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

y \u003d f (φ (x))

complex - compound function (composition of functions) then

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

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

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

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

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

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

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

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

.

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

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

Here is the ratio

It is from this formula for mutually inverse functions

and
,

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

If g(x) and f(u) are differentiable functions of their arguments, respectively, at the 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 in solving problems on derivatives is the automatic transfer of the rules for differentiating simple functions to complex functions. We will learn to avoid this mistake.

Example 2 Find the derivative of a function

Wrong solution: calculate the natural logarithm of each term in brackets and find the sum of derivatives:

Correct solution: again we determine where is the "apple" and where is the "minced meat". Here, the natural logarithm of the expression in brackets is the "apple", that is, the function on 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 table of derivatives)

In many real problems, the expression with the logarithm is 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" and where the "minced meat". Here, the cosine of the expression in brackets (formula 7 in the table of derivatives) is "apple", it is cooked in mode 1, affecting only it, and the expression in brackets (the derivative of the degree - number 3 in the table of derivatives) is "minced meat", it is cooked in mode 2, affecting only it. And as always, we connect two derivatives with a product sign. Result:

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

The first examples were for complex functions, in which the intermediate argument over the independent variable was a simple function. But in practical tasks it is often required 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 in 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 tutorials in new windows. Actions with powers and roots and Actions 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, the 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, it was obtained that the intermediate argument, which is the sum, contains a complex function as one of the terms: exponentiation is a complex function, and what is raised to a power is an intermediate argument by an independent variable x.

Therefore, we again apply the rule of differentiation of a complex function:

We transform the degree of the first factor into a root, and differentiating the second factor, we 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 the complex function required in the condition of the problem y:

Example 5 Find the derivative of a function

First, we use the rule of differentiating the sum:

Get the sum of derivatives of two complex functions. Find the first one:

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

Now we find the second term from those that form the derivative of the function y:

Here, raising the cosine to a power is a complex function f, and the cosine itself is an intermediate argument with respect to the independent variable x. Again, we use the rule of differentiation of 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 the exponential function
4. Special case of the exponential function
5. Derivative of a logarithmic function with an arbitrary positive base but
6. Derivative of a complex logarithmic function, where u is a differentiable function of the argument x
7. Sine derivative
8. Cosine derivative
9. Tangent derivative
10. Derivative of cotangent
11. Derivative of the arcsine
12. Derivative of arc cosine
13. Derivative of arc tangent
14. Derivative of the inverse tangent

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 seems, 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.

complex derivatives. Logarithmic derivative.
Derivative of exponential function

We continue to improve our differentiation technique. In this lesson, we will consolidate the material covered, consider more complex derivatives, and also get acquainted with new tricks and tricks for finding the derivative, in particular, with the logarithmic derivative.

Those readers who have a low level of preparation should refer to the article How to find the derivative? Solution examples which will allow you to raise your skills almost from scratch. Next, you need to carefully study the page Derivative of a compound function, understand and resolve all the examples I have given. This lesson is logically the third in a row, and after mastering it, you will confidently differentiate fairly complex functions. It is undesirable to stick to the position “Where else? Yes, and that's enough! ”, Since all the examples and solutions are taken from real tests and are often found in practice.

Let's start with repetition. On the lesson Derivative of a compound function we have considered a number of examples with detailed comments. In the course of studying differential calculus and other sections of mathematical analysis, you will have to differentiate very often, and it is not always convenient (and not always necessary) to paint examples in great detail. Therefore, we will practice in the oral finding of derivatives. The most suitable "candidates" for this are derivatives of the simplest of complex functions, for example:

According to the rule of differentiation of a complex function :

When studying other topics of matan in the future, such a detailed record is most often not required, it is assumed that the student is able to find similar derivatives on autopilot. Let's imagine that at 3 o'clock in the morning the phone rang, and a pleasant voice asked: "What is the derivative of the tangent of two x?". This should be followed by an almost instantaneous and polite response: .

The first example will be immediately intended for an independent solution.

Example 1

Find the following derivatives orally, in one step, for example: . To complete the task, you only need to use table of derivatives of elementary functions(if she hasn't already remembered). If you have any difficulties, I recommend re-reading the lesson Derivative of a compound function.

, , ,
, , ,
, , ,

, , ,

, , ,

, , ,

, ,

Answers at the end of the lesson

Complex derivatives

After preliminary artillery preparation, examples with 3-4-5 attachments of functions will be less scary. Perhaps the following two examples will seem complicated to some, but if they are understood (someone suffers), then almost everything else in differential calculus will seem like a child's joke.

Example 2

Find the derivative of a function

As already noted, when finding the derivative of a complex function, first of all, it is necessary right UNDERSTAND INVESTMENTS. In cases where there are doubts, I remind you of a useful trick: we take the experimental value "x", for example, and try (mentally or on a draft) to substitute this value into the "terrible expression".

1) First we need to calculate the expression, so the sum is the deepest nesting.

2) Then you need to calculate the logarithm:

4) Then cube the cosine:

5) At the fifth step, the difference:

6) And finally, the outermost function is the square root:

Complex Function Differentiation Formula are applied in reverse order, from the outermost function to the innermost. We decide:

Seems to be no error...

(1) We take the derivative of the square root.

(2) We take the derivative of the difference using the rule

(3) The derivative of the triple is equal to zero. In the second term, we take the derivative of the degree (cube).

(4) We take the derivative of the cosine.

(5) We take the derivative of the logarithm.

(6) Finally, we take the derivative of the deepest nesting .

It may seem too difficult, but this is not the most brutal example. Take, for example, Kuznetsov's collection and you will appreciate all the charm and simplicity of the analyzed derivative. I noticed that they like to give a similar thing at the exam to check whether the student understands how to find the derivative of a complex function, or does not understand.

The following example is for a standalone solution.

Example 3

Find the derivative of a function

Hint: First we apply the rules of linearity and the rule of differentiation of the product

Full solution and answer at the end of the lesson.

It's time to move on to something more compact and prettier.
It is not uncommon for a situation where the product of not two, but three functions is given in an example. How to find the derivative of the product of three factors?

Example 4

Find the derivative of a function

First, we look, but is it possible to turn the product of three functions into a product of two functions? For example, if we had two polynomials in the product, then we could open the brackets. But in this example, all functions are different: degree, exponent and logarithm.

In such cases, it is necessary successively apply the product differentiation rule twice

The trick is that for "y" we denote the product of two functions: , and for "ve" - ​​the logarithm:. Why can this be done? Is it - this is not the product of two factors and the rule does not work?! There is nothing complicated:

Now it remains to apply the rule a second time to bracket:

You can still pervert and take something out of the brackets, but in this case it is better to leave the answer in this form - it will be easier to check.

The above example can be solved in the second way:

Both solutions are absolutely equivalent.

Example 5

Find the derivative of a function

This is an example for an independent solution, in the sample it is solved in the first way.

Consider similar examples with fractions.

Example 6

Find the derivative of a function

Here you can go in several ways:

Or like this:

But the solution can be written more compactly if, first of all, we use the rule of differentiation of the quotient , taking for the whole numerator:

In principle, the example is solved, and if it is left in this form, it will not be a mistake. But if you have time, it is always advisable to check on a draft, but is it possible to simplify the answer? We bring the expression of the numerator to a common denominator and get rid of the three-story fraction:

The disadvantage of additional simplifications is that there is a risk of making a mistake not when finding a derivative, but when banal school transformations. On the other hand, teachers often reject the assignment and ask to “bring it to mind” the derivative.

A simpler example for a do-it-yourself solution:

Example 7

Find the derivative of a function

We continue to master the techniques for finding the derivative, and now we will consider a typical case when a “terrible” logarithm is proposed for differentiation

Example 8

Find the derivative of a function

Here you can go a long way, using the rule of differentiation of a complex function:

But the very first step immediately plunges you into despondency - you have to take an unpleasant derivative of a fractional degree, and then also from a fraction.

So before how to take the derivative of the “fancy” logarithm, it is previously simplified using well-known school properties:

! If you have a practice notebook handy, copy these formulas right there. If you don't have a notebook, draw them on a piece of paper, as the rest of the lesson's examples will revolve around these formulas.

The solution itself can be formulated like this:

Let's transform the function:

We find the derivative:

The preliminary transformation of the function itself greatly simplified the solution. Thus, when a similar logarithm is proposed for differentiation, it is always advisable to “break it down”.

And now a couple of simple examples for an independent solution:

Example 9

Find the derivative of a function

Example 10

Find the derivative of a function

All transformations and answers at the end of the lesson.

logarithmic derivative

If the derivative of logarithms is such sweet music, then the question arises, is it possible in some cases to organize the logarithm artificially? Can! And even necessary.

Example 11

Find the derivative of a function

Similar examples we have recently considered. What to do? One can successively apply the rule of differentiation of the quotient, and then the rule of differentiation of the product. The disadvantage of this method is that you get a huge three-story fraction, which you don’t want to deal with at all.

But in theory and practice there is such a wonderful thing as the logarithmic derivative. Logarithms can be organized artificially by "hanging" them on both sides:

Note : because function can take negative values, then, generally speaking, you need to use modules: , which disappear as a result of differentiation. However, the current design is also acceptable, where by default the complex values. But if with all rigor, then in both cases it is necessary to make a reservation that.

Now you need to “break down” the logarithm of the right side as much as possible (formulas in front of your eyes?). I will describe this process in great detail:

Let's start with the differentiation.
We conclude both parts with a stroke:

The derivative of the right side is quite simple, I will not comment on it, because if you are reading this text, you should be able to handle it with confidence.

What about the left side?

On the left side we have complex function. I foresee the question: “Why, is there one letter “y” under the logarithm?”.

The fact is that this "one letter y" - IS A FUNCTION IN ITSELF(if it is not very clear, refer to the article Derivative of a function implicitly specified). Therefore, the logarithm is an external function, and "y" is an internal function. And we use the compound function differentiation rule :

On the left side, as if by magic, we have a derivative. Further, according to the rule of proportion, we throw the “y” from the denominator of the left side to the top of the right side:

And now we remember what kind of "game"-function we talked about when differentiating? Let's look at the condition:

Final answer:

Example 12

Find the derivative of a function

This is a do-it-yourself example. Sample design of an example of this type at the end of the lesson.

With the help of the logarithmic derivative, it was possible to solve any of examples No. 4-7, another thing is that the functions there are simpler, and, perhaps, the use of the logarithmic derivative is not very justified.

Derivative of exponential function

We have not considered this function yet. An exponential function is a function that has and the degree and base depend on "x". A classic example that will be given to you in any textbook or at any lecture:

How to find the derivative of an exponential function?

It is necessary to use the technique just considered - the logarithmic derivative. We hang logarithms on both sides:

As a rule, the degree is taken out from under the logarithm on the right side:

As a result, on the right side we have a product of two functions, which will be differentiated according to the standard formula .

We find the derivative, for this we enclose both parts under strokes:

The next steps are easy:

Finally:

If some transformation is not entirely clear, please re-read the explanations of Example 11 carefully.

In practical tasks, the exponential function will always be more complicated than the considered lecture example.

Example 13

Find the derivative of a function

We use the logarithmic derivative.

On the right side we have a constant and the product of two factors - "x" and "logarithm of the logarithm of x" (another logarithm is nested under the logarithm). When differentiating a constant, as we remember, it is better to immediately take it out of the sign of the derivative so that it does not get in the way; and, of course, apply the familiar rule :

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

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

Basic definitions

Definition 1

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

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

Definition 2

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

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

Definition 3

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

Definition 4

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

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

Examples

Example 1

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

Decision

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

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

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

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

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

Hence we have that

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

The results matched.

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

Example 2

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

Decision

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

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

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

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

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

Example 3

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

Decision

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

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

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

Getting what to find

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

We combine the intermediate results and get that

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

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

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

Example 4

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

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

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

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

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

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

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

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

Example 5

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

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

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

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

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

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

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

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

Then we get that

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

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

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