Functions of a 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 can 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)) is a complex function, q 1 is a function with an exponential, 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 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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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 the 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 the detailed process for 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 we 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 the 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 obtained result 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 control work, 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.

Very easy to remember.

Well, let’s not go far, let’s immediately consider the inverse function. Which function is the inverse of the exponential function? Logarithm:

In our case, the base is the number:

Such a logarithm (that is, a logarithm with a base) is called “natural”, and we use a special notation for it: we write instead.

What is it equal to? Of course.

The derivative of the natural logarithm is also very simple:

Examples:

Find the derivative of the function.
What is the derivative of the function?

Answers: The exponential and natural logarithm are uniquely simple functions from a derivative perspective. Exponential and logarithmic functions with any other base will have a different derivative, which we will analyze later, after we go through the rules of differentiation.

Rules of differentiation

Rules of what? Again a new term, again?!...

Differentiation is the process of finding the derivative.

That's all. What else can you call this process in one word? Not derivative... The differential of mathematicians is the same increment of a function at. This term comes from the Latin differentia - difference. Here.

When deriving all these rules, we will use two functions, for example, and. We will also need formulas for their increments:

There are 5 rules in total.

The constant is taken out of the derivative sign.

If - some constant number (constant), then.

Obviously, this rule also works for the difference: .

Let's prove it. Let it be, or simpler.

Examples.

Find the derivatives of the functions:

at a point;
at a point;
at a point;
at the point.

Solutions:

(the derivative is the same at all points, since it is a linear function, remember?);

Derivative of the product

Everything is similar here: let’s introduce a new function and find its increment:

Derivative:

Examples:

Find the derivatives of the functions and;
Find the derivative of the function at a point.

Solutions:

Derivative of an exponential function

Now your knowledge is enough to learn how to find the derivative of any exponential function, and not just exponents (have you forgotten what that is yet?).

So, where is some number.

We already know the derivative of the function, so let's try to reduce our function to a new base:

To do this, we will use a simple rule: . Then:

Well, it worked. Now try to find the derivative, and don't forget that this function is complex.

Did it work?

Here, check yourself:

The formula turned out to be very similar to the derivative of an exponent: as it was, it remains the same, only a factor appeared, which is just a number, but not a variable.

Examples:
Find the derivatives of the functions:

Answers:

This is just a number that cannot be calculated without a calculator, that is, it cannot be written down in a simpler form. Therefore, we leave it in this form in the answer.

Note that here is the quotient of two functions, so we apply the corresponding differentiation rule:

In this example, the product of two functions:

Derivative of a logarithmic function

It’s similar here: you already know the derivative of the natural logarithm:

Therefore, to find an arbitrary logarithm with a different base, for example:

We need to reduce this logarithm to the base. How do you change the base of a logarithm? I hope you remember this formula:

Only now we will write instead:

The denominator is simply a constant (a constant number, without a variable). The derivative is obtained very simply:

Derivatives of exponential and logarithmic functions are almost never found in the Unified State Examination, but it will not be superfluous to know them.

Derivative of a complex function.

What is a "complex function"? No, this is not a logarithm, and not an arctangent. These functions can be difficult to understand (although if you find the logarithm difficult, read the topic “Logarithms” and you will be fine), but from a mathematical point of view, the word “complex” does not mean “difficult”.

Imagine a small conveyor belt: two people are sitting and doing some actions with some objects. For example, the first one wraps a chocolate bar in a wrapper, and the second one ties it with a ribbon. The result is a composite object: a chocolate bar wrapped and tied with a ribbon. To eat a chocolate bar, you need to do the reverse steps in reverse order.

Let's create a similar mathematical pipeline: first we will find the cosine of a number, and then square the resulting number. So, we are given a number (chocolate), I find its cosine (wrapper), and then you square what I got (tie it with a ribbon). What happened? Function. This is an example of a complex function: when, to find its value, we perform the first action directly with the variable, and then a second action with what resulted from the first.

In other words, a complex function is a function whose argument is another function: .

For our example, .

We can easily do the same steps in reverse order: first you square it, and I then look for the cosine of the resulting number: . It’s easy to guess that the result will almost always be different. An important feature of complex functions: when the order of actions changes, the function changes.

Second example: (same thing). .

The action we do last will be called "external" function, and the action performed first - accordingly "internal" function(these are informal names, I use them only to explain the material in simple language).

Try to determine for yourself which function is external and which internal:

Answers: Separating inner and outer functions is very similar to changing variables: for example, in a function

What action will we perform first? First, let's calculate the sine, and only then cube it. This means that it is an internal function, but an external one.
And the original function is their composition: .
Internal: ; external: .
Examination: .
Internal: ; external: .
Examination: .
Internal: ; external: .
Examination: .
Internal: ; external: .
Examination: .

We change variables and get a function.

Well, now we will extract our chocolate bar and look for the derivative. The procedure is always reversed: first we look for the derivative of the outer function, then we multiply the result by the derivative of the inner function. In relation to the original example, it looks like this:

Another example:

So, let's finally formulate the official rule:

Algorithm for finding the derivative of a complex function:

It seems simple, right?

Let's check with examples:

Solutions:

1) Internal: ;

External: ;

2) Internal: ;

(just don’t try to cut it by now! Nothing comes out from under the cosine, remember?)

3) Internal: ;

External: ;

It is immediately clear that this is a three-level complex function: after all, this is already a complex function in itself, and we also extract the root from it, that is, we perform the third action (put the chocolate in a wrapper and with a ribbon in the briefcase). But there is no reason to be afraid: we will still “unpack” this function in the same order as usual: from the end.

That is, first we differentiate the root, then the cosine, and only then the expression in brackets. And then we multiply it all.

In such cases, it is convenient to number the actions. That is, let's imagine what we know. In what order will we perform actions to calculate the value of this expression? Let's look at an example:

The later the action is performed, the more “external” the corresponding function will be. The sequence of actions is the same as before:

Here the nesting is generally 4-level. Let's determine the course of action.

1. Radical expression. .

2. Root. .

3. Sine. .

4. Square. .

5. Putting it all together:

DERIVATIVE. BRIEFLY ABOUT THE MAIN THINGS

Derivative of a function- the ratio of the increment of the function to the increment of the argument for an infinitesimal increment of the argument:

Basic derivatives:

Rules of differentiation:

The constant is taken out of the derivative sign:

Derivative of the sum:

Derivative of the product:

Derivative of the quotient:

Derivative of a complex function:

Algorithm for finding the derivative of a complex function:

We define the “internal” function and find its derivative.
We define the “external” function and find its derivative.
We multiply the results of the first and second points.

On which we examined the simplest derivatives, and also became acquainted with the rules of differentiation and some technical techniques for finding derivatives. Thus, if you are not very good with derivatives of functions or some points in this article are not entirely clear, then first read the above lesson. Please get in a serious mood - the material is not simple, but I will still try to present it simply and clearly.

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

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

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

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

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

To clarify the situation, consider:

Example 1

Find the derivative of a function

Under the sine we have not just the letter “X”, but an entire expression, so finding the derivative right away from the table will not work. We also notice that it is impossible to apply the first four rules here, there seems to be a difference, but the fact is that the sine cannot be “torn into pieces”:

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

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

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

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

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

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

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

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

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

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

Well, it's quite obvious that

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

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

If there is any misunderstanding, write the solution down on paper and read the explanations again.

Example 2

Find the derivative of a function

Example 3

Find the derivative of a function

As always, we write down:

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

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

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

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

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

Example 4

Find the derivative of a function

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

To consolidate your understanding of the derivative of a complex function, I will give an example without comments, try to figure it out on your own, reason where the external and where the internal function is, why the tasks are solved this way?

Example 5

a) Find the derivative of the function

b) Find the derivative of the function

Example 6

Find the derivative of a function

Here we have a root, and in order to differentiate the root, it must be represented as a power. Thus, first we bring the function into the form appropriate for differentiation:

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

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

Ready. You can also reduce the expression to a common denominator in brackets and write everything down as one fraction. It’s beautiful, of course, but when you get cumbersome long derivatives, it’s better not to do this (it’s easy to get confused, make an unnecessary mistake, and it will be inconvenient for the teacher to check).

Example 7

Find the derivative of a function

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

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

Example 8

Find the derivative of a function

Here you can use the rule of differentiation of the quotient , but it is much more profitable to find the derivative through the rule of differentiation of a complex function:

We prepare the function for differentiation - we move the minus out of the derivative sign, and raise the cosine into the numerator:

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

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

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

Example 9

Find the derivative of a function

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

So far we have looked at cases where we had only one nesting in a complex function. In practical tasks, you can often find derivatives, where, like nesting dolls, one inside the other, 3 or even 4-5 functions are nested at once.

Example 10

Find the derivative of a function

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

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

This arcsine of one should then be squared:

And finally, we raise seven to a power:

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

Let's start deciding

According to the rule First you need to take the derivative of the outer function. We look at the table of derivatives and find the derivative of the exponential function: The only difference is that instead of “x” we have a complex expression, which does not negate the validity of this formula. So, the result of applying the rule for differentiating a complex function next.

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 resolve 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. In class Derivative of a complex function We 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 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

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 is:

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 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) 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 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

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 possible – 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:

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

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 “break up” the logarithm of the right side as much as possible (the formulas in front of 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 :