The proof of the formula for the derivative of a complex function is given. Cases where a complex function depends on one or two variables are considered in detail. A generalization is made to the case of an arbitrary number of variables.
Here we present the derivation of the following formulas for the derivative of a complex function.
If , then
.
If , then
.
If , then
.
Derivative of a complex function of one variable
Let a function of a variable x be represented as a complex function in the following form:
,
where and there are some functions. The function is differentiable for some value of the variable x . The function is differentiable for the value of the variable .
Then the complex (composite) function is differentiable at the point x and its derivative is determined by the formula:
(1)
.
Formula (1) can also be written as follows:
;
.
Proof
Let us introduce the following notation.
;
.
Here there is a function of variables and , there is a function of variables and . But we will omit the arguments of these functions so as not to clutter up the calculations.
Since the functions and are differentiable at the points x and , respectively, then at these points there are derivatives of these functions, which are the following limits:
;
.
Consider the following function:
.
For a fixed value of the variable u , is a function of . It's obvious that
.
Then
.
Since the function is a differentiable function at the point , then it is continuous at that point. That's why
.
Then
.
Now we find the derivative.
.
The formula has been proven.
Consequence
If a function of variable x can be represented as a complex function of a complex function
,
then its derivative is determined by the formula
.
Here , and there are some differentiable functions.
To prove this formula, we sequentially calculate the derivative according to the rule of differentiation of a complex function.
Consider a complex function
.
Its derivative
.
Consider the original function
.
Its derivative
.
Derivative of a complex function in two variables
Now let a complex function depend on several variables. First consider case of a complex function of two variables.
Let the function depending on the variable x be represented as a complex function of two variables in the following form:
,
where
and there are differentiable functions for some value of the variable x ;
is a function of two variables, differentiable at the point , . Then the complex function is defined in some neighborhood of the point and has a derivative, which is determined by the formula:
(2)
.
Proof
Since the functions and are differentiable at the point , they are defined in some neighborhood of this point, are continuous at the point, and their derivatives at the point exist, which are the following limits:
;
.
Here
;
.
Due to the continuity of these functions at a point, we have:
;
.
Since the function is differentiable at the point , it is defined in some neighborhood of this point, is continuous at this point, and its increment can be written in the following form:
(3)
.
Here
- function increment when its arguments are incremented by the values and ;
;
- partial derivatives of the function with respect to the variables and .
For fixed values of and , and there are functions of the variables and . They tend to zero as and :
;
.
Since and , then
;
.
Function increment :
.
:
.
Substitute (3):
.
The formula has been proven.
Derivative of a complex function of several variables
The above derivation is easily generalized to the case when the number of variables of a complex function is greater than two.
For example, if f is function of three variables, then
,
where
, and there are differentiable functions for some value of the variable x ;
is a differentiable function, in three variables, at the point , , .
Then, from the definition of differentiability of the function , we have:
(4)
.
Since, due to continuity,
;
;
,
then
;
;
.
Dividing (4) by and passing to the limit , we obtain:
.
And finally, consider the most general case.
Let a function of a variable x be represented as a complex function of n variables in the following form:
,
where
there are differentiable functions for some value of the variable x ;
- differentiable function of n variables at a point
,
,
... , .
Then
.
Definition. Let the function \(y = f(x) \) be defined in some interval containing the point \(x_0 \) inside. Let's increment \(\Delta x \) to the argument so as not to leave this interval. Find the corresponding increment of the function \(\Delta y \) (when passing from the point \(x_0 \) to the point \(x_0 + \Delta x \)) and compose the relation \(\frac(\Delta y)(\Delta x) \). If there is a limit of this relation at \(\Delta x \rightarrow 0 \), then the specified limit is called derivative function\(y=f(x) \) at the point \(x_0 \) and denote \(f"(x_0) \).
$$ \lim_(\Delta x \to 0) \frac(\Delta y)(\Delta x) = f"(x_0) $$
The symbol y is often used to denote the derivative. Note that y" = f(x) is a new function, but naturally associated with the function y = f(x), defined at all points x at which the above limit exists . This function is called like this: derivative of the function y \u003d f (x).
The geometric meaning of the derivative consists of the following. If a tangent that is not parallel to the y axis can be drawn to the graph of the function y \u003d f (x) at a point with the abscissa x \u003d a, then f (a) expresses the slope of the tangent:
\(k = f"(a)\)
Since \(k = tg(a) \), the equality \(f"(a) = tg(a) \) is true.
And now we interpret the definition of the derivative in terms of approximate equalities. Let the function \(y = f(x) \) have a derivative at a particular point \(x \):
$$ \lim_(\Delta x \to 0) \frac(\Delta y)(\Delta x) = f"(x) $$
This means that near the point x, the approximate equality \(\frac(\Delta y)(\Delta x) \approx f"(x) \), i.e. \(\Delta y \approx f"(x) \cdot\Deltax\). The meaningful meaning of the obtained approximate equality is as follows: the increment of the function is “almost proportional” to the increment of the argument, and the coefficient of proportionality is the value of the derivative at a given point x. For example, for the function \(y = x^2 \) the approximate equality \(\Delta y \approx 2x \cdot \Delta x \) is valid. If we carefully analyze the definition of the derivative, we will find that it contains an algorithm for finding it.
Let's formulate it.
How to find the derivative of the function y \u003d f (x) ?
1. Fix value \(x \), find \(f(x) \)
2. Increment \(x \) argument \(\Delta x \), move to a new point \(x+ \Delta x \), find \(f(x+ \Delta x) \)
3. Find the function increment: \(\Delta y = f(x + \Delta x) - f(x) \)
4. Compose the relation \(\frac(\Delta y)(\Delta x) \)
5. Calculate $$ \lim_(\Delta x \to 0) \frac(\Delta y)(\Delta x) $$
This limit is the derivative of the function at x.
If the function y = f(x) has a derivative at the point x, then it is called differentiable at the point x. The procedure for finding the derivative of the function y \u003d f (x) is called differentiation functions y = f(x).
Let us discuss the following question: how are the continuity and differentiability of a function at a point related?
Let the function y = f(x) be differentiable at the point x. Then a tangent can be drawn to the graph of the function at the point M (x; f (x)) and, recall, the slope of the tangent is equal to f "(x). Such a graph cannot "break" at the point M, i.e., the function must be continuous at x.
It was reasoning "on the fingers". Let us present a more rigorous argument. If the function y = f(x) is differentiable at the point x, then the approximate equality \(\Delta y \approx f"(x) \cdot \Delta x \) holds. zero, then \(\Delta y \) will also tend to zero, and this is the condition for the continuity of the function at a point.
So, if a function is differentiable at a point x, then it is also continuous at that point.
The converse is not true. For example: function y = |x| is continuous everywhere, in particular at the point x = 0, but the tangent to the graph of the function at the “joint point” (0; 0) does not exist. If at some point it is impossible to draw a tangent to the function graph, then there is no derivative at this point.
One more example. The function \(y=\sqrt(x) \) is continuous on the entire number line, including at the point x = 0. And the tangent to the graph of the function exists at any point, including at the point x = 0. But at this point the tangent coincides with the y-axis, that is, it is perpendicular to the abscissa axis, its equation has the form x \u003d 0. There is no slope for such a straight line, which means that \ (f "(0) \) does not exist either
So, we got acquainted with a new property of a function - differentiability. How can you tell if a function is differentiable from the graph of a function?
The answer is actually given above. If at some point a tangent can be drawn to the graph of a function that is not perpendicular to the x-axis, then at this point the function is differentiable. If at some point the tangent to the graph of the function does not exist or it is perpendicular to the x-axis, then at this point the function is not differentiable.
Differentiation rules
The operation of finding the derivative is called differentiation. When performing this operation, you often have to work with quotients, sums, products of functions, as well as with "functions of functions", that is, complex functions. Based on the definition of the derivative, we can derive differentiation rules that facilitate this work. If C is a constant number and f=f(x), g=g(x) are some differentiable functions, then the following are true differentiation rules:
$$ f"_x(g(x)) = f"_g \cdot g"_x $$
Table of derivatives of some functions
$$ \left(\frac(1)(x) \right) " = -\frac(1)(x^2) $$ $$ (\sqrt(x)) " = \frac(1)(2\ sqrt(x)) $$ $$ \left(x^a \right) " = a x^(a-1) $$ $$ \left(a^x \right) " = a^x \cdot \ln a $$ $$ \left(e^x \right) " = e^x $$ $$ (\ln x)" = \frac(1)(x) $$ $$ (\log_a x)" = \frac (1)(x\ln a) $$ $$ (\sin x)" = \cos x $$ $$ (\cos x)" = -\sin x $$ $$ (\text(tg) x) " = \frac(1)(\cos^2 x) $$ $$ (\text(ctg) x)" = -\frac(1)(\sin^2 x) $$ $$ (\arcsin x) " = \frac(1)(\sqrt(1-x^2)) $$ $$ (\arccos x)" = \frac(-1)(\sqrt(1-x^2)) $$ $$ (\text(arctg) x)" = \frac(1)(1+x^2) $$ $$ (\text(arctg) x)" = \frac(-1)(1+x^2) $ $complex derivatives. Logarithmic derivative.
Derivative of exponential function
We continue to improve our differentiation technique. In this lesson, we will consolidate the material covered, consider more complex derivatives, and also get acquainted with new tricks and tricks for finding the derivative, in particular, with the logarithmic derivative.
Those readers who have a low level of preparation should refer to the article How to find the derivative? Solution examples which will allow you to raise your skills almost from scratch. Next, you need to carefully study the page Derivative of a compound function, understand and resolve all the examples I have given. This lesson is logically the third in a row, and after mastering it, you will confidently differentiate fairly complex functions. It is undesirable to stick to the position “Where else? Yes, and that's enough! ”, Since all the examples and solutions are taken from real tests and are often found in practice.
Let's start with repetition. On the lesson Derivative of a compound function we have considered a number of examples with detailed comments. In the course of studying differential calculus and other sections of mathematical analysis, you will have to differentiate very often, and it is not always convenient (and not always necessary) to paint examples in great detail. Therefore, we will practice in the oral finding of derivatives. The most suitable "candidates" for this are derivatives of the simplest of complex functions, for example:
According to the rule of differentiation of a complex function 
:
When studying other topics of matan in the future, such a detailed record is most often not required, it is assumed that the student is able to find similar derivatives on autopilot. Let's imagine that at 3 o'clock in the morning the phone rang, and a pleasant voice asked: "What is the derivative of the tangent of two x?". This should be followed by an almost instantaneous and polite response: 
.
The first example will be immediately intended for an independent solution.
Example 1
Find the following derivatives orally, in one step, for example: . To complete the task, you only need to use table of derivatives of elementary functions(if she hasn't already remembered). If you have any difficulties, I recommend re-reading the lesson Derivative of a compound function.
, , ,
, 
, , 
, , ,
, , 
,
, , 
,
, , ,
, 
,
Answers at the end of the lesson
Complex derivatives
After preliminary artillery preparation, examples with 3-4-5 attachments of functions will be less scary. Perhaps the following two examples will seem complicated to some, but if they are understood (someone suffers), then almost everything else in differential calculus will seem like a child's joke.
Example 2
Find the derivative of a function 
As already noted, when finding the derivative of a complex function, first of all, it is necessary right UNDERSTAND INVESTMENTS. In cases where there are doubts, I remind you of a useful trick: we take the experimental value "x", for example, and try (mentally or on a draft) to substitute this value into the "terrible expression".
1) First we need to calculate the expression, so the sum is the deepest nesting.
2) Then you need to calculate the logarithm:
4) Then cube the cosine:
5) At the fifth step, the difference:
6) And finally, the outermost function is the square root: 
Complex Function Differentiation Formula 
are applied in reverse order, from the outermost function to the innermost. We decide:
Seems to be no error...
(1) We take the derivative of the square root.
(2) We take the derivative of the difference using the rule 
(3) The derivative of the triple is equal to zero. In the second term, we take the derivative of the degree (cube).
(4) We take the derivative of the cosine.
(5) We take the derivative of the logarithm.
(6) Finally, we take the derivative of the deepest nesting .
It may seem too difficult, but this is not the most brutal example. Take, for example, Kuznetsov's collection and you will appreciate all the charm and simplicity of the analyzed derivative. I noticed that they like to give a similar thing at the exam to check whether the student understands how to find the derivative of a complex function, or does not understand.
The following example is for a standalone solution.
Example 3
Find the derivative of a function
Hint: First we apply the rules of linearity and the rule of differentiation of the product
Full solution and answer at the end of the lesson.
It's time to move on to something more compact and prettier.
It is not uncommon for a situation where the product of not two, but three functions is given in an example. How to find the derivative of the product of three factors?
Example 4
Find the derivative of a function 
First, we look, but is it possible to turn the product of three functions into a product of two functions? For example, if we had two polynomials in the product, then we could open the brackets. But in this example, all functions are different: degree, exponent and logarithm.
In such cases, it is necessary successively apply the product differentiation rule 
twice
The trick is that for "y" we denote the product of two functions: , and for "ve" - the logarithm:. Why can this be done? Is it 
- this is not the product of two factors and the rule does not work?! There is nothing complicated:
Now it remains to apply the rule a second time 
to bracket:
You can still pervert and take something out of the brackets, but in this case it is better to leave the answer in this form - it will be easier to check.
The above example can be solved in the second way: 
Both solutions are absolutely equivalent.
Example 5
Find the derivative of a function
This is an example for an independent solution, in the sample it is solved in the first way.
Consider similar examples with fractions.
Example 6
Find the derivative of a function 
Here you can go in several ways:
Or like this:
But the solution can be written more compactly if, first of all, we use the rule of differentiation of the quotient 
, taking for the whole numerator:
In principle, the example is solved, and if it is left in this form, it will not be a mistake. But if you have time, it is always advisable to check on a draft, but is it possible to simplify the answer? We bring the expression of the numerator to a common denominator and get rid of the three-story fraction:
The disadvantage of additional simplifications is that there is a risk of making a mistake not when finding a derivative, but when banal school transformations. On the other hand, teachers often reject the assignment and ask to “bring it to mind” the derivative.
A simpler example for a do-it-yourself solution:
Example 7
Find the derivative of a function
We continue to master the techniques for finding the derivative, and now we will consider a typical case when a “terrible” logarithm is proposed for differentiation
Example 8
Find the derivative of a function
Here you can go a long way, using the rule of differentiation of a complex function:
But the very first step immediately plunges you into despondency - you have to take an unpleasant derivative of a fractional degree, and then also from a fraction.
That's why before how to take the derivative of the “fancy” logarithm, it is previously simplified using well-known school properties:
! If you have a practice notebook handy, copy these formulas right there. If you don't have a notebook, draw them on a piece of paper, as the rest of the lesson's examples will revolve around these formulas.
The solution itself can be formulated like this:
Let's transform the function:
We find the derivative: 
The preliminary transformation of the function itself greatly simplified the solution. Thus, when a similar logarithm is proposed for differentiation, it is always advisable to “break it down”.
And now a couple of simple examples for an independent solution:
Example 9
Find the derivative of a function 
Example 10
Find the derivative of a function
All transformations and answers at the end of the lesson.
logarithmic derivative
If the derivative of logarithms is such sweet music, then the question arises, is it possible in some cases to organize the logarithm artificially? Can! And even necessary.
Example 11
Find the derivative of a function 
Similar examples we have recently considered. What to do? One can successively apply the rule of differentiation of the quotient, and then the rule of differentiation of the product. The disadvantage of this method is that you get a huge three-story fraction, which you don’t want to deal with at all.
But in theory and practice there is such a wonderful thing as the logarithmic derivative. Logarithms can be organized artificially by "hanging" them on both sides:
Now you need to “break down” the logarithm of the right side as much as possible (formulas in front of your eyes?). I will describe this process in great detail:
Let's start with the differentiation.
We conclude both parts with a stroke:
The derivative of the right side is quite simple, I will not comment on it, because if you are reading this text, you should be able to handle it with confidence.
What about the left side?
On the left side we have complex function. I foresee the question: “Why, is there one letter “y” under the logarithm?”.
The fact is that this "one letter y" - IS A FUNCTION IN ITSELF(if it is not very clear, refer to the article Derivative of a function implicitly specified). Therefore, the logarithm is an external function, and "y" is an internal function. And we use the compound function differentiation rule 
:
On the left side, as if by magic, we have a derivative. Further, according to the rule of proportion, we throw the “y” from the denominator of the left side to the top of the right side:
And now we remember what kind of "game"-function we talked about when differentiating? Let's look at the condition: 
Final answer: 
Example 12
Find the derivative of a function
This is a do-it-yourself example. Sample design of an example of this type at the end of the lesson.
With the help of the logarithmic derivative, it was possible to solve any of the examples No. 4-7, another thing is that the functions there are simpler, and, perhaps, the use of the logarithmic derivative is not very justified.
Derivative of exponential function
We have not considered this function yet. An exponential function is a function that has and the degree and base depend on "x". A classic example that will be given to you in any textbook or at any lecture:
How to find the derivative of an exponential function?
It is necessary to use the technique just considered - the logarithmic derivative. We hang logarithms on both sides:
As a rule, the degree is taken out from under the logarithm on the right side:
As a result, on the right side we have a product of two functions, which will be differentiated according to the standard formula 
.
We find the derivative, for this we enclose both parts under strokes:
The next steps are easy:
Finally: 
If some transformation is not entirely clear, please carefully re-read the explanations of Example #11.
In practical tasks, the exponential function will always be more complicated than the considered lecture example.
Example 13
Find the derivative of a function
We use the logarithmic derivative. 
On the right side we have a constant and the product of two factors - "x" and "logarithm of the logarithm of x" (another logarithm is nested under the logarithm). When differentiating a constant, as we remember, it is better to immediately take it out of the sign of the derivative so that it does not get in the way; and, of course, apply the familiar rule 
:
As you can see, the algorithm for applying the logarithmic derivative does not contain any special tricks or tricks, and finding the derivative of the exponential function is usually not associated with "torment".
If we follow the definition, then the derivative of a function at a point is the limit of the increment ratio of the function Δ y to the increment of the argument Δ x:
Everything seems to be clear. But try to calculate by this formula, say, the derivative of the function f(x) = x 2 + (2x+ 3) · e x sin x. If you do everything by definition, then after a couple of pages of calculations you will simply fall asleep. Therefore, there are simpler and more effective ways.
To begin with, we note that the so-called elementary functions can be distinguished from the whole variety of functions. These are relatively simple expressions, the derivatives of which have long been calculated and entered in the table. Such functions are easy enough to remember, along with their derivatives.
Derivatives of elementary functions
Elementary functions are everything listed below. The derivatives of these functions must be known by heart. Moreover, it is not difficult to memorize them - that's why they are elementary.
So, the derivatives of elementary functions:
| Name | Function | Derivative |
| Constant | f(x) = C, C ∈ R | 0 (yes, yes, zero!) |
| Degree with rational exponent | f(x) = x n | n · x n − 1 |
| Sinus | f(x) = sin x | cos x |
| Cosine | f(x) = cos x | − sin x(minus sine) |
| Tangent | f(x) = tg x | 1/cos 2 x |
| Cotangent | f(x) = ctg x | − 1/sin2 x |
| natural logarithm | f(x) = log x | 1/x |
| Arbitrary logarithm | f(x) = log a x | 1/(x ln a) |
| Exponential function | f(x) = e x | e x(nothing changed) |
If an elementary function is multiplied by an arbitrary constant, then the derivative of the new function is also easily calculated:
(C · f)’ = C · f ’.
In general, constants can be taken out of the sign of the derivative. For example:
(2x 3)' = 2 ( x 3)' = 2 3 x 2 = 6x 2 .
Obviously, elementary functions can be added to each other, multiplied, divided, and much more. This is how new functions will appear, no longer very elementary, but also differentiable according to certain rules. These rules are discussed below.
Derivative of sum and difference
Let the functions f(x) and g(x), whose derivatives are known to us. For example, you can take the elementary functions discussed above. Then you can find the derivative of the sum and difference of these functions:
- (f + g)’ = f ’ + g ’
- (f − g)’ = f ’ − g ’
So, the derivative of the sum (difference) of two functions is equal to the sum (difference) of the derivatives. There may be more terms. For example, ( f + g + h)’ = f ’ + g ’ + h ’.
Strictly speaking, there is no concept of "subtraction" in algebra. There is a concept of "negative element". Therefore, the difference f − g can be rewritten as a sum f+ (−1) g, and then only one formula remains - the derivative of the sum.
f(x) = x 2 + sinx; g(x) = x 4 + 2x 2 − 3.
Function f(x) is the sum of two elementary functions, so:
f ’(x) = (x 2+ sin x)’ = (x 2)' + (sin x)’ = 2x+ cosx;
We argue similarly for the function g(x). Only there are already three terms (from the point of view of algebra):
g ’(x) = (x 4 + 2x 2 − 3)’ = (x 4 + 2x 2 + (−3))’ = (x 4)’ + (2x 2)’ + (−3)’ = 4x 3 + 4x + 0 = 4x · ( x 2 + 1).
Answer:
f ’(x) = 2x+ cosx;
g ’(x) = 4x · ( x
2 + 1).
Derivative of a product
Mathematics is a logical science, so many people believe that if the derivative of the sum is equal to the sum of the derivatives, then the derivative of the product strike"\u003e equal to the product of derivatives. But figs to you! The derivative of the product is calculated using a completely different formula. Namely:
(f · g) ’ = f ’ · g + f · g ’
The formula is simple, but often forgotten. And not only schoolchildren, but also students. The result is incorrectly solved problems.
A task. Find derivatives of functions: f(x) = x 3 cosx; g(x) = (x 2 + 7x− 7) · e x .
Function f(x) is a product of two elementary functions, so everything is simple:
f ’(x) = (x 3 cos x)’ = (x 3)' cos x + x 3 (cos x)’ = 3x 2 cos x + x 3 (−sin x) = x 2 (3cos x − x sin x)
Function g(x) the first multiplier is a little more complicated, but the general scheme does not change from this. Obviously, the first multiplier of the function g(x) is a polynomial, and its derivative is the derivative of the sum. We have:
g ’(x) = ((x 2 + 7x− 7) · e x)’ = (x 2 + 7x− 7)' · e x + (x 2 + 7x− 7) ( e x)’ = (2x+ 7) · e x + (x 2 + 7x− 7) · e x = e x(2 x + 7 + x 2 + 7x −7) = (x 2 + 9x) · e x = x(x+ 9) · e x .
Answer:
f ’(x) = x 2 (3cos x − x sin x);
g ’(x) = x(x+ 9) · e
x
.
Note that the last step is to factorize the derivative. Formally, this is not necessary, but most derivatives are not calculated on their own, but to explore the function. This means that further the derivative will be equated to zero, its signs will be found out, and so on. For such a case, it is better to have an expression decomposed into factors.
If there are two functions f(x) and g(x), and g(x) ≠ 0 on the set of interest to us, we can define a new function h(x) = f(x)/g(x). For such a function, you can also find the derivative:
Not weak, right? Where did the minus come from? Why g 2? But like this! This is one of the most complex formulas - you can’t figure it out without a bottle. Therefore, it is better to study it with specific examples.
A task. Find derivatives of functions:
There are elementary functions in the numerator and denominator of each fraction, so all we need is the formula for the derivative of the quotient:
By tradition, we factor the numerator into factors - this will greatly simplify the answer:
A complex function is not necessarily a formula half a kilometer long. For example, it suffices to take the function f(x) = sin x and replace the variable x, say, on x 2+ln x. It turns out f(x) = sin ( x 2+ln x) is a complex function. She also has a derivative, but it will not work to find it according to the rules discussed above.
How to be? In such cases, the replacement of a variable and the formula for the derivative of a complex function help:
f ’(x) = f ’(t) · t', if x is replaced by t(x).
As a rule, the situation with the understanding of this formula is even more sad than with the derivative of the quotient. Therefore, it is also better to explain it with specific examples, with a detailed description of each step.
A task. Find derivatives of functions: f(x) = e 2x + 3 ; g(x) = sin ( x 2+ln x)
Note that if in the function f(x) instead of expression 2 x+ 3 will be easy x, then we get an elementary function f(x) = e x. Therefore, we make a substitution: let 2 x + 3 = t, f(x) = f(t) = e t. We are looking for the derivative of a complex function by the formula:
f ’(x) = f ’(t) · t ’ = (e t)’ · t ’ = e t · t ’
And now - attention! Performing a reverse substitution: t = 2x+ 3. We get:
f ’(x) = e t · t ’ = e 2x+ 3 (2 x + 3)’ = e 2x+ 3 2 = 2 e 2x + 3
Now let's look at the function g(x). Obviously needs to be replaced. x 2+ln x = t. We have:
g ’(x) = g ’(t) · t' = (sin t)’ · t' = cos t · t ’
Reverse replacement: t = x 2+ln x. Then:
g ’(x) = cos ( x 2+ln x) · ( x 2+ln x)' = cos ( x 2+ln x) · (2 x + 1/x).
That's all! As can be seen from the last expression, the whole problem has been reduced to calculating the derivative of the sum.
Answer:
f ’(x) = 2 e
2x + 3 ;
g ’(x) = (2x + 1/x) cos ( x 2+ln x).
Very often in my lessons, instead of the term “derivative”, I use the word “stroke”. For example, the stroke of the sum is equal to the sum of the strokes. Is that clearer? Well, that's good.
Thus, the calculation of the derivative comes down to getting rid of these very strokes according to the rules discussed above. As a final example, let's return to the derivative power with a rational exponent:
(x n)’ = n · x n − 1
Few know that in the role n may well be a fractional number. For example, the root is x 0.5 . But what if there is something tricky under the root? Again, a complex function will turn out - they like to give such constructions in tests and exams.
A task. Find the derivative of a function:
First, let's rewrite the root as a power with a rational exponent:
f(x) = (x 2 + 8x − 7) 0,5 .
Now we make a substitution: let x 2 + 8x − 7 = t. We find the derivative by the formula:
f ’(x) = f ’(t) · t ’ = (t 0.5)' t' = 0.5 t−0.5 t ’.
We make a reverse substitution: t = x 2 + 8x− 7. We have:
f ’(x) = 0.5 ( x 2 + 8x− 7) −0.5 ( x 2 + 8x− 7)' = 0.5 (2 x+ 8) ( x 2 + 8x − 7) −0,5 .
Finally, back to the roots:
In the "old" textbooks, it is also called the "chain" rule. So if y \u003d f (u), and u \u003d φ (x), that is
y \u003d f (φ (x))
complex - compound function (composition of functions) then
where 
, after calculation is considered at u = φ (x).
Note that here we took "different" compositions from the same functions, and the result of differentiation naturally turned out to be dependent on the order of "mixing".
The chain rule naturally extends to the composition of three or more functions. In this case, there will be three or more “links” in the “chain” that makes up the derivative, respectively. Here is an analogy with multiplication: “we have” - a table of derivatives; "there" - multiplication table; "we" - chain rule and "there" - the rule of multiplication "column". When calculating such “complex” derivatives, of course, no auxiliary arguments (u¸v, etc.) are introduced, but, having noted for themselves the number and sequence of functions participating in the composition, they “string” the corresponding links in the indicated order.
. Here, five operations are performed with "x" to obtain the value of "y", that is, a composition of five functions takes place: "external" (the last of them) - exponential - e ; then in reverse order is a power law. (♦) 2 ; trigonometric sin (); power. () 3 and finally the logarithmic ln.(). That's why
The following examples will “kill pairs of birds with one stone”: we will practice differentiating complex functions and supplement the table of derivatives of elementary functions. So:
4. For a power function - y \u003d x α - rewriting it using the well-known "basic logarithmic identity" - b \u003d e ln b - in the form x α \u003d x α ln x we get
5. For an arbitrary exponential function, using the same technique, we will have
6. For an arbitrary logarithmic function, using the well-known formula for the transition to a new base, we successively obtain
.
7. To differentiate the tangent (cotangent), we use the rule for differentiating the quotient:
To obtain derivatives of inverse trigonometric functions, we use the relation which is satisfied by the derivatives of two mutually inverse functions, that is, the functions φ (x) and f (x) connected by the relations:
Here is the ratio
It is from this formula for mutually inverse functions
and 
,
In the end, we summarize these and some other, just as easily obtained derivatives, in the following table.
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