Derivative proi. Derivative of a compound function

Derivative complex function. Solution examples

On the this lesson we will learn to find derivative of a complex function. The lesson is a logical continuation of the lesson How to find the derivative?, on which we analyzed the simplest derivatives, and also got acquainted with the rules of differentiation and some technical methods 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 will need to be done next action: , so the polynomial will be the inner 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 inner function has not changed, we do not touch it.

Well, it is quite obvious that

The final result of applying the formula 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 is the following:

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 independent decision(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 would look like a perversion funny. 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, you first need to take the derivative of the external 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 is the following:

Under the dash, we have a tricky function again! But it's already easier. It is easy to see that the inner function is the arcsine and the outer function is the degree. According to the rule of differentiation of a complex function, you first need to take the derivative of the degree.

And the theorem on the derivative of a complex function, the formulation of which is as follows:

Let 1) the function $u=\varphi (x)$ has a derivative $u_(x)"=\varphi"(x_0)$ at some point $x_0$, 2) the function $y=f(u)$ has at the corresponding point $u_0=\varphi (x_0)$ the derivative $y_(u)"=f"(u)$. Then the complex function $y=f\left(\varphi (x) \right)$ at the mentioned point will also have a derivative equal to the product of the derivatives of the functions $f(u)$ and $\varphi (x)$:

$$ \left(f(\varphi (x))\right)"=f_(u)"\left(\varphi (x_0) \right)\cdot \varphi"(x_0) $$

or, in shorter notation: $y_(x)"=y_(u)"\cdot u_(x)"$.

In the examples of this section, all functions have the form $y=f(x)$ (ie, we consider only functions of one variable $x$). Accordingly, in all examples, the derivative $y"$ is taken with respect to the variable $x$. To emphasize that the derivative is taken with respect to the variable $x$, one often writes $y"_x$ instead of $y"$.

Examples #1, #2, and #3 provide a detailed process for finding the derivative of complex functions. Example No. 4 is intended for a more complete understanding of the derivatives table and it makes sense to familiarize yourself with it.

It is advisable, after studying the material in examples No. 1-3, to move on to independently solving examples No. 5, No. 6 and No. 7. Examples #5, #6 and #7 contain a short solution so that the reader can check the correctness of his result.

Example #1

Find the derivative of the function $y=e^(\cos x)$.

We need to find the derivative of the complex function $y"$. Since $y=e^(\cos x)$, then $y"=\left(e^(\cos x)\right)"$. To find the derivative $ \left(e^(\cos x)\right)"$ use formula #6 from the table of derivatives. In order to use formula No. 6, you need to take into account that in our case $u=\cos x$. The further solution consists in a banal substitution of the expression $\cos x$ instead of $u$ into formula No. 6:

$$ y"=\left(e^(\cos x) \right)"=e^(\cos x)\cdot (\cos x)" \tag (1.1)$$

Now we need to find the value of the expression $(\cos x)"$. Again we turn to the table of derivatives, choosing formula No. 10 from it. Substituting $u=x$ into formula No. 10, we have: $(\cos x)"=-\ sin x\cdot x"$. Now we continue equality (1.1), supplementing it with the found result:

$$ y"=\left(e^(\cos x) \right)"=e^(\cos x)\cdot (\cos x)"= e^(\cos x)\cdot (-\sin x \cdot x") \tag (1.2) $$

Since $x"=1$, we continue equality (1.2):

$$ y"=\left(e^(\cos x) \right)"=e^(\cos x)\cdot (\cos x)"= e^(\cos x)\cdot (-\sin x \cdot x")=e^(\cos x)\cdot (-\sin x\cdot 1)=-\sin x\cdot e^(\cos x) \tag (1.3) $$

So, from equality (1.3) we have: $y"=-\sin x\cdot e^(\cos x)$. Naturally, explanations and intermediate equalities are usually skipped, writing the derivative in one line, as in the equality ( 1.3) So, the derivative of the complex function has been found, it remains only to write down the answer.

Answer: $y"=-\sin x\cdot e^(\cos x)$.

Example #2

Find the derivative of the function $y=9\cdot \arctg^(12)(4\cdot \ln x)$.

We need to calculate the derivative $y"=\left(9\cdot \arctg^(12)(4\cdot \ln x) \right)"$. To begin with, we note that the constant (i.e. the number 9) can be taken out of the sign of the derivative:

$$ y"=\left(9\cdot \arctg^(12)(4\cdot \ln x) \right)"=9\cdot\left(\arctg^(12)(4\cdot \ln x) \right)" \tag (2.1) $$

Now let's turn to the expression $\left(\arctg^(12)(4\cdot \ln x) \right)"$. To make it easier to select the desired formula from the table of derivatives, I will present the expression in question in this form: $\left( \left(\arctg(4\cdot \ln x) \right)^(12)\right)"$. Now it is clear that it is necessary to use formula No. 2, i.e. $\left(u^\alpha \right)"=\alpha\cdot u^(\alpha-1)\cdot u"$. Substitute $u=\arctg(4\cdot \ln x)$ and $\alpha=12$ into this formula:

Complementing equality (2.1) with the obtained result, we have:

$$ y"=\left(9\cdot \arctg^(12)(4\cdot \ln x) \right)"=9\cdot\left(\arctg^(12)(4\cdot \ln x) \right)"= 108\cdot\left(\arctg(4\cdot \ln x) \right)^(11)\cdot (\arctg(4\cdot \ln x))" \tag (2.2) $$

In this situation, a mistake is often made when the solver at the first step chooses the formula $(\arctg \; u)"=\frac(1)(1+u^2)\cdot u"$ instead of the formula $\left(u^\ alpha \right)"=\alpha\cdot u^(\alpha-1)\cdot u"$. The point is that the derivative of the external function must be found first. To understand which function will be external to the expression $\arctg^(12)(4\cdot 5^x)$, imagine that you are counting the value of the expression $\arctg^(12)(4\cdot 5^x)$ for some value of $x$. First you calculate the value of $5^x$, then multiply the result by 4 to get $4\cdot 5^x$. Now we take the arctangent from this result, getting $\arctg(4\cdot 5^x)$. Then we raise the resulting number to the twelfth power, getting $\arctg^(12)(4\cdot 5^x)$. Last action, - i.e. raising to the power of 12, - and will be an external function. And it is from it that one should begin finding the derivative, which was done in equality (2.2).

Now we need to find $(\arctg(4\cdot \ln x))"$. We use formula No. 19 of the derivatives table, substituting $u=4\cdot \ln x$ into it:

$$ (\arctg(4\cdot \ln x))"=\frac(1)(1+(4\cdot \ln x)^2)\cdot (4\cdot \ln x)" $$

Let's slightly simplify the resulting expression, taking into account $(4\cdot \ln x)^2=4^2\cdot (\ln x)^2=16\cdot \ln^2 x$.

$$ (\arctg(4\cdot \ln x))"=\frac(1)(1+(4\cdot \ln x)^2)\cdot (4\cdot \ln x)"=\frac( 1)(1+16\cdot \ln^2 x)\cdot (4\cdot \ln x)" $$

Equality (2.2) will now become:

$$ y"=\left(9\cdot \arctg^(12)(4\cdot \ln x) \right)"=9\cdot\left(\arctg^(12)(4\cdot \ln x) \right)"=\\ =108\cdot\left(\arctg(4\cdot \ln x) \right)^(11)\cdot (\arctg(4\cdot \ln x))"=108\cdot \left(\arctg(4\cdot \ln x) \right)^(11)\cdot \frac(1)(1+16\cdot \ln^2 x)\cdot (4\cdot \ln x)" \tag (2.3) $$

It remains to find $(4\cdot \ln x)"$. We take the constant (i.e. 4) out of the sign of the derivative: $(4\cdot \ln x)"=4\cdot (\ln x)"$. For In order to find $(\ln x)"$, we use formula No. 8, substituting $u=x$ into it: $(\ln x)"=\frac(1)(x)\cdot x"$. Since $x"=1$, then $(\ln x)"=\frac(1)(x)\cdot x"=\frac(1)(x)\cdot 1=\frac(1)(x )$ Substituting the result obtained into formula (2.3), we obtain:

$$ y"=\left(9\cdot \arctg^(12)(4\cdot \ln x) \right)"=9\cdot\left(\arctg^(12)(4\cdot \ln x) \right)"=\\ =108\cdot\left(\arctg(4\cdot \ln x) \right)^(11)\cdot (\arctg(4\cdot \ln x))"=108\cdot \left(\arctg(4\cdot \ln x) \right)^(11)\cdot \frac(1)(1+16\cdot \ln^2 x)\cdot (4\cdot \ln x)" =\\ =108\cdot \left(\arctg(4\cdot \ln x) \right)^(11)\cdot \frac(1)(1+16\cdot \ln^2 x)\cdot 4\ cdot \frac(1)(x)=432\cdot \frac(\arctg^(11)(4\cdot \ln x))(x\cdot (1+16\cdot \ln^2 x)).$ $

Let me remind you that the derivative of a complex function is most often in one line, as written in the last equality. Therefore, when making standard calculations or tests, it is not at all necessary to paint the solution in the same detail.

Answer: $y"=432\cdot \frac(\arctg^(11)(4\cdot \ln x))(x\cdot (1+16\cdot \ln^2 x))$.

Example #3

Find $y"$ of the function $y=\sqrt(\sin^3(5\cdot9^x))$.

First, let's slightly transform the $y$ function by expressing the radical (root) as a power: $y=\sqrt(\sin^3(5\cdot9^x))=\left(\sin(5\cdot 9^x) \right)^(\frac(3)(7))$. Now let's start finding the derivative. Since $y=\left(\sin(5\cdot 9^x)\right)^(\frac(3)(7))$, then:

$$ y"=\left(\left(\sin(5\cdot 9^x)\right)^(\frac(3)(7))\right)" \tag (3.1) $$

We use formula No. 2 from the table of derivatives, substituting $u=\sin(5\cdot 9^x)$ and $\alpha=\frac(3)(7)$ into it:

$$ \left(\left(\sin(5\cdot 9^x)\right)^(\frac(3)(7))\right)"= \frac(3)(7)\cdot \left( \sin(5\cdot 9^x)\right)^(\frac(3)(7)-1) (\sin(5\cdot 9^x))"=\frac(3)(7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) (\sin(5\cdot 9^x))" $$

We continue equality (3.1) using the obtained result:

$$ y"=\left(\left(\sin(5\cdot 9^x)\right)^(\frac(3)(7))\right)"=\frac(3)(7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) (\sin(5\cdot 9^x))" \tag (3.2) $$

Now we need to find $(\sin(5\cdot 9^x))"$. For this, we use formula No. 9 from the table of derivatives, substituting $u=5\cdot 9^x$ into it:

$$ (\sin(5\cdot 9^x))"=\cos(5\cdot 9^x)\cdot(5\cdot 9^x)" $$

Complementing equality (3.2) with the obtained result, we have:

$$ y"=\left(\left(\sin(5\cdot 9^x)\right)^(\frac(3)(7))\right)"=\frac(3)(7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) (\sin(5\cdot 9^x))"=\\ =\frac(3) (7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) \cos(5\cdot 9^x)\cdot(5\cdot 9 ^x)" \tag (3.3) $$

It remains to find $(5\cdot 9^x)"$. First, we take the constant (the number $5$) out of the sign of the derivative, i.e. $(5\cdot 9^x)"=5\cdot (9^x) "$. To find the derivative $(9^x)"$, we apply formula No. 5 of the table of derivatives, substituting $a=9$ and $u=x$ into it: $(9^x)"=9^x\cdot \ ln9\cdot x"$. Since $x"=1$, then $(9^x)"=9^x\cdot \ln9\cdot x"=9^x\cdot \ln9$. Now we can continue equality (3.3):

$$ y"=\left(\left(\sin(5\cdot 9^x)\right)^(\frac(3)(7))\right)"=\frac(3)(7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) (\sin(5\cdot 9^x))"=\\ =\frac(3) (7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) \cos(5\cdot 9^x)\cdot(5\cdot 9 ^x)"= \frac(3)(7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) \cos(5\cdot 9 ^x)\cdot 5\cdot 9^x\cdot \ln9=\\ =\frac(15\cdot \ln 9)(7)\cdot \left(\sin(5\cdot 9^x)\right) ^(-\frac(4)(7))\cdot \cos(5\cdot 9^x)\cdot 9^x. $$

You can return from powers to radicals (i.e. roots) again by writing $\left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7))$ as $\ frac(1)(\left(\sin(5\cdot 9^x)\right)^(\frac(4)(7)))=\frac(1)(\sqrt(\sin^4(5\ cdot 9^x)))$. Then the derivative will be written in the following form:

$$ y"=\frac(15\cdot \ln 9)(7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7))\cdot \cos(5\cdot 9^x)\cdot 9^x= \frac(15\cdot \ln 9)(7)\cdot \frac(\cos (5\cdot 9^x)\cdot 9^x) (\sqrt(\sin^4(5\cdot 9^x))). $$

Answer: $y"=\frac(15\cdot \ln 9)(7)\cdot \frac(\cos (5\cdot 9^x)\cdot 9^x)(\sqrt(\sin^4(5\ cdot 9^x)))$.

Example #4

Show that there are formulas No. 3 and No. 4 of the derivative table special case formula number 2 of this table.

In formula No. 2 of the table of derivatives, the derivative of the function $u^\alpha$ is written. Substituting $\alpha=-1$ into formula #2, we get:

$$(u^(-1))"=-1\cdot u^(-1-1)\cdot u"=-u^(-2)\cdot u"\tag (4.1)$$

Since $u^(-1)=\frac(1)(u)$ and $u^(-2)=\frac(1)(u^2)$, equality (4.1) can be rewritten as follows: $ \left(\frac(1)(u) \right)"=-\frac(1)(u^2)\cdot u"$. This is the formula number 3 of the derivatives table.

Let's turn again to formula No. 2 of the derivatives table. Substitute $\alpha=\frac(1)(2)$ into it:

$$\left(u^(\frac(1)(2))\right)"=\frac(1)(2)\cdot u^(\frac(1)(2)-1)\cdot u" =\frac(1)(2)u^(-\frac(1)(2))\cdot u"\tag (4.2) $$

Since $u^(\frac(1)(2))=\sqrt(u)$ and $u^(-\frac(1)(2))=\frac(1)(u^(\frac( 1)(2)))=\frac(1)(\sqrt(u))$, then equality (4.2) can be rewritten as follows:

$$ (\sqrt(u))"=\frac(1)(2)\cdot \frac(1)(\sqrt(u))\cdot u"=\frac(1)(2\sqrt(u) )\cdot u" $$

The resulting equality $(\sqrt(u))"=\frac(1)(2\sqrt(u))\cdot u"$ is formula No. 4 of the derivatives table. As you can see, formulas No. 3 and No. 4 of the derivatives table are obtained from formula No. 2 by substituting the corresponding value of $\alpha$.

How to find the derivative, how to take the derivative? In this lesson, we will learn how to find derivatives of functions. But before studying this page, I strongly recommend that you familiarize yourself with the methodological material.Hot Formulas school course mathematics. The reference manual can be opened or downloaded from the page Mathematical formulas and tables . Also from there we needDerivative table, it is better to print it, you will often have to refer to it, and not only now, but also offline.

There is? Let's get started. I have two news for you: good and very good. The good news is that in order to learn how to find derivatives, it is not at all necessary to know and understand what a derivative is. Moreover, the definition of the derivative of a function, mathematical, physical, geometric sense the derivative is more expedient to digest later, since a qualitative study of the theory, in my opinion, requires the study of a number of other topics, as well as some practical experience.

And now our task is to master these very derivatives technically. Highly good news is that learning to take derivatives is not so difficult, there is a fairly clear algorithm for solving (and explaining) this task, integrals or limits, for example, are more difficult to master.

I advise the following order of study of the topic: first, This article. Then you need to read the most important lesson Derivative of a complex function . These two basic classes will improve your skills with complete zero. Further, it will be possible to familiarize yourself with more complex derivatives in the article. complex derivatives.

logarithmic derivative. If the bar is too high, read the item first Protozoa typical tasks with derivative. In addition to the new material, the lesson covered other, more simple types derivatives, and there is a great opportunity to improve your differentiation technique. Besides, in control work almost always there are tasks for finding derivatives of functions that are specified implicitly or parametrically. There is also a tutorial for this: Derivatives of implicit and parametrically defined functions.

I will try in an accessible form, step by step, to teach you how to find derivatives of functions. All information is presented in detail, in simple words.

Actually, immediately consider an example: Example 1

Find the derivative of a function Solution:

it the simplest example, please find it in the table of derivatives of elementary functions. Now let's look at the solution and analyze what happened? And the following thing happened:

we had a function , which, as a result of the solution, turned into a function.

Quite simply, to find the derivative

functions, you need certain rules turn it into another function . Look again at the table of derivatives - there functions turn into other functions. the only

the exception is the exponential function, which

turns into itself. The operation of finding the derivative is calleddifferentiation.

Notation: The derivative is denoted or.

ATTENTION, IMPORTANT! Forgetting to put a stroke (where necessary), or drawing an extra stroke (where it is not necessary) is a GREAT MISTAKE! A function and its derivative are two different functions!

Let's return to our table of derivatives. From this table it is desirable memorize: rules of differentiation and derivatives of some elementary functions, especially:

derivative of a constant:

Where is a constant number; derivative power function:

In particular:,,.

Why memorize? This knowledge is elementary knowledge about derivatives. And if you can’t answer the teacher’s question “What is the derivative of the number?”, Then your studies at the university may end for you (I personally know two real cases from life). In addition, these are the most common formulas that we have to use almost every time we encounter derivatives.

AT In reality, simple tabular examples are rare; usually, when finding derivatives, differentiation rules are used first, and then a table of derivatives of elementary functions.

AT In this regard, we turn to the considerationdifferentiation rules:

1) A constant number can (and should) be taken out of the sign of the derivative

Where is a constant number (constant) Example 2

Find the derivative of a function

We look at the table of derivatives. The derivative of the cosine is there, but we have .

It's time to use the rule, we take out the constant factor beyond the sign of the derivative:

And now we turn our cosine according to the table:

Well, it is desirable to “comb” the result a little - put the minus in the first place, at the same time getting rid of the brackets:

2) The derivative of the sum is equal to the sum of the derivatives

Find the derivative of a function

We decide. As you probably already noticed, the first action that is always performed when finding the derivative is that we put the whole expression in brackets and put a stroke on the top right:

We apply the second rule:

Please note that for differentiation, all roots, degrees must be represented as , and if they are in the denominator, then

move them up. How to do this is discussed in my methodological materials.

Now we recall the first rule of differentiation - we take out the constant factors (numbers) outside the sign of the derivative:

Usually, during the solution, these two rules are applied simultaneously (so as not to rewrite a long expression once again).

All functions under the dashes are elementary table functions, using the table we perform the transformation:

You can leave everything in this form, since there are no more strokes, and the derivative has been found. However, expressions like this usually simplify:

It is desirable to represent all degrees of the species again as roots,

degrees from negative indicators- reset to the denominator. Although you can not do this, it will not be a mistake.

Find the derivative of a function

Try to solve this example yourself (answer at the end of the lesson).

3) Derivative of the product of functions

It seems that, by analogy, the formula suggests itself ...., but the surprise is that:

This unusual rule(as, in fact, others) follows from definitions of the derivative. But we will wait with the theory for now - now it is more important to learn how to solve:

Find the derivative of a function

Here we have the product of two functions depending on . First we apply our strange rule, and then we transform the functions according to the table of derivatives:

Difficult? Not at all, quite affordable even for a teapot.

Find the derivative of a function

This function contains the sum and product of two functions - the square trinomial and the logarithm. We remember from school that multiplication and division take precedence over addition and subtraction.

It's the same here. FIRST we use the product differentiation rule:

Now for the bracket we use the first two rules:

As a result of applying the rules of differentiation under the strokes, we have only elementary functions left, according to the table of derivatives we turn them into other functions:

With some experience in finding derivatives, simple derivatives do not seem to need to be described in such detail. In general, they are usually resolved verbally, and it is immediately recorded that .

Find the derivative of a function This is an example for self-solving (answer at the end of the lesson)

4) Derivative of private functions

A hatch has opened in the ceiling, don't be scared, it's a glitch. And here is the harsh reality:

Find the derivative of a function

What is not here - the sum, difference, product, fraction .... What should I start with?! There are doubts, no doubts, but, IN ANY CASE, first draw brackets and put a stroke at the top right:

Now we look at the expression in brackets, how would we simplify it? In this case, we notice a factor, which, according to the first rule, it is advisable to take it out of the sign of the derivative:

At the same time, we get rid of the brackets in the numerator, which are no longer needed. Generally speaking, the constant factors in finding the derivative

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 power function higher order. Examples of calculating derivatives.

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 variable x . Namely, let a be rational number. Then it can be represented as an irreducible fraction:
,
where m and n are integers without 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 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:
.

Solution

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:
.

Definition. Let the function \(y = f(x) \) be defined in some interval containing the point \(x_0 \) inside. Let's increment \(\Delta x \) to the argument so as not to leave this interval. Find the corresponding increment of the function \(\Delta y \) (when passing from the point \(x_0 \) to the point \(x_0 + \Delta x \)) and compose the relation \(\frac(\Delta y)(\Delta x) \). If there is a limit of this relation at \(\Delta x \rightarrow 0 \), then the indicated limit is called derivative function\(y=f(x) \) at the point \(x_0 \) and denote \(f"(x_0) \).

$$ \lim_(\Delta x \to 0) \frac(\Delta y)(\Delta x) = f"(x_0) $$

The symbol y is often used to denote the derivative. Note that y" = f(x) is a new function, but naturally associated with the function y = f(x), defined at all points x at which the above limit exists . This function is called like this: derivative of the function y \u003d f (x).

The geometric meaning of the derivative consists of the following. If a tangent that is not parallel to the y axis can be drawn to the graph of the function y \u003d f (x) at a point with the abscissa x \u003d a, then f (a) expresses the slope of the tangent:
\(k = f"(a)\)

Since \(k = tg(a) \), the equality \(f"(a) = tg(a) \) is true.

And now we interpret the definition of the derivative in terms of approximate equalities. Let the function \(y = f(x) \) have a derivative at a particular point \(x \):
$$ \lim_(\Delta x \to 0) \frac(\Delta y)(\Delta x) = f"(x) $$
This means that near the point x, the approximate equality \(\frac(\Delta y)(\Delta x) \approx f"(x) \), i.e. \(\Delta y \approx f"(x) \cdot\Deltax\). The meaningful meaning of the obtained approximate equality is as follows: the increment of the function is “almost proportional” to the increment of the argument, and the coefficient of proportionality is the value of the derivative in given point X. For example, for the function \(y = x^2 \) the approximate equality \(\Delta y \approx 2x \cdot \Delta x \) is valid. If we carefully analyze the definition of the derivative, we will find that it contains an algorithm for finding it.

Let's formulate it.

How to find the derivative of the function y \u003d f (x) ?

1. Fix value \(x \), find \(f(x) \)
2. Increment \(x \) argument \(\Delta x \), move to a new point \(x+ \Delta x \), find \(f(x+ \Delta x) \)
3. Find the function increment: \(\Delta y = f(x + \Delta x) - f(x) \)
4. Compose the relation \(\frac(\Delta y)(\Delta x) \)
5. Calculate $$ \lim_(\Delta x \to 0) \frac(\Delta y)(\Delta x) $$
This limit is the derivative of the function at x.

If the function y = f(x) has a derivative at the point x, then it is called differentiable at the point x. The procedure for finding the derivative of the function y \u003d f (x) is called differentiation functions y = f(x).

Let us discuss the following question: how are the continuity and differentiability of a function at a point related?

Let the function y = f(x) be differentiable at the point x. Then a tangent can be drawn to the graph of the function at the point M (x; f (x)) and, recall, the slope of the tangent is equal to f "(x). Such a graph cannot "break" at the point M, i.e., the function must be continuous at x.

It was reasoning "on the fingers". Let us present a more rigorous argument. If the function y = f(x) is differentiable at the point x, then the approximate equality \(\Delta y \approx f"(x) \cdot \Delta x \) holds. zero, then \(\Delta y \) will also tend to zero, and this is the condition for the continuity of the function at a point.

So, if a function is differentiable at a point x, then it is also continuous at that point.

The converse is not true. For example: function y = |x| is continuous everywhere, in particular at the point x = 0, but the tangent to the graph of the function at the “joint point” (0; 0) does not exist. If at some point it is impossible to draw a tangent to the function graph, then there is no derivative at this point.

One more example. The function \(y=\sqrt(x) \) is continuous on the entire number line, including at the point x = 0. And the tangent to the graph of the function exists at any point, including at the point x = 0. But at this point the tangent coincides with the y-axis, that is, it is perpendicular to the abscissa axis, its equation has the form x \u003d 0. There is no slope for such a straight line, which means that \ (f "(0) \) does not exist either

So, we got acquainted with a new property of a function - differentiability. How can you tell if a function is differentiable from the graph of a function?

The answer is actually given above. If at some point a tangent can be drawn to the graph of a function that is not perpendicular to the x-axis, then at this point the function is differentiable. If at some point the tangent to the graph of the function does not exist or it is perpendicular to the x-axis, then at this point the function is not differentiable.

Differentiation rules

The operation of finding the derivative is called differentiation. When performing this operation, you often have to work with quotients, sums, products of functions, as well as with "functions of functions", that is, complex functions. Based on the definition of the derivative, we can derive differentiation rules that facilitate this work. If C is a constant number and f=f(x), g=g(x) are some differentiable functions, then the following are true differentiation rules:

Table of derivatives of some functions