Since you came here, you probably already saw this formula in the textbook
and make a face like this:
Friend, don't worry! In fact, everything is simply outrageous. You will definitely understand everything. Just one request - read the article slowly, try to understand every step. I wrote as simply and clearly as possible, but you still need to understand the idea. And be sure to solve the tasks from the article.
What is a complex function?
Imagine that you are moving to another apartment and therefore packing things into large boxes. Suppose you need to collect some small items, for example, school writing materials. If you just throw them into a huge box, they will get lost among other things. To avoid this, you first put them, for example, in a bag, which you then put in a large box, after which you seal it. This “complex” process is presented in the diagram below:
It would seem, what does mathematics have to do with it? Yes, despite the fact that a complex function is formed in EXACTLY THE SAME way! Only we “pack” not notebooks and pens, but \(x\), while the “packages” and “boxes” are different.
For example, let's take x and “pack” it into a function:
As a result, we get, of course, \(\cosx\). This is our “bag of things”. Now let’s put it in a “box” - pack it, for example, into a cubic function.
What will happen in the end? Yes, that’s right, there will be a “bag of things in a box,” that is, “cosine of X cubed.”
The resulting design is a complex function. It differs from simple one in that SEVERAL “influences” (packages) are applied to one X in a row and it turns out as if “function from function” - “packaging within packaging”.
IN school course There are very few types of these “packages”, only four:
Let's now “pack” X first into an exponential function with base 7, and then into a trigonometric function. We get:
\(x → 7^x → tg(7^x)\)
Now let’s “pack” X twice into trigonometric functions, first in , and then in:
\(x → sinx → cotg (sinx)\)
Simple, right?
Now write the functions yourself, where x:
- first it is “packed” into a cosine, and then into an exponential function with base \(3\);
- first to the fifth power, and then to the tangent;
- first to the logarithm to the base \(4\)
, then to the power \(-2\).
Find the answers to this task at the end of the article.
Can we “pack” X not two, but three times? No problem! And four, and five, and twenty-five times. Here, for example, is a function in which x is “packed” \(4\) times:
\(y=5^(\log_2(\sin(x^4)))\)
But such formulas will not be found in school practice (students are luckier - theirs may be more complicated☺).
"Unpacking" a complex function
Look at the previous function again. Can you figure out the “packing” sequence? What X was stuffed into first, what then, and so on until the very end. That is, which function is nested within which? Take a piece of paper and write down what you think. You can do this with a chain with arrows as we wrote above or in any other way.
Now the correct answer is: first, x was “packed” into the \(4\)th power, then the result was packed into a sine, it, in turn, was placed into the logarithm to the base \(2\), and in the end this whole construction was stuffed into a power fives.
That is, you need to unwind the sequence IN REVERSE ORDER. And here’s a hint on how to do it easier: immediately look at the X – you should dance from it. Let's look at a few examples.
For example, here is the following function: \(y=tg(\log_2x)\). We look at X - what happens to it first? Taken from him. And then? The tangent of the result is taken. The sequence will be the same:
\(x → \log_2x → tg(\log_2x)\)
Another example: \(y=\cos((x^3))\). Let's analyze - first we cubed X, and then took the cosine of the result. This means the sequence will be: \(x → x^3 → \cos((x^3))\). Pay attention, the function seems to be similar to the very first one (where it has pictures). But this is a completely different function: here in the cube is x (that is, \(\cos((x·x·x)))\), and there in the cube is the cosine \(x\) (that is, \(\cos x·\cosx·\cosx\)). This difference arises from different "packing" sequences.
The last example (with important information in it): \(y=\sin((2x+5))\). It's clear what they did here first arithmetic operations with x, then took the sine of the result: \(x → 2x+5 → \sin((2x+5))\). And this important point: despite the fact that arithmetic operations are not functions in themselves, here they also act as a way of “packing”. Let's delve a little deeper into this subtlety.
As I said above, in simple functions x is “packed” once, and in complex functions - two or more. Moreover, any combination of simple functions (that is, their sum, difference, multiplication or division) is also simple function. For example, \(x^7\) is a simple function and so is \(ctg x\). This means that all their combinations are simple functions:
\(x^7+ ctg x\) - simple,
\(x^7· cot x\) – simple,
\(\frac(x^7)(ctg x)\) – simple, etc.
However, if one more function is applied to such a combination, it will become a complex function, since there will be two “packages”. See diagram:
Okay, go ahead now. Write the sequence of “wrapping” functions:
\(y=cos((sinx))\)
\(y=5^(x^7)\)
\(y=arctg(11^x)\)
\(y=log_2(1+x)\)
The answers are again at the end of the article.
Internal and external functions
Why do we need to understand function nesting? What does this give us? The fact is that without such an analysis we will not be able to reliably find derivatives of the functions discussed above.
And in order to move on, we will need two more concepts: internal and external functions. This is very simple thing, moreover, in fact, we have already analyzed them above: if we recall our analogy at the very beginning, then the internal function is a “package”, and the external function is a “box”. Those. what X is “wrapped” in first is an internal function, and what the internal function is “wrapped” in is already external. Well, it’s clear why - she’s outside, that means external.
In this example: \(y=tg(log_2x)\), the function \(\log_2x\) is internal, and 
- external.
And in this: \(y=\cos((x^3+2x+1))\), \(x^3+2x+1\) is internal, and 
- external.
Complete the last practice of analyzing complex functions, and let's finally move on to what we were all started for - we will find derivatives of complex functions:
Fill in the blanks in the table:
Derivative of a complex function
Bravo to us, we finally got to the “boss” of this topic - actually, a derivative complex function, and specifically, to that very terrible formula from the beginning of the article.☺
\((f(g(x)))"=f"(g(x))\cdot g"(x)\)
This formula reads like this:
The derivative of a complex function is equal to the product of the derivative of the external function with respect to a constant internal function and the derivative of the internal function.
And immediately look at the parsing diagram, according to the words, so that you understand what to do with what:
I hope the terms “derivative” and “product” do not cause any difficulties. “Complex function” - we have already sorted it out. The catch in the “derivative” external function according to an unchanged internal one.” What it is?
Answer: This is the usual derivative of an external function, in which only the external function changes, and the internal one remains the same. Still not clear? Okay, let's use an example.
Let us have a function \(y=\sin(x^3)\). It is clear that the internal function here is \(x^3\), and the external 
. Let us now find the derivative of the exterior with respect to the constant interior.
Definition. Let the function \(y = f(x) \) be defined in a certain interval containing the point \(x_0\) within itself. Let's give the argument an increment \(\Delta x \) such that it does not leave this interval. Let's find the corresponding increment of the function \(\Delta y \) (when moving 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 to this ratio at \(\Delta x \rightarrow 0\), then the specified limit is called derivative of a 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 new feature, 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 = f(x).
Geometric meaning of derivative is as follows. If it is possible to draw a tangent to the graph of the function y = f(x) at the point with abscissa x=a, which is not parallel to the y-axis, then f(a) expresses the slope of the tangent:
\(k = f"(a)\)
Since \(k = tg(a) \), then the equality \(f"(a) = tan(a) \) is true.
Now let’s interpret the definition of derivative from the point of view of approximate equalities. Let the function \(y = f(x)\) have a derivative at a specific 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. \(\Delta y \approx f"(x) \cdot\Delta x\). The meaningful meaning of the resulting 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 in 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 a 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 = f(x)?
1. Fix the value of \(x\), find \(f(x)\)
2. Give the argument \(x\) an increment \(\Delta x\), go to new point\(x+ \Delta x \), find \(f(x+ \Delta x) \)
3. Find the increment of the function: \(\Delta y = f(x + \Delta x) - f(x) \)
4. Create 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 point x.
If a function y = f(x) has a derivative at a point x, then it is called differentiable at a point x. The procedure for finding the derivative of the function y = f(x) is called differentiation functions y = f(x).
Let us discuss the following question: how are continuity and differentiability of a function at a point related to each other?
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 point M(x; f(x)), and, recall, the angular coefficient of the tangent is equal to f "(x). Such a graph cannot “break” at point M, i.e. the function must be continuous at point x.
These were “hands-on” arguments. Let us give a more rigorous reasoning. 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. If in this equality \(\Delta x \) tends to zero, then \(\Delta y \) will 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 continuous at that point.
The reverse statement 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 “junction point” (0; 0) does not exist. If at some point a tangent cannot be drawn to the graph of a function, then the derivative does not exist at that 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, i.e., it is perpendicular to the abscissa axis, its equation has the form x = 0. Slope coefficient such a line does not have, which means that \(f"(0) \) does not exist either
So, we got acquainted with a new property of a function - differentiability. How can one conclude from the graph of a function that it is differentiable?
The answer is actually given above. If at some point it is possible to draw a tangent to the graph of a function that is not perpendicular to the abscissa axis, then at this point the function is differentiable. If at some point the tangent to the graph of a function does not exist or it is perpendicular to the abscissa axis, then at this point the function is not differentiable.
Rules of differentiation
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 “functions of functions,” that is, complex functions. Based on the definition of derivative, we can derive differentiation rules that make this work easier. If C - 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(arcctg) x)" = \frac(-1)(1+x^2) $ $On which we analyzed the simplest derivatives, and also got acquainted with the rules of differentiation and some technical methods 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 apply 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 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.
When 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 find the derivative of the external function (sine), look at the table of derivatives elementary functions and we notice that . All table formulas are also applicable if “x” is replaced with a complex expression, V 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:
Constant multiplier 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 exponentiation performed, therefore, 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. Looking for in the table the required formula: . We repeat again: any tabular formula valid not only for “x”, but also for complex expressions. 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 independent decision(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 give the expression in parentheses to common denominator 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 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, with the innermost function being the arcsine and the outermost function being 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 exponential function: The only difference is that instead of “X” we have complex expression, which does not negate the validity of this formula. So, the result of applying the rule for differentiating a complex function 
next.
After preliminary artillery preparation, examples with 3-4-5 nestings of functions will be less scary. Perhaps the following two examples will seem complicated to some, but if you understand them (someone will suffer), then almost everything else in differential calculus It 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 useful trick: we take the experimental meaning of “x”, for example, and try (mentally or in a draft) to substitute this meaning into the “terrible expression”.
1) First we need to calculate the expression, which means the sum is the deepest embedding.
2) Then you need to calculate the logarithm:
4) Then cube the cosine:
5) At the fifth step the difference:
6) And finally, the outermost function is the square root: 
Formula for differentiating a complex function 
will be used in reverse order, from the outermost function to the innermost. We decide:
It seems without errors:
1) Take the derivative of the square root.
2) Take the derivative of the difference using the rule 
3) The derivative of a triple is zero. In the second term we take the derivative of the degree (cube).
4) Take the derivative of the cosine.
6) And finally, we take the derivative of the deepest embedding.
It may seem too difficult, but this is not the most brutal example. Take, for example, Kuznetsov’s collection and you will appreciate all the beauty and simplicity of the analyzed derivative. I noticed that they like to give a similar thing in an exam to check whether a student understands how to find the derivative of a complex function or does not understand.
The following example is for 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 products of three multipliers?
Example 4
Find the derivative of a function 
First we look, is it possible to turn the product of three functions into the product of two functions? For example, if we had two polynomials in the product, then we could open the brackets. But in the example under consideration, all the functions are different: degree, exponent and logarithm.
In such cases it is necessary sequentially apply the product differentiation rule 
twice
The trick is that by “y” we denote the product of two functions: , and by “ve” we denote the logarithm: . Why can this be done? Is it really 
- this is not a product of two factors and the rule does not work?! There is nothing complicated:
Now it remains to apply the rule a second time to bracket:
You can 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's reduce the expression of the numerator to a common denominator and get rid of the three-story structure of the fraction:
The disadvantage of additional simplifications is that there is a risk of making a mistake not when finding the derivative, but during banal school transformations. On the other hand, teachers often reject the assignment and ask to “bring it to mind” the derivative.
A simpler example to solve on your own:
Example 7
Find the derivative of a function
We continue to master the methods of finding the derivative, and now we will consider a typical case when the “terrible” logarithm is proposed for differentiation
In this article we will talk about such an important mathematical concept as a complex function, and learn how to find the derivative of a complex function.
Before learning to find the derivative of a complex function, let's understand the concept of a complex function, what it is, “what it is eaten with,” and “how to cook it correctly.”
Let's consider arbitrary function, for example, like this:
Note that the argument on the right and left sides of the function equation is the same number, or expression.
Instead of a variable, we can put, for example, the following expression: . And then we get the function
Let's call the expression an intermediate argument, and the function an outer function. It's not strict mathematical concepts, but they help to understand the meaning of the concept of a complex function.
A strict definition of the concept of a complex function sounds like this:
Let a function be defined on a set and be the set of values of this function. Let the set (or its subset) be the domain of definition of the function. Let's assign a number to each of them. Thus, the function will be defined on the set. It is called function composition or complex function.
In this definition, if we use our terminology, an external function is an intermediate argument.
The derivative of a complex function is found according to the following rule:
To make it more clear, I like to write this rule as follows:
In this expression, using denotes an intermediate function.
So. To find the derivative of a complex function, you need
1. Determine which function is external and find the corresponding derivative from the table of derivatives.
2. Define an intermediate argument.
In this procedure, the greatest difficulty is finding the external function. A simple algorithm is used for this:
A. Write down the equation of the function.
b. Imagine that you need to calculate the value of a function for some value of x. To do this, you substitute this x value into the function equation and perform arithmetic. The last action you do is the external function.
For example, in the function
The last action is exponentiation.
Let's find the derivative of this function. To do this, we write an intermediate argument