Derivative of a compound root. Complex derivatives. Logarithmic derivative. Derivative of a power-exponential function. More complex examples

And the theorem on the derivative of a complex function, the formulation of which is as follows:

Let 1) the function $u=\varphi (x)$ have at some point $x_0$ the derivative $u_(x)"=\varphi"(x_0)$, 2) the function $y=f(u)$ have at the corresponding at the point $u_0=\varphi (x_0)$ the derivative $y_(u)"=f"(u)$. Then complex function$y=f\left(\varphi (x) \right)$ at the mentioned point will also have a derivative equal to the product of the derivatives of the functions $f(u)$ and $\varphi (x)$:

$$ \left(f(\varphi (x))\right)"=f_(u)"\left(\varphi (x_0) \right)\cdot \varphi"(x_0) $$

or, in shorter notation: $y_(x)"=y_(u)"\cdot u_(x)"$.

In the examples in this section, all functions have the form $y=f(x)$ (i.e., we consider only functions of one variable $x$). Accordingly, in all examples the derivative $y"$ is taken with respect to the variable $x$. To emphasize that the derivative is taken with respect to the variable $x$, $y"_x$ is often written instead of $y"$.

Examples No. 1, No. 2 and No. 3 outline detailed process finding the derivative of complex functions. Example No. 4 is intended for a more complete understanding of the derivative table and it makes sense to familiarize yourself with it.

It is advisable, after studying the material in examples No. 1-3, to move on to independently solving examples No. 5, No. 6 and No. 7. Examples #5, #6 and #7 contain a short solution so that the reader can check the correctness of his result.

Example No. 1

Find the derivative of the function $y=e^(\cos x)$.

We need to find the derivative of a complex function $y"$. Since $y=e^(\cos x)$, then $y"=\left(e^(\cos x)\right)"$. To find the derivative $ \left(e^(\cos x)\right)"$ we use formula No. 6 from the table of derivatives. In order to use formula No. 6, we need to take into account that in our case $u=\cos x$. The further solution consists in simply substituting the expression $\cos x$ instead of $u$ into formula No. 6:

$$ y"=\left(e^(\cos x) \right)"=e^(\cos x)\cdot (\cos x)" \tag (1.1)$$

Now we need to find the value of the expression $(\cos x)"$. We turn again to the table of derivatives, choosing formula No. 10 from it. Substituting $u=x$ into formula No. 10, we have: $(\cos x)"=-\ sin x\cdot x"$. Now let's continue equality (1.1), supplementing it with the result found:

$$ y"=\left(e^(\cos x) \right)"=e^(\cos x)\cdot (\cos x)"= e^(\cos x)\cdot (-\sin x \cdot x") \tag (1.2) $$

Since $x"=1$, we continue equality (1.2):

$$ y"=\left(e^(\cos x) \right)"=e^(\cos x)\cdot (\cos x)"= e^(\cos x)\cdot (-\sin x \cdot x")=e^(\cos x)\cdot (-\sin x\cdot 1)=-\sin x\cdot e^(\cos x) \tag (1.3) $$

So, from equality (1.3) we have: $y"=-\sin x\cdot e^(\cos x)$. Naturally, explanations and intermediate equalities are usually skipped, writing down the finding of the derivative in one line, as in the equality ( 1.3) So, the derivative of a complex function has been found, all that remains is to write down the answer.

Answer: $y"=-\sin x\cdot e^(\cos x)$.

Example No. 2

Find the derivative of the function $y=9\cdot \arctg^(12)(4\cdot \ln x)$.

We need to calculate the derivative $y"=\left(9\cdot \arctg^(12)(4\cdot \ln x) \right)"$. To begin with, we note that the constant (i.e. the number 9) can be taken out of the derivative sign:

$$ y"=\left(9\cdot \arctg^(12)(4\cdot \ln x) \right)"=9\cdot\left(\arctg^(12)(4\cdot \ln x) \right)" \tag (2.1) $$

Now let's turn to the expression $\left(\arctg^(12)(4\cdot \ln x) \right)"$. To make it easier to select the desired formula from the table of derivatives, I will present the expression in question in this form: $\left( \left(\arctg(4\cdot \ln x) \right)^(12)\right)"$. Now it is clear that it is necessary to use formula No. 2, i.e. $\left(u^\alpha \right)"=\alpha\cdot u^(\alpha-1)\cdot u"$. Let’s substitute $u=\arctg(4\cdot \ln x)$ and $\alpha=12$ into this formula:

Supplementing equality (2.1) with the result obtained, we have:

$$ y"=\left(9\cdot \arctg^(12)(4\cdot \ln x) \right)"=9\cdot\left(\arctg^(12)(4\cdot \ln x) \right)"= 108\cdot\left(\arctg(4\cdot \ln x) \right)^(11)\cdot (\arctg(4\cdot \ln x))" \tag (2.2) $$

In this situation, a mistake is often made when the solver at the first step chooses the formula $(\arctg \; u)"=\frac(1)(1+u^2)\cdot u"$ instead of the formula $\left(u^\ alpha \right)"=\alpha\cdot u^(\alpha-1)\cdot u"$. The point is that the derivative of the external function must come first. To understand which function will be external to the expression $\arctg^(12)(4\cdot 5^x)$, imagine that you are calculating the value of the expression $\arctg^(12)(4\cdot 5^x)$ at some value $x$. First you will calculate the value of $5^x$, then multiply the result by 4, getting $4\cdot 5^x$. Now we take the arctangent from this result, obtaining $\arctg(4\cdot 5^x)$. Then we raise the resulting number to the twelfth power, getting $\arctg^(12)(4\cdot 5^x)$. The last action, i.e. raising to the power of 12 will be an external function. And it is from this that we must begin to find the derivative, which was done in equality (2.2).

Now we need to find $(\arctg(4\cdot \ln x))"$. We use formula No. 19 of the derivatives table, substituting $u=4\cdot \ln x$ into it:

$$ (\arctg(4\cdot \ln x))"=\frac(1)(1+(4\cdot \ln x)^2)\cdot (4\cdot \ln x)" $$

Let's simplify the resulting expression a little, taking into account $(4\cdot \ln x)^2=4^2\cdot (\ln x)^2=16\cdot \ln^2 x$.

$$ (\arctg(4\cdot \ln x))"=\frac(1)(1+(4\cdot \ln x)^2)\cdot (4\cdot \ln x)"=\frac( 1)(1+16\cdot \ln^2 x)\cdot (4\cdot \ln x)" $$

Equality (2.2) will now become:

It remains to find $(4\cdot \ln x)"$. Let's take the constant (i.e. 4) out of the derivative sign: $(4\cdot \ln x)"=4\cdot (\ln x)"$. For In order to find $(\ln x)"$ we use formula No. 8, substituting $u=x$ into it: $(\ln x)"=\frac(1)(x)\cdot x"$. Since $x"=1$, then $(\ln x)"=\frac(1)(x)\cdot x"=\frac(1)(x)\cdot 1=\frac(1)(x )$ Substituting the result obtained into formula (2.3), we obtain:

$$ y"=\left(9\cdot \arctg^(12)(4\cdot \ln x) \right)"=9\cdot\left(\arctg^(12)(4\cdot \ln x) \right)"=\\ =108\cdot\left(\arctg(4\cdot \ln x) \right)^(11)\cdot (\arctg(4\cdot \ln x))"=108\cdot \left(\arctg(4\cdot \ln x) \right)^(11)\cdot \frac(1)(1+16\cdot \ln^2 x)\cdot (4\cdot \ln x)" =\\ =108\cdot \left(\arctg(4\cdot \ln x) \right)^(11)\cdot \frac(1)(1+16\cdot \ln^2 x)\cdot 4\ cdot \frac(1)(x)=432\cdot \frac(\arctg^(11)(4\cdot \ln x))(x\cdot (1+16\cdot \ln^2 x)).$ $

Let me remind you that the derivative of a complex function is most often found in one line, as written in the last equality. Therefore, when preparing standard calculations or tests It is not at all necessary to describe the solution in such detail.

Answer: $y"=432\cdot \frac(\arctg^(11)(4\cdot \ln x))(x\cdot (1+16\cdot \ln^2 x))$.

Example No. 3

Find $y"$ of the function $y=\sqrt(\sin^3(5\cdot9^x))$.

First, let's slightly transform the function $y$, expressing the radical (root) as a power: $y=\sqrt(\sin^3(5\cdot9^x))=\left(\sin(5\cdot 9^x) \right)^(\frac(3)(7))$. Now let's start finding the derivative. Since $y=\left(\sin(5\cdot 9^x)\right)^(\frac(3)(7))$, then:

$$ y"=\left(\left(\sin(5\cdot 9^x)\right)^(\frac(3)(7))\right)" \tag (3.1) $$

Let's use formula No. 2 from the table of derivatives, substituting $u=\sin(5\cdot 9^x)$ and $\alpha=\frac(3)(7)$ into it:

$$ \left(\left(\sin(5\cdot 9^x)\right)^(\frac(3)(7))\right)"= \frac(3)(7)\cdot \left( \sin(5\cdot 9^x)\right)^(\frac(3)(7)-1) (\sin(5\cdot 9^x))"=\frac(3)(7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) (\sin(5\cdot 9^x))" $$

Let us continue equality (3.1) using the result obtained:

$$ y"=\left(\left(\sin(5\cdot 9^x)\right)^(\frac(3)(7))\right)"=\frac(3)(7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) (\sin(5\cdot 9^x))" \tag (3.2) $$

Now we need to find $(\sin(5\cdot 9^x))"$. For this we use formula No. 9 from the table of derivatives, substituting $u=5\cdot 9^x$ into it:

$$ (\sin(5\cdot 9^x))"=\cos(5\cdot 9^x)\cdot(5\cdot 9^x)" $$

Having supplemented equality (3.2) with the result obtained, we have:

It remains to find $(5\cdot 9^x)"$. First, let's take the constant (the number $5$) outside the derivative sign, i.e. $(5\cdot 9^x)"=5\cdot (9^x) "$. To find the derivative $(9^x)"$, apply formula No. 5 of the table of derivatives, substituting $a=9$ and $u=x$ into it: $(9^x)"=9^x\cdot \ ln9\cdot x"$. Since $x"=1$, then $(9^x)"=9^x\cdot \ln9\cdot x"=9^x\cdot \ln9$. Now we can continue equality (3.3):

$$ y"=\left(\left(\sin(5\cdot 9^x)\right)^(\frac(3)(7))\right)"=\frac(3)(7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) (\sin(5\cdot 9^x))"=\\ =\frac(3) (7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) \cos(5\cdot 9^x)\cdot(5\cdot 9 ^x)"= \frac(3)(7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) \cos(5\cdot 9 ^x)\cdot 5\cdot 9^x\cdot \ln9=\\ =\frac(15\cdot \ln 9)(7)\cdot \left(\sin(5\cdot 9^x)\right) ^(-\frac(4)(7))\cdot \cos(5\cdot 9^x)\cdot 9^x. $$

We can again return from powers to radicals (i.e., roots), writing $\left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7))$ in the form $\ frac(1)(\left(\sin(5\cdot 9^x)\right)^(\frac(4)(7)))=\frac(1)(\sqrt(\sin^4(5\ cdot 9^x)))$. Then the derivative will be written in this form:

$$ y"=\frac(15\cdot \ln 9)(7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7))\cdot \cos(5\cdot 9^x)\cdot 9^x= \frac(15\cdot \ln 9)(7)\cdot \frac(\cos (5\cdot 9^x)\cdot 9^x) (\sqrt(\sin^4(5\cdot 9^x))).$$

Answer: $y"=\frac(15\cdot \ln 9)(7)\cdot \frac(\cos (5\cdot 9^x)\cdot 9^x)(\sqrt(\sin^4(5\ cdot 9^x)))$.

Example No. 4

Show that formulas No. 3 and No. 4 of the table of derivatives are a special case of formula No. 2 of this table.

Formula No. 2 of the table of derivatives contains the derivative of the function $u^\alpha$. Substituting $\alpha=-1$ into formula No. 2, we get:

$$(u^(-1))"=-1\cdot u^(-1-1)\cdot u"=-u^(-2)\cdot u"\tag (4.1)$$

Since $u^(-1)=\frac(1)(u)$ and $u^(-2)=\frac(1)(u^2)$, then equality (4.1) can be rewritten as follows: $ \left(\frac(1)(u) \right)"=-\frac(1)(u^2)\cdot u"$. This is formula No. 3 of the table of derivatives.

Let us turn again to formula No. 2 of the table of derivatives. Let's substitute $\alpha=\frac(1)(2)$ into it:

$$\left(u^(\frac(1)(2))\right)"=\frac(1)(2)\cdot u^(\frac(1)(2)-1)\cdot u" =\frac(1)(2)u^(-\frac(1)(2))\cdot u"\tag (4.2) $$

Since $u^(\frac(1)(2))=\sqrt(u)$ and $u^(-\frac(1)(2))=\frac(1)(u^(\frac( 1)(2)))=\frac(1)(\sqrt(u))$, then equality (4.2) can be rewritten as follows:

$$ (\sqrt(u))"=\frac(1)(2)\cdot \frac(1)(\sqrt(u))\cdot u"=\frac(1)(2\sqrt(u) )\cdot u" $$

The resulting equality $(\sqrt(u))"=\frac(1)(2\sqrt(u))\cdot u"$ is formula No. 4 of the table of derivatives. As you can see, formulas No. 3 and No. 4 of the derivative table are obtained from formula No. 2 by substituting the corresponding $\alpha$ value.

After preliminary artillery preparation, examples with 3-4-5 nestings of functions will be less scary. The following two examples may seem complicated to some, but if you understand them (someone will suffer), 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 your investments. In cases where there are doubts, I remind you of a useful technique: we take the experimental value of “x”, for example, and try (mentally or in a draft) to substitute this value into the “terrible expression”.

1) First we need to calculate the expression, which means the sum is the deepest embedding.

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:

Formula for differentiating a complex function are applied in reverse order, from the outermost function to the innermost. We decide:

It seems without errors:

1) Take the derivative of square root.

2) Take the derivative of the difference using the rule

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

4) Take the derivative of the cosine.

6) And finally, we take the derivative of the deepest embedding.

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 beauty and simplicity of the analyzed derivative. I noticed that they like to give a similar thing in an exam to check whether a student understands how to find the derivative of a complex function or does not understand.

The following example is for independent decision.

Example 3

Find the derivative of a function

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

Full solution and answer at the end of the lesson.

It's time to move on to something smaller and nicer.
It is not uncommon for an example to show the 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 we look, is it 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 open the brackets. But in the example under consideration, all the functions are different: degree, exponent and logarithm.

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

The trick is that by “y” we denote the product of two functions: , and by “ve” we denote the logarithm: . Why can this be done? Is it really - 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 bracket:

You can still be perverted and take something out of brackets, but in 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 using the first method.

Let's look at similar examples with fractions.

Example 6

Find the derivative of a function

There are several ways you can go here:

Or like this:

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

In principle, the example is solved, and if it is left as is, it will not be an error. But if you have time, it is always advisable to check on a draft to see if the answer can be simplified?

Let's reduce the expression of the numerator to a common denominator and get rid of the three-story structure of the fraction:

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

A simpler example to solve on your own:

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

Functions complex type do not always fit the definition of a complex function. If there is a function of the form y = 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 = 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 derivative table and differentiation rules significantly reduces the time for finding the derivative.

Basic definitions

Definition 1

A complex function is one 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 it is a cotangent function, then g(x) = ln x is the natural logarithm function. We find 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) = x 2 + 2 x - 3 is considered an entire rational function, we obtain that f (g (x)) = (x 2 + 2 x - 3) 4 .

Obviously g(x) can be complex. From the example y = sin 2 x + 1 x 3 - 5 it is clear that the value of g has the cube root of the fraction. This expression allowed to be denoted as y = f (f 1 (f 2 (x))) . From where we have that f is a sine function, and f 1 is a function located under the square root, f 2 (x) = 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 = 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 conditions of the problem. To solve, use 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.

Solution

The condition shows that f is a squaring function, and g(x) = 2 x + 1 is considered a linear function.

Let's 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) = (2 x + 1) " = (2 x) " + 1 " = 2 x " + 0 = 2 1 x 1 - 1 = 2 ⇒ (f (g (x))) " = f " (g (x)) g " (x) = 2 (2 x + 1) 2 = 8 x + 4

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

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

From here we have that

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

The results were the same.

When solving problems of this type, 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 = sin 2 x and y = sin x 2.

Solution

The first function notation 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 denotes a power function. It follows that we write the product of a complex function as

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

The formula for the derivative y = 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))) )) · . . . fn "(x)

Example 3

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

Solution

This example shows the difficulty of writing and determining the location of functions. Then y = 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, function with logarithm and base e, arctangent and linear function.

From the formula for defining 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)

We get what we need to find

f " (f 1 (f 2 (f 3 (f 4 (x))))) as the derivative of the sine according to 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)) .
f 1 " (f 2 (f 3 (f 4 (x)))) as the 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) .
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) .
f 3 " (f 4 (x)) as the derivative of the arctangent, then f 3 " (f 4 (x)) = 1 1 + (2 x) 2 = 1 1 + 4 x 2.
When finding the derivative f 4 (x) = 2 x, remove 2 from 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) = (2 x) " = 2 x " = 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)

Analysis of such functions is reminiscent of nesting dolls. Differentiation rules cannot always be applied explicitly using a derivative table. Often you need to use a formula for finding derivatives of complex functions.

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

Example 4

It is necessary to consider giving 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 use the formula for a complex derivative:

f " (g (x)) = (g 2 (x) + 3 g (x) + 1) " = (g 2 (x)) " + (3 g (x)) " + 1 " = = 2 · g 2 - 1 (x) + 3 g " (x) + 0 = 2 g (x) + 3 1 g 1 - 1 (x) = = 2 g (x) + 3 = 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 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) = x 2 and f, which is a tangent function. To do this, differentiate by 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 = 2 x ⇒ (t g x 2) " = f " (g (x)) g " (x) = 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

Functions of a complex type can be included in complex functions, and complex functions themselves can be components of functions of a complex type.

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 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) = 3 cos 3 (2 x + 1) , where p (x) = 3 p 1 (p 2 (p 3 (x))) is a complex function with numerical coefficient 3, and p 1 is a cube function, p 2 by a cosine function, p 3 (x) = 2 x + 1 by a 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)) - complex function, q 1 - function with exponent, q 2 (x) = x 2 - 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 moving to an expression of the form k (x) = ln 2 x · (x 2 + 1) = s (x) · t (x), it is clear that the function is presented in the form of a complex s (x) = ln 2 x = s 1 ( s 2 (x)) with a rational integer 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)

Based on the structures of the function, it became clear how and what formulas need to be used to simplify the expression when differentiating it. To become familiar 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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It is not entirely correct to call functions of a complex type the term “complex function”. For example, it looks very impressive, but this function is not complicated, unlike.

In this article we will understand the concept of a complex function, learn how to identify it as part of elementary functions, we will give a formula for finding its derivative and consider in detail the solution of typical examples.

When solving examples, we will constantly use the table of derivatives and differentiation rules, so keep them before your eyes.

Complex function is a function whose argument is also a function.

From our point of view, this definition is the most understandable. Conventionally, it can be denoted as f(g(x)) . That is, g(x) is like an argument of the function f(g(x)) .

For example, let f be the arctangent function and g(x) = lnx be the natural logarithm function, then the complex function f(g(x)) is arctan(lnx) . Another example: f is the function of raising to the fourth power, and is an entire rational function (see ), then .

In turn, g(x) can also be a complex function. For example, . Conventionally, such an expression can be denoted as . Here f is the sine function, is the square root function, - fractional rational function. It is logical to assume that the degree of nesting of functions can be any finite natural number.

You can often hear a complex function called composition of functions.

Formula for finding the derivative of a complex function.

Example.

Find the derivative of a complex function.

Solution.

In this example, f is the squaring function, and g(x) = 2x+1 – linear function.

Here is the detailed solution using the complex function derivative formula:

Let's find this derivative by first simplifying the form of the original function.

Hence,

As you can see, the results are the same.

Try not to confuse which function is f and which is g(x) .

Let's illustrate this with an example to show your attention.

Example.

Find derivatives of complex functions and .

Solution.

In the first case, f is the squaring function and g(x) is the sine function, so
.

In the second case, f is a sine function, and is a power function. Therefore, by the formula for the product of a complex function we have

The derivative formula for a function has the form

Example.

Differentiate function .

Solution.

In this example, the complex function can be conventionally written as , where is the sine function, the third power function, the base e logarithm function, the arctangent function and the linear function, respectively.

According to the formula for the derivative of a complex function

Now we find

Let's put together the obtained intermediate results:

There is nothing scary, analyze complex functions like nesting dolls.

This could be the end of the article, if not for one thing...

It is advisable to clearly understand when to apply the rules of differentiation and the table of derivatives, and when to apply the formula for the derivative of a complex function.

BE EXTREMELY CAREFUL NOW. We will talk about the difference between complex functions and complex functions. Your success in finding derivatives will depend on how much you see this difference.

Let's start with simple examples. Function can be considered as complex: g(x) = tanx , . Therefore, you can immediately apply the formula for the derivative of a complex function

And here is the function It can no longer be called complex.

This function is the sum of three functions, 3tgx and 1. Although - is a complex function: - a power function (quadratic parabola), and f is a tangent function. Therefore, first we apply the sum differentiation formula:

It remains to find the derivative of the complex function:

That's why .

We hope you get the gist.

If we look more broadly, it can be argued that functions of a complex type can be part of complex functions, and complex functions can be components of functions of a complex type.

As an example, let us analyze the function into its component parts .

Firstly, this is a complex function that can be represented as , where f is the base 3 logarithm function, and g(x) is the sum of two functions And . That is, .

Secondly, let's deal with the function h(x) . It represents a relationship to .

This is the sum of two functions and , Where - a complex function with a numerical coefficient of 3. - cube function, - cosine function, - linear function.

This is the sum of two functions and , where - complex function, - exponential function, - power function.

Thus, .

Third, go to , which is the product of a complex function and the whole rational function

The squaring function is the logarithm function to base e.

Hence, .

Let's summarize:

Now the structure of the function is clear and it has become clear which formulas and in what sequence to apply when differentiating it.

In the section on differentiating a function (finding the derivative) you can familiarize yourself with the solution to similar problems.

Complex derivatives. Logarithmic derivative.
Derivative of a power-exponential function

We continue to improve our differentiation technique. In this lesson, we will consolidate the material we have covered, look at more complex derivatives, and also get acquainted with new techniques and tricks for finding a 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? Examples of solutions, which will allow you to 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 fairly complex functions. It is undesirable to take the position of “Where else? That’s enough!”, since all examples and solutions are taken from real tests and are often encountered in practice.

Let's start with repetition. At the lesson Derivative of a complex function We looked at a number of examples with detailed comments. During the study of differential calculus and other sections mathematical analysis– you will have to differentiate very often, and it is not always convenient (and not always necessary) to describe 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:

According to the rule of differentiation of complex functions :

When studying other matan topics in the future, such a detailed record is most often not required; it is assumed that the student knows how to find such 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’s?” This should be followed by an almost instant and polite response: .

The first example will be immediately intended for independent solution.

Example 1

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

, , ,
, , ,
, , ,

, , ,

, ,

Answers at the end of the lesson

Complex derivatives

Example 2

Find the derivative of a function

1) First we need to calculate the expression, which means the sum is the deepest embedding.

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:

Formula for differentiating a complex function are applied in reverse order, from the outermost function to the innermost. We decide:

There seem to be no errors...

(1) Take the derivative of the square root.

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

(3) The derivative of a 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) And finally, we take the derivative of the deepest embedding.

The following example is for you to solve on your own.

Example 3

Find the derivative of a function

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

Full solution and answer at the end of the lesson.

It's time to move on to something smaller and nicer.
It is not uncommon for an example to show the 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

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

The trick is that by “y” we denote the product of two functions: , and by “ve” we denote the logarithm: . Why can this be done? Is it really – 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 bracket:

You can also get twisted and put something out of brackets, but in this case it’s better to leave the answer exactly 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 using the first method.

Let's look at similar examples with fractions.

Example 6

Find the derivative of a function

There are several ways you can go here:

Or like this:

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

In principle, the example is solved, and if it is left as is, it will not be an error. But if you have time, it is always advisable to check on a draft to see if the answer can be simplified? Let us reduce the expression of the numerator to a common denominator and let's get rid of the three-story fraction:

A simpler example to solve on your own:

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 the long way, using the rule for differentiating a complex function:

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

That's why before how to take the derivative of a “sophisticated” logarithm, it is first simplified using well-known school properties:

! If you have a practice notebook at hand, copy these formulas directly there. If you don't have a notebook, copy them onto a piece of paper, since the remaining examples of the lesson will revolve around these formulas.

The solution itself can be written something like this:

Let's transform the function:

Finding the derivative:

Pre-converting 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 you to solve on your own:

Example 9

Find the derivative of a function

Example 10

Find the derivative of a function

All transformations and answers are 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 recently looked at similar examples. What to do? You can sequentially 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 end up with 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 a function can take negative values, then, generally speaking, you need to use modules: , which will disappear as a result of differentiation. However, the current design is also acceptable, where by default it is taken into account complex meanings. But if in all rigor, then in both cases a reservation should be made that.

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

Let's start with differentiation.
We conclude both parts under the prime:

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 be able to handle it confidently.

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 game” - IS ITSELF A FUNCTION(if it is not very clear, refer to the article Derivative of a function specified implicitly). Therefore, the logarithm is an external function, and the “y” is an internal function. And we use the rule for differentiating a complex function :

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

And now let’s remember what kind of “player”-function we talked about during differentiation? Let's look at the condition:

Final answer:

Example 12

Find the derivative of a function

This is an example for you to solve on your own. A sample design of an example of this type is at the end of the lesson.

Using 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 a power-exponential function

We have not considered this function yet. A power-exponential function is a function for which both the degree and the base depend on the “x”. A classic example that will be given to you in any textbook or lecture:

How to find the derivative of a power-exponential function?

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

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

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

We find the derivative; to do this, we enclose both parts under strokes:

Further actions are simple:

Finally:

If any conversion is not entirely clear, please re-read the explanations of Example No. 11 carefully.

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

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 logarithm x” (another logarithm is nested under the logarithm). When differentiating, as we remember, it is better to immediately move the constant out of the derivative sign so that it does not get in the way; and, of course, we apply the familiar rule :