The derivative of a complex function is equal to. Derivative of a complex function. Derivative of a power-exponential function

Solving physical problems or examples in mathematics is completely impossible without knowledge of the derivative and methods for calculating it. The derivative is one of the most important concepts in mathematical analysis. We decided to devote today’s article to this fundamental topic. What is a derivative, what is its physical and geometric meaning, how to calculate the derivative of a function? All these questions can be combined into one: how to understand the derivative?

Geometric and physical meaning of derivative

Let there be a function f(x) , specified in a certain interval (a, b) . Points x and x0 belong to this interval. When x changes, the function itself changes. Changing the argument - the difference in its values x-x0 . This difference is written as delta x and is called argument increment. A change or increment of a function is the difference between the values ​​of a function at two points. Definition of derivative:

The derivative of a function at a point is the limit of the ratio of the increment of the function at a given point to the increment of the argument when the latter tends to zero.

Otherwise it can be written like this:

What's the point of finding such a limit? Here's what it is:

the derivative of a function at a point is equal to the tangent of the angle between the OX axis and the tangent to the graph of the function at a given point.

Physical meaning of the derivative: the derivative of the path with respect to time is equal to the speed of rectilinear motion.

Indeed, since school days everyone knows that speed is a particular path x=f(t) and time t . Average speed over a certain period of time:

To find out the speed of movement at a moment in time t0 you need to calculate the limit:

Rule one: set a constant

The constant can be taken out of the derivative sign. Moreover, this must be done. When solving examples in mathematics, take it as a rule - If you can simplify an expression, be sure to simplify it .

Example. Let's calculate the derivative:

Rule two: derivative of the sum of functions

The derivative of the sum of two functions is equal to the sum of the derivatives of these functions. The same is true for the derivative of the difference of functions.

We will not give a proof of this theorem, but rather consider a practical example.

Find the derivative of the function:

Rule three: derivative of the product of functions

The derivative of the product of two differentiable functions is calculated by the formula:

Example: find the derivative of a function:

Solution:

It is important to talk about calculating derivatives of complex functions here. The derivative of a complex function is equal to the product of the derivative of this function with respect to the intermediate argument and the derivative of the intermediate argument with respect to the independent variable.

In the above example we come across the expression:

In this case, the intermediate argument is 8x to the fifth power. In order to calculate the derivative of such an expression, we first calculate the derivative of the external function with respect to the intermediate argument, and then multiply by the derivative of the intermediate argument itself with respect to the independent variable.

Rule four: derivative of the quotient of two functions

Formula for determining the derivative of the quotient of two functions:

We tried to talk about derivatives for dummies from scratch. This topic is not as simple as it seems, so be warned: there are often pitfalls in the examples, so be careful when calculating derivatives.

With any questions on this and other topics, you can contact the student service. In a short time, we will help you solve the most difficult test and understand the tasks, even if you have never done derivative calculations before.

Very easy to remember.

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

In our case, the base is the number:

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

What is it equal to? Of course, .

The derivative of the natural logarithm is also very simple:

Examples:

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

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

Rules of differentiation

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

Differentiation is the process of finding the derivative.

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

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

There are 5 rules in total.

The constant is taken out of the derivative sign.

If - some constant number (constant), then.

Obviously, this rule also works for the difference: .

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

Examples.

Find the derivatives of the functions:

1. at a point;
2. at a point;
3. at a point;
4. at the point.

Solutions:

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

Derivative of the product

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

Derivative:

Examples:

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

Solutions:

Derivative of an exponential function

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

So, where is some number.

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

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

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

Happened?

Here, check yourself:

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

Examples:
Find the derivatives of the functions:

Answers:

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

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

In this example, the product of two functions:

Derivative of a logarithmic function

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

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

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

Only now we will write instead:

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

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

Derivative of a complex function.

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

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

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

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

For our example, .

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

Second example: (same thing). .

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

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

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

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

We change variables and get a function.

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

Another example:

So, let's finally formulate the official rule:

Algorithm for finding the derivative of a complex function:

It seems simple, right?

Let's check with examples:

Solutions:

1) Internal: ;

External: ;

2) Internal: ;

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

3) Internal: ;

External: ;

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

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

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

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

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

1. Radical expression. .

2. Root. .

3. Sine. .

4. Square. .

5. Putting it all together:

DERIVATIVE. BRIEFLY ABOUT THE MAIN THINGS

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

Basic derivatives:

Rules of differentiation:

The constant is taken out of the derivative sign:

Derivative of the sum:

Derivative of the product:

Derivative of the quotient:

Derivative of a complex function:

Algorithm for finding the derivative of a complex function:

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

Examples are given of calculating derivatives using the formula for the derivative of a complex function.

Content

See also: Proof of the formula for the derivative of a complex function

Basic formulas

Here we give examples of calculating derivatives of the following functions:
; ; ; ; .

If a function can be represented as a complex function in the following form:
,
then its derivative is determined by the formula:
.
In the examples below, we will write this formula as follows:
.
Where .
Here, the subscripts or , located under the derivative sign, denote the variables by which differentiation is performed.

Usually, in tables of derivatives, derivatives of functions from the variable x are given. However, x is a formal parameter. The variable x can be replaced by any other variable. Therefore, when differentiating a function from a variable, we simply change, in the table of derivatives, the variable x to the variable u.

Simple examples

Example 1

Find the derivative of a complex function
.

Let's write the given function in equivalent form:
.
In the table of derivatives we find:
;
.

According to the formula for the derivative of a complex function, we have:
.
Here .

Example 2

Find the derivative
.

We take the constant 5 out of the derivative sign and from the table of derivatives we find:
.

.
Here .

Example 3

Find the derivative
.

We take out a constant -1 for the sign of the derivative and from the table of derivatives we find:
;
From the table of derivatives we find:
.

We apply the formula for the derivative of a complex function:
.
Here .

More complex examples

In more complex examples, we apply the rule for differentiating a complex function several times. In this case, we calculate the derivative from the end. That is, we break the function into its component parts and find the derivatives of the simplest parts using table of derivatives. We also use rules for differentiating sums, products and fractions. Then we make substitutions and apply the formula for the derivative of a complex function.

Example 4

Find the derivative
.

Let's select the simplest part of the formula and find its derivative. .

.
Here we have used the notation
.

We find the derivative of the next part of the original function using the results obtained. We apply the rule for differentiating the sum:
.

Once again we apply the rule of differentiation of complex functions.

.
Here .

Example 5

Find the derivative of the function
.

Let's select the simplest part of the formula and find its derivative from the table of derivatives. .

We apply the rule of differentiation of complex functions.
.
Here
.

Let's differentiate the next part using the results obtained.
.
Here
.

Let's differentiate the next part.

.
Here
.

Now we find the derivative of the desired function.

.
Here
.

See also:

This lesson is devoted to the topic “Differentiation of complex functions. A problem from the practice of preparing for the Unified State Exam in mathematics.” This lesson explores differentiating complex functions. A table of derivatives of a complex function is compiled. In addition, an example of solving a problem from the practice of preparing for the Unified State Exam in mathematics is considered.

Topic: Derivative

Lesson: Differentiating a complex function. A practice task for preparing for the Unified State Exam in mathematics

Complexfunction we have already differentiated, but the argument was a linear function, namely, we know how to differentiate the function . For example, . Now, in the same way, we will find derivatives of a complex function, where instead of a linear function there may be another function.

Let's start with the function

So, we found the derivative of the sine from a complex function, where the argument of the sine was a quadratic function.

If you need to find the value of the derivative at a specific point, then this point must be substituted into the found derivative.

So, in two examples we saw how the rule works differentiation complex functions.

2.

3. . Let us remind you that .

7.

8. .

Thus, we will finish the table of differentiation of complex functions at this stage. Further, of course, it will be generalized even more, but now let’s move on to specific problems on the derivative.

In the practice of preparing for the Unified State Exam, the following tasks are proposed.

Find the minimum of a function .

ODZ: .

Let's find the derivative. Let us recall that, .

Let's equate the derivative to zero. The dot is included in the ODZ.

Let us find the intervals of constant sign of the derivative (intervals of monotonicity of the function) (see Fig. 1).

Rice. 1. Monotonicity intervals for a function .

Let's look at a point and find out whether it is an extremum point. A sufficient sign of an extremum is that the derivative changes sign when passing through a point. In this case, the derivative changes sign, which means it is an extremum point. Since the derivative changes sign from “-” to “+”, then this is the minimum point. Let's find the value of the function at the minimum point: . Let's draw a diagram (see Fig. 2).

Fig.2. Extremum of the function .

On the interval - the function decreases, on - the function increases, the extremum point is unique. The function takes its smallest value only at point .

During the lesson we looked at the differentiation of complex functions, compiled a table and looked at the rules for differentiating a complex function, and gave an example of using a derivative from the practice of preparing for the Unified State Exam.

1. Algebra and beginning of analysis, grade 10 (in two parts). Textbook for general education institutions (profile level), ed. A. G. Mordkovich. -M.: Mnemosyne, 2009.

2. Algebra and beginning of analysis, grade 10 (in two parts). Problem book for educational institutions (profile level), ed. A. G. Mordkovich. -M.: Mnemosyne, 2007.

3. Vilenkin N.Ya., Ivashev-Musatov O.S., Shvartsburd S.I. Algebra and mathematical analysis for grade 10 (textbook for students of schools and classes with in-depth study of mathematics). - M.: Prosveshchenie, 1996.

4. Galitsky M.L., Moshkovich M.M., Shvartsburd S.I. In-depth study of algebra and mathematical analysis.-M.: Education, 1997.

5. Collection of problems in mathematics for applicants to higher educational institutions (edited by M.I. Skanavi). - M.: Higher School, 1992.

6. Merzlyak A.G., Polonsky V.B., Yakir M.S. Algebraic simulator.-K.: A.S.K., 1997.

7. Zvavich L.I., Shlyapochnik L.Ya., Chinkina Algebra and the beginnings of analysis. 8-11 grades: A manual for schools and classes with in-depth study of mathematics (didactic materials). - M.: Bustard, 2002.

8. Sahakyan S.M., Goldman A.M., Denisov D.V. Problems on algebra and principles of analysis (a manual for students in grades 10-11 of general education institutions). - M.: Prosveshchenie, 2003.

9. Karp A.P. Collection of problems on algebra and principles of analysis: textbook. allowance for 10-11 grades. with depth studied Mathematics.-M.: Education, 2006.

10. Glazer G.I. History of mathematics at school. Grades 9-10 (manual for teachers).-M.: Education, 1983

Additional web resources

2. Portal of Natural Sciences ().

Make it at home

Nos. 42.2, 42.3 (Algebra and beginnings of analysis, grade 10 (in two parts). Problem book for general education institutions (profile level) edited by A. G. Mordkovich. - M.: Mnemosyne, 2007.)

If you follow the definition, then the derivative of a function at a point is the limit of the ratio of the increment of the function Δ y to the argument increment Δ x:

Everything seems to be clear. But try using this formula to calculate, 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 from the entire variety of functions we can distinguish the so-called elementary functions. These are relatively simple expressions, the derivatives of which have long been calculated and tabulated. Such functions are quite easy to remember - along with their derivatives.

Derivatives of elementary functions

Elementary functions are all those listed below. The derivatives of these functions must be known by heart. Moreover, it is not at all difficult to memorize them - that’s why they are elementary.

So, derivatives of elementary functions:

Name	Function	Derivative
Constant	f(x) = C, C ∈ R	0 (yes, zero!)
Power 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/sin 2 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 particularly elementary, but also differentiated according to certain rules. These rules are discussed below.

Derivative of sum and difference

Let the functions be given f(x) And g(x), the derivatives of which 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:

1. (f + g)’ = f ’ + g ’
2. (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 + sin x; g(x) = x 4 + 2x 2 − 3.

Function f(x) is the sum of two elementary functions, therefore:

f ’(x) = (x 2 + sin x)’ = (x 2)’ + (sin x)’ = 2x+ cos x;

We reason 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+ cos x;
g ’(x) = 4x · ( x 2 + 1).

Derivative of the product

Mathematics is a logical science, so many people believe that if the derivative of a sum is equal to the sum of derivatives, then the derivative of the product strike">equal to the product of derivatives. But screw you! The derivative of a product is calculated using a completely different formula. Namely:

(f · g) ’ = f ’ · g + f · g ’

The formula is simple, but it is often forgotten. And not only schoolchildren, but also students. The result is incorrectly solved problems.

Task. Find derivatives of functions: f(x) = x 3 cos x; g(x) = (x 2 + 7x− 7) · e x .

Function f(x) is the 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. Obviously, the first factor 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 .

Please note that in the last step the derivative is factorized. Formally, this does not need to be done, but most derivatives are not calculated on their own, but to examine the function. This means that further the derivative will be equated to zero, its signs will be determined, and so on. For such a case, it is better to have an expression factorized.

If there are two functions f(x) And g(x), and g(x) ≠ 0 on the set we are interested in, 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? And 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.

Task. Find derivatives of functions:

The numerator and denominator of each fraction contain elementary functions, so all we need is the formula for the derivative of the quotient:

According to tradition, let's factorize the numerator - this will greatly simplify the answer:

A complex function is not necessarily a half-kilometer-long formula. For example, it is enough to take the function f(x) = sin x and replace the variable x, say, on x 2 + ln x. It will work out f(x) = sin ( x 2 + ln x) - this is a complex function. It also has a derivative, but it will not be possible to find it using the rules discussed above.

What should I do? In such cases, replacing a variable and formula for the derivative of a complex function helps:

f ’(x) = f ’(t) · t', If x is replaced by t(x).

As a rule, the situation with understanding this formula is even more sad than with the derivative of the quotient. Therefore, it is also better to explain it using specific examples, with a detailed description of each step.

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 replacement: let 2 x + 3 = t, f(x) = f(t) = e t. We look for the derivative of a complex function using the formula:

f ’(x) = f ’(t) · t ’ = (e t)’ · t ’ = e t · t ’

And now - attention! We perform the reverse replacement: 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 it 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 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 “prime.” For example, the stroke of the sum is equal to the sum of the strokes. Is that clearer? Well, that's good.

Thus, calculating the derivative comes down to getting rid of these same 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 people know that in the role n may well be a fractional number. For example, the root is x 0.5. What if there is something fancy under the root? Again, the result will be a complex function - they like to give such constructions in tests and exams.

Task. Find the derivative of the 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 replacement: let x 2 + 8x − 7 = t. We find the derivative using the formula:

f ’(x) = f ’(t) · t ’ = (t 0.5)’ · t’ = 0.5 · t−0.5 · t ’.

Let's do the reverse replacement: 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: