How to calculate derivatives of complex functions. Derivative of a complex function

This lesson is devoted to the topic “Differentiation of complex functions. A task from the practice of preparing for the Unified State Examination in mathematics. In this lesson, we study the differentiation of 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 USE in mathematics is considered.

Theme: Derivative

Lesson: Differentiating a complex function. Task from the practice of preparing for the 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 of a complex function, where the argument of the sine was a quadratic function.

If it is necessary to find the value of the derivative at a particular 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. . Recall that .

7.

8. .

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

In the practice of preparing for the exam, the following tasks are proposed.

Find the minimum of a function .

ODZ: .

Let's find the derivative. Recall that, .

Let's equate the derivative to zero. Point - 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. Intervals of monotonicity for a function .

Consider a point and find out if 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 - the minimum point. Find the value of the function at the minimum point: . Let's draw a diagram (see Fig. 2).

Fig.2. Function extremum .

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

In the lesson, we considered the differentiation of complex functions, compiled a table and examined the rules for differentiating a complex function, gave an example of using a derivative from the practice of preparing for the exam.

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

2. Algebra and the beginning of analysis, grade 10 (in two parts). Task 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 .: Education, 1996.

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

5. Collection of problems in mathematics for applicants to technical universities (under the editorship of M.I.Skanavi).-M.: Higher School, 1992.

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

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

8. Saakyan S.M., Goldman A.M., Denisov D.V. Tasks in Algebra and the Beginnings of Analysis (a manual for students in grades 10-11 of general educational institutions).-M .: Education, 2003.

9. Karp A.P. Collection of problems in algebra and the beginnings of analysis: textbook. allowance for 10-11 cells. with a deep study mathematics.-M.: Education, 2006.

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

Additional web resources

2. Portal of Natural Sciences ().

do at home

No. 42.2, 42.3 (Algebra and the beginnings of analysis, grade 10 (in two parts). A task book for educational institutions (profile level) edited by A. G. Mordkovich. - M .: Mnemozina, 2007.)

It's very easy to remember.

Well, we will not go far, we will immediately consider the inverse function. What is the inverse of the exponential function? Logarithm:

In our case, the base is a number:

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

What is 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 exponent and the natural logarithm are functions that are uniquely simple in terms of the derivative. 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.

Differentiation rules

What rules? Another new term, again?!...

Differentiation is the process of finding the derivative.

Only and everything. What is another word for this process? Not proizvodnovanie... The differential of mathematics is called the very increment of the 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 sign of the derivative.

If - some constant number (constant), then.

Obviously, this rule also works for the difference: .

Let's prove it. Let, or easier.

Examples.

Find derivatives of functions:

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

Solutions:

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

Derivative of a product

Everything is similar here: we introduce a new function and find its increment:

Derivative:

Examples:

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

Solutions:

Derivative of exponential function

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

So where is some number.

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

To do this, we 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 the exponent: as it was, it remains, only a factor appeared, which is just a number, but not a variable.

Examples:
Find derivatives of functions:

Answers:

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

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

In this example, the product of two functions:

Derivative of a logarithmic function

Here it is similar: you already know the derivative of the natural logarithm:

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

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

Only now instead of we will write:

The denominator turned out to be just a constant (a constant number, without a variable). The derivative is very simple:

Derivatives of the exponential and logarithmic functions are almost never found in the exam, 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 arc tangent. These functions can be difficult to understand (although if the logarithm seems difficult to you, read the topic "Logarithms" and everything will work out), but in terms of mathematics, the word "complex" does not mean "difficult".

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

Let's create a similar mathematical pipeline: first we will find the cosine of a number, and then we will square the resulting number. So, they give us 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, in order to find its value, we do the first action directly with the variable, and then another second action with what happened as a result of the first.

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

For our example, .

We may well do the same actions in reverse order: first you square, and then I look for the cosine of the resulting number:. It is 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). .

The last action we do will be called "external" function, and the action performed first - respectively "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 is internal:

Answers: The separation of inner and outer functions is very similar to changing variables: for example, in the function

1. What action will we take first? First we calculate the sine, and only then we raise it to a cube. So it's an internal function, not 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 - 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. For 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 to be simple, right?

Let's check with examples:

Solutions:

1) Internal: ;

External: ;

2) Internal: ;

(just don’t try to reduce by now! Nothing is taken out from under the cosine, remember?)

3) Internal: ;

External: ;

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

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

1. Radical expression. .

2. Root. .

3. Sinus. .

4. Square. .

5. Putting it all together:

DERIVATIVE. BRIEFLY ABOUT THE MAIN

Function derivative- the ratio of the increment of the function to the increment of the argument with an infinitesimal increment of the argument:

Basic derivatives:

Differentiation rules:

The constant is taken out of the sign of the derivative:

Derivative of sum:

Derivative 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, find its derivative.
2. We define the "external" function, find its derivative.
3. We multiply the results of the first and second points.

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:

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:

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 + 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 in the last step, the derivative is factorized. 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:

On which we analyzed the simplest derivatives, and also got acquainted with the rules of differentiation and some techniques for finding derivatives. Thus, if you are not very good with derivatives of functions or some points of this article are not entirely clear, then first read the above lesson. Please tune in to a serious mood - the material is not easy, 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 in the table at the rule (No. 5) for differentiating a complex function:

We understand. First of all, let's take a look at the notation. Here we have two functions - and , and the function, figuratively speaking, is nested in the function . A function of this kind (when one function is nested within another) is called a complex function.

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

! These definitions are not theoretical and should not appear in the final design of assignments. I use the 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 the whole expression, so finding the derivative immediately 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 it is impossible to “tear apart” the sine:

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

First step, which must be performed when finding the derivative of a complex function is to understand which function is internal and which is external.

In the case of simple examples, it seems clear that a polynomial is nested under the sine. But what if it's not obvious? How to determine exactly which function is external and which is internal? To do this, I propose to use the following technique, which can be carried out mentally or on a draft.

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

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

Secondly you will need to find, so the sine - will be an external function:

After we UNDERSTAND with inner and outer functions, it's time to apply the compound function differentiation rule .

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

First we find the derivative of the external function (sine), look at the table of derivatives of elementary functions and notice that . All tabular formulas are applicable even if "x" is replaced by a complex expression, in this case:

Note that the inner function has not changed, we do not touch it.

Well, it is quite obvious that

The result of applying the formula clean looks like this:

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

If there is any misunderstanding, write down the decision 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:

We figure out where we have an external function, and where is an internal one. To do this, we try (mentally or on a draft) to calculate the value of the expression for . What needs to be done first? First of all, you need to calculate what the base is equal to:, which means that the polynomial is the internal function:

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

According to the formula , first you need to find the derivative of the external function, in this case, the degree. We are looking for the desired formula in the table:. We repeat again: any tabular formula is valid not only for "x", but also for a complex expression. Thus, the result of applying the rule of differentiation of a complex function next:

I emphasize again that when we take the derivative of the outer function, the inner function does not change:

Now it remains to find a very simple derivative of the inner function and “comb” the result a little:

Example 4

Find the derivative of a function

This is an example for self-solving (answer at the end of the lesson).

To consolidate the 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 is the external and where is the internal function, why are the tasks solved that way?

Example 5

a) Find the derivative of a 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 degree. Thus, we first bring the function into the proper form for differentiation:

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

The degree is again represented 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 bring the expression to a common denominator in brackets and write everything as one fraction. It’s beautiful, of course, but when cumbersome long derivatives are obtained, 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 self-solving (answer at the end of the lesson).

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

Example 8

Find the derivative of a function

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

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

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

We find the derivative of the inner function, reset the cosine back down:

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

Example 9

Find the derivative of a function

This is an example for self-solving (answer at the end of the lesson).

So far, we have considered 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

We understand the attachments of this function. We try to evaluate the expression using the experimental value . How would we count on a calculator?

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

This arcsine of unity should then be squared:

And finally, we raise the seven to the power:

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

We start to decide

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

Derivation of the formula for the derivative of a power function (x to the power of a). Derivatives of roots from x are considered. The formula for the derivative of a higher order power function. Examples of calculating derivatives.

See also: Power function and roots, formulas and graph
Power Function Plots

Basic Formulas

The derivative of x to the power of a is a times x to the power of a minus one:
(1) .

The derivative of the nth root of x to the mth power is:
(2) .

Derivation of the formula for the derivative of a power function

Case x > 0

Consider a power function of variable x with exponent a :
(3) .
Here a is an arbitrary real number. Let's consider the case first.

To find the derivative of the function (3), we use the properties of the power function and transform it to the following form:
.

Now we find the derivative by applying:
;
.
Here .

Formula (1) is proved.

Derivation of the formula for the derivative of the root of the degree n of x to the degree m

Now consider a function that is the root of the following form:
(4) .

To find the derivative, we convert the root to a power function:
.
Comparing with formula (3), we see that
.
Then
.

By formula (1) we find the derivative:
(1) ;
;
(2) .

In practice, there is no need to memorize formula (2). It is much more convenient to first convert the roots to power functions, and then find their derivatives using formula (1) (see examples at the end of the page).

Case x = 0

If , then the exponential function is also defined for the value of the variable x = 0 . Let us find the derivative of function (3) for x = 0 . To do this, we use the definition of a derivative:
.

Substitute x = 0 :
.
In this case, by derivative we mean the right-hand limit for which .

So we found:
.
From this it can be seen that at , .
At , .
At , .
This result is also obtained by formula (1):
(1) .
Therefore, formula (1) is also valid for x = 0 .

case x< 0

Consider function (3) again:
(3) .
For some values ​​of the constant a , it is also defined for negative values ​​of the variable x . Namely, let a be a rational number. Then it can be represented as an irreducible fraction:
,
where m and n are integers with no common divisor.

If n is odd, then the exponential function is also defined for negative values ​​of the variable x. For example, for n = 3 and m = 1 we have the cube root of x :
.
It is also defined for negative values ​​of x .

Let us find the derivative of the power function (3) for and for rational values ​​of the constant a , for which it is defined. To do this, we represent x in the following form:
.
Then ,
.
We find the derivative by taking the constant out of the sign of the derivative and applying the rule of differentiation of a complex function:

.
Here . But
.
Since , then
.
Then
.
That is, formula (1) is also valid for:
(1) .

Derivatives of higher orders

Now we find the higher order derivatives of the power function
(3) .
We have already found the first order derivative:
.

Taking the constant a out of the sign of the derivative, we find the second-order derivative:
.
Similarly, we find derivatives of the third and fourth orders:
;

.

From here it is clear that derivative of an arbitrary nth order has the following form:
.

notice, that if a is a natural number, , then the nth derivative is constant:
.
Then all subsequent derivatives are equal to zero:
,
at .

Derivative Examples

Example

Find the derivative of the function:
.

Let's convert the roots to powers:
;
.
Then the original function takes the form:
.

We find derivatives of degrees:
;
.
The derivative of a constant is zero:
.