Complex derivatives. Logarithmic derivative.
The derivative of the 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 techniques and tricks for finding the derivative, in particular, with the logarithmic derivative.

Those readers with a low level of training should refer to the article How to find a derivative? Solution examples, which will raise your skills almost from scratch. Next, you need to carefully study the page Derivative of a complex function, understand and solve all the examples I gave. This lesson is logically the third in a row, and after mastering it, you will confidently differentiate rather complex functions. It is undesirable to adhere to the position “Where else? And that's enough! ”Because all examples and solutions are taken from real tests and are often found in practice.

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

By the rule of differentiation of a complex function :

When studying other topics of matan in the future, such a detailed note is often not required, it is assumed that the student is able to find similar derivatives on the automatic autopilot. Imagine that at 3 am the phone rang, and a pleasant voice asked: "What is the derivative of the tangent of two Xs?" This should be followed by an almost instant 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 need to use only table of derivatives of elementary functions (if it is not remembered yet). If you have any difficulties, I recommend rereading the lesson. Derivative of a complex function.

, , ,
, , ,
, , ,

, , ,

, , ,

, , ,

, ,

Answers at the end of the lesson

Complex derivatives

After preliminary artillery preparation, examples with 3-4-5 function attachments will be less scary. Perhaps the following two examples will seem difficult to some, but if you understand them (someone will suffer), then almost everything else in the differential calculus will seem like a childish 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 rightUNDERSTAND attachments. In cases where there are doubts, I recall a useful technique: we take the experimental value of "X", for example, and try (mentally or on a draft) to substitute this value in the "terrible expression".

1) First we need to calculate the expression, which means that the amount is the deepest investment.

2) Then you need to calculate the logarithm:

4) Then raise the cosine to a cube:

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:

It seems without mistakes….

(1) Take the derivative of the square root.

(2) Take the derivative of the difference using the rule

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

(4) Take the derivative of the cosine.

(5) Take the derivative of the logarithm.

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

It may sound 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 on the exam to check whether the student understands how to find the derivative of a complex function, or does not understand.

The next example is for an independent solution.

Example 3

Find the derivative of a function

Hint: First, apply the linearity rules and the product differentiation rule

Complete solution and answer at the end of the tutorial.

Now is the time to move on to something more compact and cute.
It is not uncommon for an example to give a product of not two, but three functions. How to find the derivative of the product of three factors?

Example 4

Find the derivative of a function

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

In such cases, it is necessary consistentlyapply product differentiation rule twice

The trick is that for "y" we denote the product of two functions:, and for "ve" - \u200b\u200bthe logarithm:. Why can this be done? Is it - this is not a 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 the parenthesis:

You can still be perverted and put something outside the brackets, but in this case it is better to leave the answer in this form - it will be easier to check.

The considered 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.

Let's look at 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 will be written more compactly if first of all we use the rule for differentiating the quotient , taking for the entire numerator:

In principle, the example is solved, and if you leave it as it is, it will not be an error. But if you have time, it is always advisable to check on a draft, but can the answer be simplified? Let us reduce 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 in finding the derivative, but in banal school transformations. On the other hand, teachers often reject an assignment and ask to "bring 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 methods of finding the derivative, and now we will consider a typical case when the "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 differentiating a complex function:

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

therefore before how to take the derivative of the "fancy" logarithm, it is preliminarily simplified using the well-known school properties:

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

The solution itself can be structured like this:

Let's transform the function:

Find the derivative:

Preconfiguring the function itself made the solution much easier. Thus, when such a logarithm is proposed for differentiation, it is always advisable to "break up" it.

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

We have seen similar examples recently. What to do? You can consistently apply the rule for differentiating the quotient, and then the rule for differentiating the work. 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 : since function can take negative values, then, generally speaking, you need to use modules: that will disappear as a result of differentiation. However, the current design is also acceptable, where the defaults are taken into account complex values. But if with all the severity, then in both cases, a reservation should be made that.

Now you need to maximally "break up" the logarithm of the right side (formulas before your eyes?). I will describe this process in great detail:

Actually, we proceed to differentiation.
We enclose both parts under the stroke:

The derivative of the right-hand side is quite simple, I will not comment on it, because if you are reading this text, you should confidently cope with it.

What about the left side?

On the left we have complex function... I foresee the question: "Why, there is also one letter" igrek "under the logarithm?"

The fact is that this "one letter igrek" - ITSELF IS A FUNCTION (if not very clear, refer to the article Derived from an Implicit Function). Therefore, the logarithm is an external function, and the "game" is an internal function. And we use the rule of differentiating a complex function :

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

And now we recall what kind of “game” -function we discussed in differentiation? We look at the condition:

Final answer:

Example 12

Find the derivative of a function

This is an example for a do-it-yourself solution. A sample of the 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 the 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.

The derivative of the exponential function

We have not considered this function yet. An exponential function is a function in which and the degree and base depend on "x"... A classic example that will be given to you in any textbook or in 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-hand side we have a product of two functions, which will be differentiated according to the standard formula .

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

Further actions are simple:

Finally:

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

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

If a 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 xand is found by the formula

A typical mistake when solving derivative problems 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 parentheses and look for the sum of the derivatives:

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

Then (using formula 14 from the table of derivatives)

In many real-life problems, the expression with the logarithm is somewhat more complicated, so there is a lesson

Example 3.Find the derivative of a function

Wrong solution:

Correct solution. Once again, we determine where is "apple" and where is "minced meat". Here, the cosine of the expression in brackets (formula 7 in the table of derivatives) is "apple", it is prepared in mode 1, affecting only it, and the expression in parentheses (the derivative of the power is number 3 in the table of derivatives) is "minced meat", it prepares with mode 2, which affects only it. And, as always, we connect the two derivatives with a work sign. Result:

The derivative of a complex logarithmic function is a frequent assignment in test papers, 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 on 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 a complex function itself 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.

It is also helpful 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 opening tutorials in new windows Actions with powers and roots and Fraction actions .

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 a root, therefore

Thus, we obtained 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 raised to a power is an intermediate argument with respect to the independent variable x.

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

We transform the degree of the first factor into a root, and 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 condition y:

Example 5.Find the derivative of a function

First, let's use the sum differentiation rule:

Received the sum of the derivatives of two complex functions. We find the first of them:

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

Now we find the second term from the generators of 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... Let's use the rule of differentiation of a complex function again:

The result is the required derivative:

Derivative table of some complex functions

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

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

If we 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 increment of the argument Δ x:

Everything seems to be clear. But try to calculate using this formula, say, the derivative of a 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 just fall asleep. Therefore, there are simpler and more effective ways.

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

Derivatives of elementary functions

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

So, the derivatives of elementary functions:

Name	Function	Derivative
Constant	f(x) = C, C ∈ R	0 (yes, zero!)
Rational grade	f(x) = x n	n · x n − 1
Sinus	f(x) \u003d sin x	cos x
Cosine	f(x) \u003d cos x	- sin x (minus sine)
Tangent	f(x) \u003d tg x	1 / cos 2 x
Cotangent	f(x) \u003d ctg x	- 1 / sin 2 x
Natural logarithm	f(x) \u003d ln x	1/x
Arbitrary logarithm	f(x) \u003d log a x	1/(x Ln a)
Exponential function	f(x) = e x	e x (nothing changed)

If the 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 moved outside the derivative sign. For example:

(2x 3) ’\u003d 2 · ( x 3) '\u003d 2 3 x 2 = 6x 2 .

Obviously, elementary functions can be added to each other, multiplied, divided - and much more. So new functions will appear, which are no longer particularly elementary, but also differentiable according to certain rules. These rules are discussed below.

Derivative of sum and difference

Let functions 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 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 a work

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

(f · g) ’ = f ’ · g + f · g ’

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

A 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)

The function g(x) the first factor is a little more complicated, but the general scheme does not change from this. 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 .

Note that in the last step, the derivative is factorized. Formally, this is not necessary, but most derivatives are not calculated by themselves, but in order to investigate the function. This means that further the derivative will be equated to zero, its signs will be clarified, and so on. For such a case it is better to have a factorized expression.

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

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

A task. Find derivatives of functions:

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

By tradition, factoring the numerator into factors 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) \u003d sin x and replace the variable xlet's say on x 2 + ln x... It will turn out f(x) \u003d sin ( x 2 + ln x) Is a complex function. It also has a derivative, but it will not work to find it according to the rules discussed above.

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

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

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

A task. Find derivatives of functions: f(x) = e 2x + 3 ; g(x) \u003d sin ( x 2 + ln x)

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

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

And now - attention! We carry out the reverse replacement: t = 2x + 3. We get:

f ’(x) = e t · t ’ = e 2x + 3 (2 x + 3)’ = e 2x + 3 2 \u003d 2 e 2x + 3

Now let's deal with the function g(x). Obviously, you need to replace x 2 + ln x = t... We have:

g ’(x) = g ’(t) · t ’\u003d (Sin t)’ · t ’\u003d Cos t · t ’

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

g ’(x) \u003d cos ( x 2 + ln x) · ( x 2 + ln x) ’\u003d Cos ( x 2 + ln x) · (2 x + 1/x).

That's all! As can be seen from the last expression, the whole problem was reduced to calculating the derived sum.

Answer:
f ’(x) \u003d 2 e 2x + 3 ;
g ’(x) = (2x + 1/x) Cos ( x 2 + ln x).

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

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

(x n)’ = n · x n − 1

Few know what 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, a complex function will turn out - they like to give such constructions on tests and exams.

A task. Find the derivative of a function:

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

f(x) = (x 2 + 8x − 7) 0,5 .

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

f ’(x) = f ’(t) · t ’ = (t 0.5) ' t '\u003d 0.5 t −0.5 t ’.

We do the reverse replacement: t = x 2 + 8x - 7. We have:

f ’(x) \u003d 0.5 ( x 2 + 8x - 7) −0.5 ( x 2 + 8x - 7) ’\u003d 0.5 · (2 x + 8) ( x 2 + 8x − 7) −0,5 .

Finally, back to the roots:

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 rule of differentiation significantly reduces 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 in this way: f (g (x)). We have that the function g (x) is considered an argument to f (g (x)).

Definition 2

If there is a function f and is a cotangent function, then g (x) \u003d ln x is a function of the natural logarithm. We get that the complex function f (g (x)) will be written as arctan (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 obtain 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, you can see that the value of g has a cube root with a fraction. This expression is allowed to be denoted as y \u003d f (f 1 (f 2 (x))). Whence we have that f is a sine function, and f 1 is a function located under square root, f 2 (x) \u003d 2 x + 1 x 3 - 5 is a fractional rational function.

Definition 3

The degree of nesting is determined by any natural number and is written as y \u003d f (f 1 (f 2 (f 3 (... (f n (x)))))).

Definition 4

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

(f (g (x))) "\u003d f" (g (x)) g "(x)

Examples of

Example 1

Find the derivative of a complex function of the form y \u003d (2 x + 1) 2.

Decision

By the condition, you can see that f is a squaring function, and g (x) \u003d 2 x + 1 is considered a linear function.

Let's apply the derivative formula for a complex function and write:

f "(g (x)) \u003d ((g (x)) 2)" \u003d 2 · (g (x)) 2 - 1 \u003d 2 · g (x) \u003d 2 · (2 \u200b\u200bx + 1); g "(x) \u003d (2 x + 1)" \u003d (2 x) "+ 1" \u003d 2 x "+ 0 \u003d 2 1 x 1 - 1 \u003d 2 ⇒ (f (g (x))) "\u003d f" (g (x)) g "(x) \u003d 2 (2 x + 1) 2 \u003d 8 x + 4

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

y \u003d (2 x + 1) 2 \u003d 4 x 2 + 4 x + 1

Hence we have that

y "\u003d (4 x 2 + 4 x + 1)" \u003d (4 x 2) "+ (4 x)" + 1 "\u003d 4 · (x 2)" + 4 · (x) "+ 0 \u003d \u003d 4 2 x 2 - 1 + 4 1 x 1 - 1 \u003d 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 notation of the function says that f is a squaring function and g (x) is a sine function. Then we get that

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

The second entry shows that f is a sine function, and g (x) \u003d x 2 we denote a power function. Hence 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 (... (Fn (x)))))) can be written as y "\u003d f" (f 1 (f 2 (f 3 (.. ( fn (x)))))) f 1 "(f 2 (f 3 (... (fn (x))))) f 2" (f 3 (.. (fn (x)) )) ·. ... ... · F n "(x)

Example 3

Find the derivative of the function y \u003d sin (ln 3 a r c t g (2 x)).

Decision

This example shows the complexity of writing and locating 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 a sine function, a function of raising in 3 degree, function with logarithm and base e, arctangent function and linear.

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

y "\u003d 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)

We get what to find

1. f "(f 1 (f 2 (f 3 (f 4 (x))))) as the sine derivative according to the table of derivatives, then f" (f 1 (f 2 (f 3 (f 4 (x)))) ) \u003d cos (ln 3 arctan (2 x)).
2. f 1 "(f 2 (f 3 (f 4 (x)))) as the derivative of the power function, then f 1" (f 2 (f 3 (f 4 (x)))) \u003d 3 ln 3 - 1 arctan (2 x) \u003d 3 ln 2 arctan (2 x).
3. f 2 "(f 3 (f 4 (x))) as the derivative of the logarithmic, then f 2" (f 3 (f 4 (x))) \u003d 1 a r c t g (2 x).
4. f 3 "(f 4 (x)) as the derivative of the arctangent, then f 3" (f 4 (x)) \u003d 1 1 + (2 x) 2 \u003d 1 1 + 4 x 2.
5. When finding the derivative f 4 (x) \u003d 2 x, subtract 2 outside the sign of the derivative using the formula for the derivative of a power function with an exponent equal to 1, then f 4 "(x) \u003d (2 x)" \u003d 2 x "\u003d 2 1 x 1 - 1 \u003d 2.

We combine the intermediate results and get that

y "\u003d 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) \u003d \u003d cos (ln 3 arctan (2 x)) 3 ln 2 arctan (2 x) 1 arctan (2 x) 1 1 + 4 x 2 2 \u003d \u003d 6 cos (ln 3 arctan (2 x)) ln 2 arctan (2 x) arctan (2 x) (1 + 4 x 2)

Parsing such functions resembles matryoshka dolls. Differentiation rules cannot always be applied explicitly using a table of derivatives. It is often necessary to use a formula for finding derivatives of complex functions.

There are some differences between complex and complex functions. With an obvious ability to distinguish this, finding derivatives will be especially easy.

Example 4

It is necessary to consider giving a similar example. If there is a function of the form y \u003d t g 2 x + 3 t g x + 1, then it can be considered as a complex form g (x) \u003d t g x, f (g) \u003d g 2 + 3 g + 1. Obviously, it is necessary to apply a formula for a complex derivative:

f "(g (x)) \u003d (g 2 (x) + 3 g (x) + 1)" \u003d (g 2 (x)) "+ (3 g (x))" + 1 "\u003d \u003d 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 tgx + 3; g "(x) \u003d (tgx)" \u003d 1 cos 2 x ⇒ y "\u003d (f (g (x)))" \u003d f "(g (x)) g" (x) \u003d (2 tgx + 3 ) 1 cos 2 x \u003d 2 tgx + 3 cos 2 x

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

y "\u003d (tgx 2 + 3 tgx + 1)" \u003d (tgx 2) "+ (3 tgx)" + 1 "\u003d \u003d (tgx 2)" + 3 · (tgx) "+ 0 \u003d (tgx 2)" + 3 cos 2 x

We proceed to finding the derivative of a complex function (t g x 2) ":

f "(g (x)) \u003d (tg (g (x)))" \u003d 1 cos 2 g (x) \u003d 1 cos 2 (x 2) g "(x) \u003d (x 2)" \u003d 2 x 2 - 1 \u003d 2 x ⇒ (tgx 2) "\u003d f" (g (x)) g "(x) \u003d 2 x cos 2 (x 2)

We get that y "\u003d (t g x 2 + 3 t g x + 1)" \u003d (t g x 2) "+ 3 cos 2 x \u003d 2 x cos 2 (x 2) + 3 cos 2 x

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

Example 5

For example, consider a complex function of the form y \u003d 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 \u003d f (g (x)), where f is a function of the logarithm to base 3, and g (x) is considered the sum of two functions of the form h (x) \u003d x 2 + 3 cos 3 (2 x + 1) + 7 ex 2 + 3 3 and k (x) \u003d ln 2 x (x 2 + 1). Obviously, y \u003d f (h (x) + k (x)).

Consider the function h (x). This is the ratio l (x) \u003d x 2 + 3 cos 3 (2 x + 1) + 7 to m (x) \u003d e x 2 + 3 3

We have that l (x) \u003d x 2 + 3 cos 2 (2 x + 1) + 7 \u003d n (x) + p (x) is the sum of two functions n (x) \u003d 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 3, and p 1 is a cubing function, p 2 as a cosine function, p 3 (x) \u003d 2 x + 1 - a linear function.

We got that m (x) \u003d ex 2 + 3 3 \u003d q (x) + r (x) is the sum of two functions q (x) \u003d ex 2 and r (x) \u003d 3 3, where q (x) \u003d q 1 (q 2 (x)) is a complex function, q 1 is a function with exponential function, q 2 (x) \u003d x 2 is a power function.

This shows that h (x) \u003d l (x) m (x) \u003d n (x) + p (x) q (x) + r (x) \u003d 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 can be seen that the function is represented as a complex function s (x) \u003d ln 2 x \u003d s 1 ( s 2 (x)) with rational integer t (x) \u003d x 2 + 1, where s 1 is the squaring function and s 2 (x) \u003d ln x is logarithmic with base e.

Hence it follows that the expression takes the form k (x) \u003d s (x) t (x) \u003d s 1 (s 2 (x)) t (x).

Then we get that

y \u003d log 3 x 2 + 3 cos 3 (2 x + 1) + 7 ex 2 + 3 3 + ln 2 x (x 2 + 1) \u003d \u003d fn (x) + 3 p 1 (p 2 (p 3 (x))) q 1 (q 2 (x)) \u003d r (x) + s 1 (s 2 (x)) t (x)

By function structures, it became clear how and what formulas should be used to simplify an expression when differentiating it. To familiarize yourself with such problems and for the concept of their solution, it is necessary to turn to the point of differentiating a function, that is, finding its derivative.

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

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

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

Understanding. First of all, let's pay attention to the recording. Here we have two functions - and, moreover, the function, figuratively speaking, is embedded in the function. A function of this kind (when one function is nested within another) is called a complex function.

I will call the function external functionand the function - an internal (or nested) function.

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

In order 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 integer expression, so it will not be possible to find the derivative immediately 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 you cannot “tear apart” a sine:

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

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

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

Imagine that we need to calculate the value of an expression at on a calculator (instead of one, there can be any number).

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

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

After we Figured out with internal and external functions, it's time to apply the rule of differentiation of a complex function .

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

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

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

Well, it is quite obvious that

The result of applying the formula in the final design looks like this:

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

If there is any misunderstanding, write the solution down 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 is an internal one. To do this, try (mentally or on a draft) to calculate the value of the expression at. What should be done first? First of all, you need to calculate what the base is equal to: which means that the polynomial is the internal function:

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

According to the formula , first you need to find the derivative of the external function, in this case, from the degree. We are looking 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 of differentiation of a complex function following:

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

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

Example 4

Find the derivative of a function

This is an example for a stand alone solution (answer at the end of the tutorial).

To consolidate the understanding of the derivative of a complex function, I will give an example without comments, try to figure it out on your own, speculate where is the external and where is the internal function, why are the tasks 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 degree. Thus, first we bring the function into a form appropriate for differentiation:

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

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

Done. You can also bring the expression to a common denominator in brackets and write everything down in one fraction. Nice, of course, but when cumbersome long derivatives are obtained, it is better not to do this (it is 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 a stand alone solution (answer at the end of the tutorial).

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

Example 8

Find the derivative of a function

Here you can use the rule of differentiation of the private , 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 outside the sign of the derivative, and raise the cosine to the numerator:

Cosine is an internal function, exponentiation is an external function.
We use our rule :

Find the derivative of the internal function, reset the cosine back down:

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

Example 9

Find the derivative of a function

This is an example for a stand alone solution (answer at the end of the tutorial).

So far, we've looked at cases where we only had one attachment in a complex function. In practical tasks, you can often find derivatives, where, like nesting dolls, one into another, 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. Trying to evaluate the expression using the test value. How would we count on a calculator?

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

Then this arcsine of one should be squared:

And finally, raise the 7 to the power:

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

We start to solve

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