoid this mistake.

Example 2. Find the derivative of a function

Wrong solution: calculate the natural logarithm of each term in parentheses and look for the sum of the derivatives:

Correct solution: again we determine where the “apple” is and where the “minced meat” is. Here the natural logarithm of the expression in parentheses is an “apple”, that is, a function over the intermediate argument u, and the expression in brackets is “minced meat”, that is, an intermediate argument u by independent variable x.

Then (using formula 14 from the derivatives table)

In many real-life problems, the expression with a logarithm can be somewhat more complicated, which is why there is a lesson

Example 3. Find the derivative of a function

Wrong solution:

Correct solution. Once again we determine where the “apple” is and where the “mincemeat” is. Here, the cosine of the expression in brackets (formula 7 in the table of derivatives) is an “apple”, it is prepared in mode 1, which affects only it, and the expression in brackets (the derivative of the degree is number 3 in the table of derivatives) is “minced meat”, it is prepared under mode 2, which affects only it. And as always, we connect two derivatives with the product sign. Result:

The derivative of a complex logarithmic function is a frequent task in tests, so we strongly recommend that you attend the lesson “Derivative of a logarithmic function.”

The first examples were on complex functions, in which the intermediate argument on the independent variable was a simple function. But in practical tasks it is often necessary to find the derivative of a complex function, where the intermediate argument is either itself a complex function or contains such a function. What to do in such cases? Find derivatives of such functions using tables and differentiation rules. When the derivative of the intermediate argument is found, it is simply substituted into the right place in the formula. Below are two examples of how this is done.

In addition, it is useful to know the following. If a complex function can be represented as a chain of three functions

then its derivative should be found as the product of the derivatives of each of these functions:

Many of your homework assignments may require you to open your guides in new windows. Actions with powers and roots And Operations with fractions .

Example 4. Find the derivative of a function

We apply the rule of differentiation of a complex function, not forgetting that in the resulting product of derivatives there is an intermediate argument with respect to the independent variable x does not change:

We prepare the second factor of the product and apply the rule for differentiating the sum:

The second term is the root, so

Thus, we found that the intermediate argument, which is a sum, contains a complex function as one of the terms: raising to a power is a complex function, and what is being raised to a power is an intermediate argument with respect to the independent variable x.

Therefore, we again apply the rule for differentiating a complex function:

We transform the degree of the first factor into a root, and when differentiating the second factor, do not forget that the derivative of the constant is equal to zero:

Now we can find the derivative of the intermediate argument needed to calculate the derivative of a complex function required in the problem statement y:

Example 5. Find the derivative of a function

First, we use the rule for differentiating the sum:

We obtained the sum of the derivatives of two complex functions. Let's find the first one:

Here, raising the sine to a power is a complex function, and the sine itself is an intermediate argument for the independent variable x. Therefore, we will use the rule of differentiation of a complex function, along the way taking the factor out of brackets :

Now we find the second term of the derivatives of the function y:

Here raising the cosine to a power is a complex function f, and the cosine itself is an intermediate argument in the independent variable x. Let us again use the rule for differentiating a complex function:

The result is the required derivative:

Table of derivatives of some complex functions

For complex functions, based on the rule of differentiation of a complex function, the formula for the derivative of a simple function takes a different form.

1. Derivative of a complex power function, where u x
2. Derivative of the root of the expression
3. Derivative of an exponential function
4. Special case of exponential function
5. Derivative of a logarithmic function with an arbitrary positive base A
6. Derivative of a complex logarithmic function, where u– differentiable function of the argument x
7. Derivative of sine
8. Derivative of cosine
9. Derivative of tangent
10. Derivative of cotangent
11. Derivative of arcsine
12. Derivative of arccosine
13. Derivative of arctangent
14. Derivative of arc cotangent

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. These are relatively simple expressions, the derivatives of which have long been calculated and entered into 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 differentiated according 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 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 + 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 multiplier is a little more complicated, but the general scheme does not 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 a new function h(x) = f(x)/g(x). For such a function you can also find the derivative:

Not weak, huh? 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 with 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 using specific examples, with a detailed description of each 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 we get an 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 the last expression, the whole problem has been 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, the stroke of the sum is equal to the sum of the 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 a final 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 be 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 in 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: