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.

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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
.
Because , 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:
.

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 steps 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.

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 complex 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 complex 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 complex 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 task 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.

So 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:

Note : because function can take negative values, then, generally speaking, you need to use modules: , which disappear as a result of differentiation. However, the current design is also acceptable, where by default the complex values. But if with all rigor, then in both cases it is necessary to make a reservation that.

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 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 re-read the explanations of Example 11 carefully.

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 :

An exponential function is a function that has the form of a power function
y = u v ,
whose base u and exponent v are some functions of the variable x :
u = u (x); v=v (x).
This function is also called exponential-power or .

Note that the exponential function can be represented in exponential form:
.
Therefore, it is also called complex exponential function.

Derivative of exponential function

Calculation using the logarithmic derivative

Find the derivative of the exponential function
(2) ,
where and are functions of the variable .
To do this, we take the logarithm of equation (2), using the property of the logarithm:
.
Differentiate with respect to x :
(3) .
Apply rules for differentiating a compound function and works:
;
.

Substitute in (3):
.
From here
.

So, we found the derivative of the exponential function:
(1) .
If the exponent is constant, then . Then the derivative is equal to the derivative of the compound power function:
.
If the base of the degree is constant, then . Then the derivative is equal to the derivative of the compound exponential function:
.
When and are functions of x, then the derivative of the exponential function is equal to the sum of the derivatives of the compound power and exponential functions.

Calculation of the derivative by reduction to a complex exponential function

Now we find the derivative of the exponential function
(2) ,
representing it as a complex exponential function:
(4) .

Let's differentiate the product:
.
We apply the rule for finding the derivative of a complex function:

.
And we again got the formula (1).

Example 1

Find the derivative of the following function:
.

We calculate using the logarithmic derivative. We take the logarithm of the original function:
(P1.1) .

From the table of derivatives we find:
;
.
According to the formula for the derivative of a product, we have:
.
We differentiate (A1.1):
.
Insofar as
,
then
.

Here is a summary table for convenience and clarity when studying the topic.

Constanty=C Power function y = x p (x p)" = p x p - 1	Exponential functiony = x (a x)" = a x ln a In particular, whena = ewe have y = e x (e x)" = e x
logarithmic function (log a x) " = 1 x ln a In particular, whena = ewe have y = log x (ln x)" = 1 x	Trigonometric functions (sin x) "= cos x (cos x)" = - sin x (t g x) " = 1 cos 2 x (c t g x)" = - 1 sin 2 x
Inverse trigonometric functions (a r c sin x) " = 1 1 - x 2 (a r c cos x) " = - 1 1 - x 2 (a r c t g x) " = 1 1 + x 2 (a r c c t g x) " = - 1 1 + x 2	Hyperbolic functions (s h x) " = c h x (c h x) " = s h x (t h x) " = 1 c h 2 x (c t h x) " = - 1 s h 2 x

Let us analyze how the formulas of the specified table were obtained, or, in other words, we will prove the derivation of formulas for derivatives for each type of function.

Derivative of a constant

Proof 1

In order to derive this formula, we take as a basis the definition of the derivative of a function at a point. We use x 0 = x, where x takes on the value of any real number, or, in other words, x is any number from the domain of the function f (x) = C . Let's write the limit of the ratio of the increment of the function to the increment of the argument as ∆ x → 0:

lim ∆ x → 0 ∆ f (x) ∆ x = lim ∆ x → 0 C - C ∆ x = lim ∆ x → 0 0 ∆ x = 0

Please note that the expression 0 ∆ x falls under the limit sign. It is not the uncertainty of “zero divided by zero”, since the numerator contains not an infinitesimal value, but zero. In other words, the increment of a constant function is always zero.

So, the derivative of the constant function f (x) = C is equal to zero over the entire domain of definition.

Example 1

Given constant functions:

f 1 (x) = 3 , f 2 (x) = a , a ∈ R , f 3 (x) = 4 . 13 7 22 , f 4 (x) = 0 , f 5 (x) = - 8 7

Decision

Let us describe the given conditions. In the first function we see the derivative of the natural number 3 . In the following example, you need to take the derivative of a, where a- any real number. The third example gives us the derivative of the irrational number 4 . 13 7 22 , the fourth - the derivative of zero (zero is an integer). Finally, in the fifth case, we have the derivative of the rational fraction - 8 7 .

Answer: the derivatives of the given functions are zero for any real x(over the entire domain of definition)

f 1 " (x) = (3) " = 0 , f 2 " (x) = (a) " = 0 , a ∈ R , f 3 " (x) = 4 . 13 7 22 " = 0 , f 4 " (x) = 0 " = 0 , f 5 " (x) = - 8 7 " = 0

Power function derivative

We turn to the power function and the formula for its derivative, which has the form: (x p) " = p x p - 1, where the exponent p is any real number.

Proof 2

Here is the proof of the formula when the exponent is a natural number: p = 1 , 2 , 3 , …

Again, we rely on the definition of a derivative. Let's write the limit of the ratio of the increment of the power function to the increment of the argument:

(x p) " = lim ∆ x → 0 = ∆ (x p) ∆ x = lim ∆ x → 0 (x + ∆ x) p - x p ∆ x

To simplify the expression in the numerator, we use Newton's binomial formula:

(x + ∆ x) p - x p = C p 0 + x p + C p 1 x p - 1 ∆ x + C p 2 x p - 2 (∆ x) 2 + . . . + + C p p - 1 x (∆ x) p - 1 + C p p (∆ x) p - x p = = C p 1 x p - 1 ∆ x + C p 2 x p - 2 (∆ x) 2 + . . . + C p p - 1 x (∆ x) p - 1 + C p p (∆ x) p

Thus:

(x p) " = lim ∆ x → 0 ∆ (x p) ∆ x = lim ∆ x → 0 (x + ∆ x) p - x p ∆ x = = lim ∆ x → 0 (C p 1 x p - 1 ∆ x + C p 2 x p - 2 (∆ x) 2 + . . . + C p p - 1 x (∆ x) p - 1 + C p p (∆ x) p) ∆ x = = lim ∆ x → 0 (C p 1 x p - 1 + C p 2 x p - 2 ∆ x + . . . + C p p - 1 x (∆ x) p - 2 + C p p (∆ x) p - 1) = = C p 1 x p - 1 + 0 + 0 + . . . + 0 = p! 1! (p - 1)! x p - 1 = p x p - 1

So, we proved the formula for the derivative of a power function when the exponent is a natural number.

Proof 3

To give proof for the case when p- any real number other than zero, we use the logarithmic derivative (here we should understand the difference from the derivative of the logarithmic function). To have a more complete understanding, it is desirable to study the derivative of the logarithmic function and additionally deal with the derivative of an implicitly given function and the derivative of a complex function.

Consider two cases: when x positive and when x are negative.

So x > 0 . Then: x p > 0 . We take the logarithm of the equality y \u003d x p to the base e and apply the property of the logarithm:

y = x p ln y = ln x p ln y = p ln x

At this stage, an implicitly defined function has been obtained. Let's define its derivative:

(ln y) " = (p ln x) 1 y y " = p 1 x ⇒ y " = p y x = p x p x = p x p - 1

Now we consider the case when x- a negative number.

If the indicator p is an even number, then the power function is also defined for x< 0 , причем является четной: y (x) = - y ((- x) p) " = - p · (- x) p - 1 · (- x) " = = p · (- x) p - 1 = p · x p - 1

Then xp< 0 и возможно составить доказательство, используя логарифмическую производную.

If a p is an odd number, then the power function is defined for x< 0 , причем является нечетной: y (x) = - y (- x) = - (- x) p . Тогда x p < 0 , а значит логарифмическую производную задействовать нельзя. В такой ситуации возможно взять за основу доказательства правила дифференцирования и правило нахождения производной сложной функции:

y "(x) \u003d (- (- x) p) " \u003d - ((- x) p) " \u003d - p (- x) p - 1 (- x) " = \u003d p (- x) p - 1 = p x p - 1

The last transition is possible because if p is an odd number, then p - 1 either an even number or zero (for p = 1), therefore, for negative x the equality (- x) p - 1 = x p - 1 is true.

So, we have proved the formula for the derivative of a power function for any real p.

Example 2

Given functions:

f 1 (x) = 1 x 2 3 , f 2 (x) = x 2 - 1 4 , f 3 (x) = 1 x log 7 12

Determine their derivatives.

Decision

We transform part of the given functions into a tabular form y = x p , based on the properties of the degree, and then use the formula:

f 1 (x) \u003d 1 x 2 3 \u003d x - 2 3 ⇒ f 1 "(x) \u003d - 2 3 x - 2 3 - 1 \u003d - 2 3 x - 5 3 f 2 "(x) \u003d x 2 - 1 4 = 2 - 1 4 x 2 - 1 4 - 1 = 2 - 1 4 x 2 - 5 4 f 3 (x) = 1 x log 7 12 = x - log 7 12 ⇒ f 3 "( x) = - log 7 12 x - log 7 12 - 1 = - log 7 12 x - log 7 12 - log 7 7 = - log 7 12 x - log 7 84

Derivative of exponential function

Proof 4

We derive the formula for the derivative, based on the definition:

(a x) " = lim ∆ x → 0 a x + ∆ x - a x ∆ x = lim ∆ x → 0 a x (a ∆ x - 1) ∆ x = a x lim ∆ x → 0 a ∆ x - 1 ∆ x = 0 0

We got uncertainty. To expand it, we write a new variable z = a ∆ x - 1 (z → 0 as ∆ x → 0). In this case a ∆ x = z + 1 ⇒ ∆ x = log a (z + 1) = ln (z + 1) ln a . For the last transition, the formula for the transition to a new base of the logarithm is used.

Let's perform a substitution in the original limit:

(a x) " = a x lim ∆ x → 0 a ∆ x - 1 ∆ x = a x ln a lim ∆ x → 0 1 1 z ln (z + 1) = = a x ln a lim ∆ x → 0 1 ln (z + 1) 1 z = a x ln a 1 ln lim ∆ x → 0 (z + 1) 1 z

Recall the second wonderful limit and then we get the formula for the derivative of the exponential function:

(a x) " = a x ln a 1 ln lim z → 0 (z + 1) 1 z = a x ln a 1 ln e = a x ln a

Example 3

The exponential functions are given:

f 1 (x) = 2 3 x , f 2 (x) = 5 3 x , f 3 (x) = 1 (e) x

We need to find their derivatives.

Decision

We use the formula for the derivative of the exponential function and the properties of the logarithm:

f 1 "(x) = 2 3 x" = 2 3 x ln 2 3 = 2 3 x (ln 2 - ln 3) f 2 "(x) = 5 3 x" = 5 3 x ln 5 1 3 = 1 3 5 3 x ln 5 f 3 "(x) = 1 (e) x" = 1 e x " = 1 e x ln 1 e = 1 e x ln e - 1 = - 1 e x

Derivative of a logarithmic function

Proof 5

We present the proof of the formula for the derivative of the logarithmic function for any x in the domain of definition and any valid values ​​of the base a of the logarithm. Based on the definition of the derivative, we get:

(log a x) " = lim ∆ x → 0 log a (x + ∆ x) - log a x ∆ x = lim ∆ x → 0 log a x + ∆ x x ∆ x = = lim ∆ x → 0 1 ∆ x log a 1 + ∆ x x = lim ∆ x → 0 log a 1 + ∆ x x 1 ∆ x = = lim ∆ x → 0 log a 1 + ∆ x x 1 ∆ x x x = lim ∆ x → 0 1 x log a 1 + ∆ x x x ∆ x = = 1 x log a lim ∆ x → 0 1 + ∆ x x x ∆ x = 1 x log a e = 1 x ln e ln a = 1 x ln a

It can be seen from the specified chain of equalities that the transformations were built on the basis of the logarithm property. The equality lim ∆ x → 0 1 + ∆ x x x ∆ x = e is true in accordance with the second remarkable limit.

Example 4

Logarithmic functions are given:

f 1 (x) = log log 3 x , f 2 (x) = log x

It is necessary to calculate their derivatives.

Decision

Let's apply the derived formula:

f 1 "(x) = (log ln 3 x)" = 1 x ln (ln 3) ; f 2 "(x) \u003d (ln x)" \u003d 1 x ln e \u003d 1 x

So the derivative of the natural logarithm is one divided by x.

Derivatives of trigonometric functions

Proof 6

We use some trigonometric formulas and the first wonderful limit to derive the formula for the derivative of a trigonometric function.

According to the definition of the derivative of the sine function, we get:

(sin x) " = lim ∆ x → 0 sin (x + ∆ x) - sin x ∆ x

The formula for the difference of sines will allow us to perform the following actions:

(sin x) " = lim ∆ x → 0 sin (x + ∆ x) - sin x ∆ x = = lim ∆ x → 0 2 sin x + ∆ x - x 2 cos x + ∆ x + x 2 ∆ x = = lim ∆ x → 0 sin ∆ x 2 cos x + ∆ x 2 ∆ x 2 = = cos x + 0 2 lim ∆ x → 0 sin ∆ x 2 ∆ x 2

Finally, we use the first wonderful limit:

sin "x = cos x + 0 2 lim ∆ x → 0 sin ∆ x 2 ∆ x 2 = cos x

So the derivative of the function sin x will cos x.

We will also prove the formula for the cosine derivative in the same way:

cos "x = lim ∆ x → 0 cos (x + ∆ x) - cos x ∆ x = = lim ∆ x → 0 - 2 sin x + ∆ x - x 2 sin x + ∆ x + x 2 ∆ x = = - lim ∆ x → 0 sin ∆ x 2 sin x + ∆ x 2 ∆ x 2 = = - sin x + 0 2 lim ∆ x → 0 sin ∆ x 2 ∆ x 2 = - sin x

Those. the derivative of the function cos x will be – sin x.

We derive the formulas for the derivatives of the tangent and cotangent based on the rules of differentiation:

t g "x = sin x cos x" = sin "x cos x - sin x cos "x cos 2 x = = cos x cos x - sin x (- sin x) cos 2 x = sin 2 x + cos 2 x cos 2 x = 1 cos 2 x c t g "x = cos x sin x" = cos "x sin x - cos x sin "x sin 2 x = = - sin x sin x - cos x cos x sin 2 x = - sin 2 x + cos 2 x sin 2 x = - 1 sin 2 x

Derivatives of inverse trigonometric functions

The section on the derivative of inverse functions provides comprehensive information on the proof of the formulas for the derivatives of the arcsine, arccosine, arctangent and arccotangent, so we will not duplicate the material here.

Derivatives of hyperbolic functions

Proof 7

We can derive formulas for the derivatives of the hyperbolic sine, cosine, tangent and cotangent using the differentiation rule and the formula for the derivative of the exponential function:

s h "x = e x - e - x 2" = 1 2 e x "- e - x" == 1 2 e x - - e - x = e x + e - x 2 = c h x c h "x = e x + e - x 2" = 1 2 e x "+ e - x" == 1 2 e x + - e - x = e x - e - x 2 = s h x t h "x = s h x c h x" = s h "x c h x - s h x c h "x c h 2 x = c h 2 x - s h 2 x c h 2 x = 1 c h 2 x c t h "x = c h x s h x" = c h "x s h x - c h x s h "x s h 2 x = s h 2 x - c h 2 x s h 2 x = - 1 s h 2 x

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