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 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 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:
$$ y"=\left(9\cdot \arctg^(12)(4\cdot \ln x) \right)"=9\cdot\left(\arctg^(12)(4\cdot \ln x) \right)"=\\ =108\cdot\left(\arctg(4\cdot \ln x) \right)^(11)\cdot (\arctg(4\cdot \ln x))"=108\cdot \left(\arctg(4\cdot \ln x) \right)^(11)\cdot \frac(1)(1+16\cdot \ln^2 x)\cdot (4\cdot \ln x)" \tag (2.3) $$
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:
$$ y"=\left(9\cdot \arctg^(12)(4\cdot \ln x) \right)"=9\cdot\left(\arctg^(12)(4\cdot \ln x) \right)"=\\ =108\cdot\left(\arctg(4\cdot \ln x) \right)^(11)\cdot (\arctg(4\cdot \ln x))"=108\cdot \left(\arctg(4\cdot \ln x) \right)^(11)\cdot \frac(1)(1+16\cdot \ln^2 x)\cdot (4\cdot \ln x)" =\\ =108\cdot \left(\arctg(4\cdot \ln x) \right)^(11)\cdot \frac(1)(1+16\cdot \ln^2 x)\cdot 4\ cdot \frac(1)(x)=432\cdot \frac(\arctg^(11)(4\cdot \ln x))(x\cdot (1+16\cdot \ln^2 x)). $
Let me remind you that the derivative of a complex function is most often found in one line, as written in the last equality. Therefore, when preparing standard calculations or tests It is not at all necessary to describe the solution in such detail.
Answer: $y"=432\cdot \frac(\arctg^(11)(4\cdot \ln x))(x\cdot (1+16\cdot \ln^2 x))$.
Example No. 3
Find $y"$ of the function $y=\sqrt(\sin^3(5\cdot9^x))$.
First, let's slightly transform the function $y$, expressing the radical (root) as a power: $y=\sqrt(\sin^3(5\cdot9^x))=\left(\sin(5\cdot 9^x) \right)^(\frac(3)(7))$. Now let's start finding the derivative. Since $y=\left(\sin(5\cdot 9^x)\right)^(\frac(3)(7))$, then:
$$ 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)" $$
Having supplemented equality (3.2) with the result obtained, we have:
$$ 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)" \tag (3.3) $$
It remains to find $(5\cdot 9^x)"$. First, let's take the constant (the number $5$) outside the derivative sign, i.e. $(5\cdot 9^x)"=5\cdot (9^x) "$. To find the derivative $(9^x)"$, apply formula No. 5 of the table of derivatives, substituting $a=9$ and $u=x$ into it: $(9^x)"=9^x\cdot \ ln9\cdot x"$. Since $x"=1$, then $(9^x)"=9^x\cdot \ln9\cdot x"=9^x\cdot \ln9$. Now we can continue equality (3.3):
$$ 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 derivatives table.
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.
If g(x) And f(u) – differentiable functions of their arguments, respectively, at points x And u= g(x), then the complex function is also differentiable at the point x and is found by the formula
A typical mistake when solving problems on derivatives is mechanical transfer of differentiation rules simple functions for complex functions. Let's learn to avoid this mistake.
Example 2. Find the derivative of a function
Wrong solution: calculate natural logarithm each term in brackets and look for the sum of the derivatives:
Correct solution: again we determine where the “apple” is and where the “minced meat” is. Here the natural logarithm of the expression in parentheses is an “apple”, that is, a function over the intermediate argument u, and the expression in brackets is “minced meat”, that is, an intermediate argument u by independent variable x.
Then (using formula 14 from the derivatives table)
In many real-life problems, the expression with a logarithm can be somewhat more complicated, which is why there is a lesson
Example 3. Find the derivative of a function
Wrong solution:
The right decision. Once again we determine where the “apple” is and where the “mincemeat” is. Here, the cosine of the expression in brackets (formula 7 in the table of derivatives) is an “apple”, it is prepared in mode 1, which affects only it, and the expression in brackets (the derivative of the degree is number 3 in the table of derivatives) is “minced meat”, it is prepared under mode 2, which affects only it. And as always, we connect two derivatives with the product sign. Result:
Derivative of complex logarithmic function- a frequent task on tests, so we strongly recommend that you attend the lesson “Derivative of a logarithmic function.”
The first examples were on complex functions, in which the intermediate argument on the independent variable was a simple function. But in practical tasks It is often necessary to find the derivative of a complex function, where the intermediate argument is either itself a complex function or contains such a function. What to do in such cases? Find derivatives of such functions using tables and differentiation rules. When the derivative of the intermediate argument is found, it is simply substituted into right place formulas. Below are two examples of how this is done.
In addition, it is useful to know the following. If a complex function can be represented as a chain of three functions
then its derivative should be found as the product of the derivatives of each of these functions:
Many of your homework assignments may require you to open your guides in new windows. Actions with powers and roots And Operations with fractions .
Example 4. Find the derivative of a function
We apply the rule of differentiation of a complex function, not forgetting that in the resulting product of derivatives there is an intermediate argument with respect to the independent variable x does not change:
We prepare the second factor of the product and apply the rule for differentiating the sum:
The second term is the root, so
Thus, we found that the intermediate argument, which is a sum, contains a complex function as one of the terms: raising to a power is a complex function, and what is being raised to a power is an intermediate argument with respect to the independent variable x.
Therefore, we again apply the rule for differentiating a complex function:
We transform the degree of the first factor into a root, and when differentiating the second factor, do not forget that the derivative of the constant is equal to zero:
Now we can find the derivative of the intermediate argument needed to calculate the derivative of a complex function required in the problem statement y:
Example 5. Find the derivative of a function
First, we use the rule for differentiating the sum:
We obtained the sum of the derivatives of two complex functions. Let's find the first one:
Here, raising the sine to a power is a complex function, and the sine itself is an intermediate argument for the independent variable x. Therefore, we will use the rule of differentiation of a complex function, along the way taking the factor out of brackets :
Now we find the second term of the derivatives of the function y:
Here raising the cosine to a power is a complex function f, and the cosine itself is an intermediate argument in the independent variable x. Let us again use the rule for differentiating a complex function:
The result is the required derivative:
Table of derivatives of some complex functions
For complex functions, based on the rule of differentiation of a complex function, the formula for the derivative of a simple function takes a different form.
| 1. Derivative of a complex power function, Where u x |  |
| 2. Derivative of the root of the expression | |
| 3. Derivative exponential function |  |
| 4. Special case exponential function | |
| 5. Derivative of a logarithmic function with an arbitrary positive base A |  |
| 6. Derivative of a complex logarithmic function, where u– differentiable function of the argument x | |
| 7. Derivative of sine |  |
| 8. Derivative of cosine |  |
| 9. Derivative of tangent |  |
| 10. Derivative of cotangent |  |
| 11. Derivative of arcsine |  |
| 12. Derivative of arc cosine |  |
| 13. Derivative of arctangent |  |
| 14. Derivative of arc cotangent |  |
In “old” textbooks it is also called the “chain” rule. So if y = f (u), and u = φ (x), that is
y = f (φ (x))
complex - composite function (composition of functions) then
Where 
, after calculation is considered at u = φ (x).
Note that here we took “different” compositions from the same functions, and the result of differentiation naturally turned out to depend on the order of “mixing”.
The chain rule naturally extends to compositions of three or more functions. In this case, there will be three or more “links” in the “chain” that makes up the derivative. Here is an analogy with multiplication: “we have” a table of derivatives; “there” - multiplication table; “with us” is the chain rule and “there” is the “column” multiplication rule. When calculating such “complex” derivatives, no auxiliary arguments (u¸v, etc.), of course, are introduced, but, having noted for themselves the number and sequence of functions involved in the composition, the corresponding links are “strung” in the indicated order.
. Here, with the “x” to obtain the value of the “y”, five operations are performed, that is, there is a composition of five functions: “external” (the last of them) - exponential - e ; then in reverse order, power. (♦) 2 ; trigonometric sin(); sedate. () 3 and finally logarithmic ln.(). That's why
With the following examples we will “kill a couple of birds with one stone”: we will practice differentiating complex functions and add to the table of derivatives of elementary functions. So:
4. For a power function - y = x α - rewriting it using the well-known “basic logarithmic identity” - b=e ln b - in the form x α = x α ln x we obtain
5. For an arbitrary exponential function, using the same technique we will have
6. For an arbitrary logarithmic function, using the well-known formula for transition to a new base, we consistently obtain
.
7. To differentiate the tangent (cotangent), we use the rule of differentiation of quotients:
To obtain derivatives of inverse trigonometric functions, we use the relation that is satisfied by the derivatives of two mutually inverse functions, that is, the functions φ (x) and f (x) related by the relations:
This is the ratio
It is from this formula for mutually inverse functions
And 
,
Finally, let us summarize these and some other derivatives that are also easily obtained in the following table.
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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, 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. 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 decision.
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 given value into a "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:
There seem to be no errors...
(1) We take the derivative of 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 .
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, let’s see if it’s possible to turn the product of three functions into the product of two functions? For example, if we had two polynomials in the product, 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 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 the 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 function can accept 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 internal function. And we use the rule for differentiating a complex function 
:
On the left side, as if by magic magic wand 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 
:
The operation of finding the derivative is called differentiation.
As a result of solving problems of finding derivatives of the simplest (and not very simple) functions by defining the derivative as the limit of the ratio of the increment to the increment of the argument, a table of derivatives and precisely defined rules of differentiation appeared. The first to work in the field of finding derivatives were Isaac Newton (1643-1727) and Gottfried Wilhelm Leibniz (1646-1716).
Therefore, in our time, to find the derivative of any function, you do not need to calculate the above-mentioned limit of the ratio of the increment of the function to the increment of the argument, but you only need to use the table of derivatives and the rules of differentiation. The following algorithm is suitable for finding the derivative.
To find the derivative, you need an expression under the prime sign break down simple functions into components and determine what actions (product, sum, quotient) these functions are related. Next, we find the derivatives of elementary functions in the table of derivatives, and the formulas for the derivatives of the product, sum and quotient - in the rules of differentiation. The derivative table and differentiation rules are given after the first two examples.
Example 1. Find the derivative of a function
Solution. From the rules of differentiation we find out that the derivative of a sum of functions is the sum of derivative functions, i.e.
From the table of derivatives we find out that the derivative of “X” is equal to one, and the derivative of sine is equal to cosine. We substitute these values into the sum of derivatives and find the derivative required by the condition of the problem:
Example 2. Find the derivative of a function
Solution. We differentiate as a derivative of a sum in which the second term has a constant factor; it can be taken out of the derivative sign:
If questions still arise about where something comes from, they are usually cleared up after familiarization with the table of derivatives and the simplest rules of differentiation. We are moving on to them right now.
Table of derivatives of simple functions
| 1. Derivative of a constant (number). Any number (1, 2, 5, 200...) that is in the function expression. Always equal to zero. This is very important to remember, as it is required very often | |
| 2. Derivative of the independent variable. Most often "X". Always equal to one. This is also important to remember for a long time | |
| 3. Derivative of degree. When solving problems, you need to convert non-square roots into powers. | |
| 4. Derivative of a variable to the power -1 | |
| 5. Derivative of square root | |
| 6. Derivative of sine | |
| 7. Derivative of cosine |  |
| 8. Derivative of tangent |  |
| 9. Derivative of cotangent |  |
| 10. Derivative of arcsine |  |
| 11. Derivative of arccosine |  |
| 12. Derivative of arctangent |  |
| 13. Derivative of arc cotangent |  |
| 14. Derivative of the natural logarithm | |
| 15. Derivative of a logarithmic function |  |
| 16. Derivative of the exponent | |
| 17. Derivative of an exponential function |
Rules of differentiation
| 1. Derivative of a sum or difference |  |
| 2. Derivative of the product |  |
| 2a. Derivative of an expression multiplied by a constant factor | |
| 3. Derivative of the quotient |  |
| 4. Derivative of a complex function |  |
Rule 1.If the functions
are differentiable at some point, then the functions are differentiable at the same point
and
those. the derivative of an algebraic sum of functions is equal to the algebraic sum of the derivatives of these functions.
Consequence. If two differentiable functions differ by a constant term, then their derivatives are equal, i.e.
Rule 2.If the functions
are differentiable at some point, then their product is differentiable at the same point
and
those. The derivative of the product of two functions is equal to the sum of the products of each of these functions and the derivative of the other.
Corollary 1. The constant factor can be taken out of the sign of the derivative:
Corollary 2. The derivative of the product of several differentiable functions is equal to the sum of the products of the derivative of each factor and all the others.
For example, for three multipliers:
Rule 3.If the functions
differentiable at some point And , then at this point their quotient is also differentiableu/v , and
those. the derivative of the quotient of two functions is equal to a fraction, the numerator of which is the difference between the products of the denominator and the derivative of the numerator and the numerator and the derivative of the denominator, and the denominator is the square of the former numerator.
Where to look for things on other pages
When finding the derivative of a product and a quotient in real problems, it is always necessary to apply several differentiation rules at once, so there are more examples on these derivatives in the article"Derivative of the product and quotient of functions".
Comment. You should not confuse a constant (that is, a number) as a term in a sum and as a constant factor! In the case of a term, its derivative is equal to zero, and in the case of a constant factor, it is taken out of the sign of the derivatives. This typical mistake, which occurs on initial stage studying derivatives, but as they solve several one- and two-part examples, the average student no longer makes this mistake.
And if, when differentiating a product or quotient, you have a term u"v, in which u- a number, for example, 2 or 5, that is, a constant, then the derivative of this number will be equal to zero and, therefore, the entire term will be equal to zero (this case is discussed in example 10).
Another common mistake is mechanically solving the derivative of a complex function as the derivative of a simple function. That's why derivative of a complex function a separate article is devoted. But first we will learn to find derivatives of simple functions.
Along the way, you can’t do without transforming expressions. To do this, you may need to open the manual in new windows. Actions with powers and roots And Operations with fractions .
If you are looking for solutions to derivatives of fractions with powers and roots, that is, when the function looks like 
, then follow the lesson “Derivative of sums of fractions with powers and roots.”
If you have a task like 
, then you will take the lesson “Derivatives of simple trigonometric functions”.
Step-by-step examples - how to find the derivative
Example 3. Find the derivative of a function
Solution. We define the parts of the function expression: the entire expression represents a product, and its factors are sums, in the second of which one of the terms contains a constant factor. We apply the product differentiation rule: the derivative of the product of two functions is equal to the sum of the products of each of these functions by the derivative of the other:
Next, we apply the rule of sum differentiation: the derivative of an algebraic sum of functions is equal to the algebraic sum of the derivatives of these functions. In our case, in each sum the second term has a minus sign. In each sum we see both an independent variable, the derivative of which is equal to one, and a constant (number), the derivative of which is equal to zero. So, “X” turns into one, and minus 5 turns into zero. In the second expression, "x" is multiplied by 2, so we multiply two by the same unit as the derivative of "x". We obtain the following derivative values:
We substitute the found derivatives into the sum of products and obtain the derivative of the entire function required by the condition of the problem:
And you can check the solution to the derivative problem on.
Example 4. Find the derivative of a function
Solution. We are required to find the derivative of the quotient. We apply the formula for differentiating the quotient: the derivative of the quotient of two functions is equal to a fraction, the numerator of which is the difference between the products of the denominator and the derivative of the numerator and the numerator and the derivative of the denominator, and the denominator is the square of the former numerator. We get:
We have already found the derivative of the factors in the numerator in example 2. Let us also not forget that the product, which is the second factor in the numerator in the current example, is taken with a minus sign:
If you are looking for solutions to problems in which you need to find the derivative of a function, where there is a continuous pile of roots and powers, such as, for example, 
, then welcome to class "Derivative of sums of fractions with powers and roots" .
If you need to learn more about the derivatives of sines, cosines, tangents and others trigonometric functions, that is, when the function looks like , then a lesson for you "Derivatives of simple trigonometric functions" .
Example 5. Find the derivative of a function
Solution. In this function we see a product, one of the factors of which is the square root of the independent variable, the derivative of which we familiarized ourselves with in the table of derivatives. Using the rule for differentiating the product and the tabular value of the derivative of the square root, we obtain:
You can check the solution to the derivative problem at online derivatives calculator .
Example 6. Find the derivative of a function
Solution. In this function we see a quotient whose dividend is the square root of the independent variable. Using the rule for differentiating quotients, which we repeated and applied in example 4, and the tabular value of the derivative of the square root, we obtain:
To get rid of a fraction in the numerator, multiply the numerator and denominator by .
