Derivative calculation - One of the most important operations in differential calculus. Below is a table of finding derivatives of simple functions. More complex Differentiation Rules, see other lessons:

• Table of derivatives of exponential and logarithmic functions

Limited formulas use as reference values. They will help in solving differential equations and tasks. In the picture, in the table of derivatives of simple functions, a "cheat sheet" of the basic cases of the derivative of the derivative in the formative for use, there are explanations for each case next to it.

Derivatives of simple functions

1. The derivative of the number is zero
c'\u003d 0.
Example:
5'\u003d 0.

Explanation:
The derivative shows the speed of changing the value of the function when the argument changes. Since the number does not change in no circumstances - the speed of its change is always zero.

2. Derivative of the variable equal to unity
x'\u003d 1.

Explanation:
With each increment of the argument (x) per unit, the value of the function (the result of the calculations) increases on the same size. Thus, the rate of change value of the function y \u003d x is exactly equal to the rate of change of the value of the argument.

3. The derivative of the variable and the multiplier is equal to this factor
cX' \u003d S.
Example:
(3x) '\u003d 3
(2x) '\u003d 2
Explanation:
In this case, with each change of the function argument ( h.) Its value (y) is growing in from time. Thus, the rate of change value of the function with respect to the rate of change of the argument is exactly equal from.

From where it follows that
(CX + B) "\u003d C
that is, the differential of the linear function y \u003d kx + B is equal to the angular coefficient of the tilt (K).

4. Module derivative equal to the private one variable to its module
| x | "\u003d x / | x | provided that x ≠ 0
Explanation:
Since the variable derivative (see Formula 2) is equal to unit, the derivative of the module is distinguished only by the fact that the value of the function of the function changes is changing to the opposite when the point of origin of the origin is crossed (try drawing a function of the Y \u003d | X | and make sure that it yourself. value and returns expression x / | x |. When x< 0 оно равно (-1), а когда x > 0 - unity. That is, with the negative values \u200b\u200bof the variable x each time an argument change, the value of the function is reduced to exactly the same value, and with positive - on the contrary, it increases, but exactly the same meaning.

5. Degree derivative equal to the product of the number of this degree and variable to the degree reduced by one
(x C) "\u003d CX C-1, provided that x C and CX C-1 are defined and C ≠ 0
Example:
(x 2) "\u003d 2x
(x 3) "\u003d 3x 2
To memorize formula:
Make the degree of the "down" variable as a multiplier, and then reduce the degree of the degree per unit. For example, for X 2 - two turned out to be ahead of the ICA, and then a reduced degree (2-1 \u003d 1) simply gave us 2x. The same thing happened for the X 3 - the top three "descend down", we reduce it per unit and instead of the cube we have a square, that is, 3x 2. A little "not scientifically", but very easy to remember.

6. Derived 1 / H.
(1 / x) "\u003d - 1 / x 2
Example:
Since the fraction can be represented as the construction of a negative degree
(1 / x) "\u003d (x -1)", then you can apply the formula from the rule 5 of the derivative table
(x -1) "\u003d -1x -2 \u003d - 1 / x 2

7. Derived with a variable degree in denominator
(1 / x c) "\u003d - C / X C + 1
Example:
(1 / x 2) "\u003d - 2 / x 3

8. The root derivative (variable derivative under square root)
(√x) "\u003d 1 / (2√x) or 1/2 x -1/2
Example:
(√x) "\u003d (x 1/2)" So you can apply the formula from rule 5
(x 1/2) "\u003d 1/2 x -1/2 \u003d 1 / (2√x)

9. Derivative variable under random degree
(n √x) "\u003d 1 / (n n √x n-1)

Operation of finding a derivative is called differentiation.

As a result of solving problems of finding derivatives from the simplest (and not very simple) functions to determine the derivative as a limit of the attitude towards an argument, a table of derivatives and precisely defined differentiation rules appeared. Isaac Newton (1643-1727) and Gottfried Wilhelm Leibniz (1646-1716) were first for the field of findings of derivatives.

Therefore, in our time, to find a derivative of any function, it is not necessary to calculate the above limit of the ratio of the increment of the function to increments the argument, and you only need to use the Table of Derivatives and Differentiation Rules. To find the derivative, the following algorithm is suitable.

To find a derivative, it is necessary for expression under the sign of the stroke disassemble the components of simple functions and determine what actions (Work, amount, private) These functions are connected. Next, derivatives of elementary functions are found in the table of derivatives, and formulas of derivatives, amounts and private - in the differentiation rules. Table of derivatives and differentiation rules are given after the first two examples.

Example 1. Find a derivative function

Decision. From the rules of differentiation, we find out that the derivative of the function of functions is the amount of derivatives, i.e.

From the table of derivatives, we find out that the derivative of the "ICCA" is equal to one, and the sinus derivative is cosine. We substitute these values \u200b\u200bin the amount of derivatives and we find the required condition of the task derivative:

Example 2. Find a derivative function

Decision. Differentiating as a derivative sum in which the second term with a constant factor can be reached by a derivative sign:

If there are questions yet, from where that it is taken, they are usually clarifying after familiarization with the table derivatives and the simplest differentiation rules. We go to them right now.

Table of derived simple functions

1. Derivative constant (numbers). Any number (1, 2, 5, 200 ...), which is in the expression of the function. Always equal to zero. It is very important to remember since it is necessary very often
2. The derivative of an independent variable. Most often "IKSA". Always equal to one. It is also important to remember for a long time.
3. derived degree. The degree in solving tasks you need to convert unquadant roots.
4. variable derivative to degree -1
5. Square root derivative
6. Sinus derivative
7. Cosine derivative
8. Derivative Tangent
9. The derivative of Kotangens
10. Arksinus derivative
11. ArcKosinus derivative
12. Arctangen derivative
13. Arkkotangen derivative
14. Derivative of natural logarithm
15. Derivative Logarithmic Function
16. Exhibit derivative
17. Derivative indicative function

Differentiation rules

1. Derivative amount or difference
2. Derivative work
2a. The derivative of the expression multiplied by the constant multiplier
3. Private derivative
4. Derivative complex function

Rule 1. If functions

differentially at some point, then at the same point differentiate and functions

and

those. The derivative of the algebraic amount of functions is equal to the algebraic amount of derivatives of these functions.

Corollary. If two differentiable functions differ on a permanent term, their derivatives are equal.

Rule 2.If functions

differentially at some point, then at the same point differently and their work

and

those. The derivative of the two functions is equal to the amount of the works of each of these functions on the different derivative.

Corollary 1. Permanent multiplier can be made for a derivative mark:

Corollary 2. The derivative of the work of several differentiable functions is equal to the amount of the products of the derivative of each of the factors to all other.

For example, for three multipliers:

Rule 3.If functions

differential at some point and , then at this point differently and their privateu / V, and

those. The derivative of the private two functions is equal to the fraction, the numerator of which is the difference in the products of the denominator on the derivative of the numerator and the numerator on the denominator derivative, and the denominator is the square of the previous numerator.

Where what to look for on other pages

When finding a derivative of the work and private in real tasks, several differentiation rules can always be applied, so more examples for these derivatives - in the article"Derivative work and private functions".

Comment.It should not be confused by a constant (that is, the number) as the term in the amount and as a constant multiplier! In the case of the foundation, its derivative is zero, and in the case of a constant multiplier, it is submitted for the sign of derivatives. This is a typical error that meets on initial stage study of derivatives, but as several single-volume examples have already solved medium student This error no longer does.

And if, with the differentiation of the work or private, you have a term appeared u."v. , in which u. - A number, for example, 2 or 5, that is, a constant, the derivative of this number will be zero and, therefore, the entire term will be zero (such a case is disassembled in Example 10).

Another frequent error is a mechanical solution of a derivative complex function as a derivative of a simple function. therefore derivative complex function Dedicated separate article. But first we will learn to find derivatives of simple functions.

In the course, do not do without transformations of expressions. To do this, you may need to open the benefits in new windows. Actions with degrees and roots and Actions with fractions .

If you are looking for solutions of derivatives with degrees and roots, that is, when the function is like a kind , Follow the occupation "Derivative of fractions with degrees and roots."

If you have a task like , then you are on the "derivatives of simple trigonometric functions".

Step-by-step examples - how to find a derivative

Example 3. Find a derivative function

Decision. We determine the part of the expression of the function: the whole expression represents the work, and its factors are sums, in the second of which one of the terms contains a permanent multiplier. We use a derivation of the product: a derivative of the work of two functions is equal to the amount of works of each of these functions on the different derivative:

Next, apply the amount of differentiation amount: the derivative of the algebraic amount of functions is equal to the algebraic amount of derivatives of these functions. In our case, every sum is the second term with a minus sign. In each sum, we see and an independent variable, the derivative of which is equal to one, and the constant (number), the derivative of which is zero. So, "X" we turn into one, and minus 5 - in zero. In the second expression "X" is multiplied by 2, so the two is multiplied by the same unit as a derivative of "IKSA". We obtain the following values \u200b\u200bof derivatives:

We substitute the found derivatives in the amount of works and obtain the required condition for the problem of the derivative of the entire function:

Example 4. Find a derivative function

Decision. We need to find a private derivative. Using the formula for differentiation of private: the derivative of the private two functions is equal to the fraction, the numerator of which is the difference of the products of the denominator on the derivative of the numerator and the numerator on the denominator derivative, and the denominator is the square of the previous numerator. We get:

We have already found a derivative of the factors in the Numertel in the example 2. I will not even forget that the work that is the second factory in the Numerator in the current example is taken with a minus sign:

If you are looking for solutions to such tasks in which it is necessary to find a derivative function, where the solid races of the roots and degrees, such as, for example, , then welcome to occupation "The derivative of fractions with degrees and roots" .

If you need to learn more about derivatives of sinuses, cosine, tangents and other trigonometric functions, that is, when the function seems like then you are on the lesson "Derivatives of simple trigonometric functions" .

Example 5. Find a derivative function

Decision. In this feature, we see the work, one of the factors of which - square root From an independent variable, with the derivative of which we got acquainted in the Table of Derivatives. According to the derivation of the product and the table value of the square root derivative, we get:

Example 6. Find a derivative function

Decision. In this feature, we see private, which is a square root from an independent variable. According to the rule of differentiation of the private, which we repeated and applied in Example 4, we obtain the tabletable value of the square root derivative:

To get rid of the fraction in the numerator, multiply the numerator and denominator on.

First level

Derived function. Exhaustive guide (2019)

Imagine a straight road passing through a hilly area. That is, it goes up, then down, but right or left does not turn. If the axis is directed along the road horizontally, and - vertically, then the line of the road will be very similar to a schedule of some continuous function:

The axis is a certain level of zero height, we use the level of the sea as it.

Moving forward on such a road, we also move up or down. We can also say: when the argument is changed (advanced along the abscissa axis) the value of the function changes (movement along the ordinate axis). And now let's think about how to determine the "steepness" of our road? What could it be for the magnitude? Very simple: how much will the height change when moving forward for a certain distance. After all, at different parts of the road, moving forward (along the abscissa axis) for one kilometer, we will rise or fall on a different number of meters relative to the sea level (along the ordinate axis).

Promotion forward to be denoted (read "Delta X").

The Greek letter (Delta) in mathematics is usually used as a prefix meaning "change". That is - this is a change in the value - change; Then what is? That's right, change value.

Important: Expression is a single integer, one variable. You can never tear off the "Delta" from "IKSA" or any other letter! That is, for example.

So, we advanced forward, horizontally, on. If the line of the road we compare the function with a graph, then how do we designate the rise? Sure, . That is, when moving forward on we rising above.

It is easy to calculate the amount: if at the beginning we were at the height, and after moving were at the height, then. If the end point turned out to be lower than the initial, it will be negative - this means that we do not go up, but let down.

Let's go back to the "steepness": this is the value that shows how much strongly (cool) increases the height when moving forward per unit distance:

Suppose that on a site of the path when moving at km the road rises upwards at km. Then the steepness in this place is equal. And if the road when promoting on m sank to km? Then the steep is equal to.

Now consider the top of some hill. If you take the beginning of the site for half a kilometer to the top, and the end - after half a kilometer after it, it can be seen that the height is almost the same.

That is, in our logic it turns out that the steepness here is almost equal to zero, which is clearly not true. Just at a distance in km can change a lot. It is necessary to consider smaller sections for a more adequate and accurate assessment of the steepness. For example, if you measure the change in the height when moving to one meter, the result will be much more accurate. But this accuracy may not be enough for us - because if there is a pillar in the middle of the road, we can simply slip it. What distance then choose? Centimeter? Millimeter? Less is better!

IN real life Measure the distance with an accuracy to the milimeter - more than enough. But mathematicians always strive for perfection. Therefore, the concept was invented infinitely smallThat is, the magnitude of the module is less than any number that can only be called. For example, you say: one trillion! Where is less? And you filed this number on - and it will be even less. Etc. If we want to write that the magnitude is infinitely small, we write like this: (I read "X is striving for zero"). It is very important to understand that this number is not zero! But very close to it. This means that it can be divided into it.

The concept opposite is infinitely small - infinitely large (). You already probably snapped with him when I was engaged in inequalities: this is the number of module more than any number that can be invented. If you came up with the largest of the possible numbers, just multiply it to two, and it will turn out even more. And infinity even more than what happens. In fact, the infinitely large and infinitely small reversed each other, that is, when, and on the contrary: when.

Now back to our road. The perfectly counted steepness is a bengeon, calculated for an infinitely small segment of the path, that is:

I note that with an infinitely small movement, the change in the height will also be infinitely small. But I remind you, infinitely small - does not mean equal to zero. If you share infinitely small numbers each other, it may be quite a common number, for example. That is, one low value can be exactly more than once more.

What is all this? The road, steepness ... We are not going to go to the rally, and we learn mathematics. And in mathematics everything is just the same, only called differently.

The concept of derivative

The derivative of the function is the ratio of the increment of the function to the increment of the argument with an infinitely small increment of the argument.

Increment In mathematics call change. How much the argument changed () when moving along the axis is called increment of argument and referred to how much the function changed (height) when moving forward along the axis is called, called increment of function and is denoted.

So, the derived function is attitude to when. We indicate the derivative of the same letter as the function, only with the stroke on the right: or simply. So, we will write the derivative formula using these notation:

As in analogy with expensive here, with an increase in the function, the derivative is positive, and when decreasing is negative.

Does the derivative happen to zero? Sure. For example, if we are going along a flat horizontal road, the steep is zero. And the truth is, the height is not entirely changing. So with the derivative: the derivative of the constant function (constant) is zero:

since the increment of such a function is zero at any.

Let's remember the example from the hill. It turned out that it was possible that you can position the ends of the segment along different directions from the vertex that the height at the ends turns out to be the same, that is, the segment is located in parallel axis:

But large segments are a sign of inaccurate measurement. We will raise our cut up parallel to yourself, then its length will decrease.

In the end, when we are infinitely close to the top, the length of the segment will become infinitely small. But at the same time, it remained parallel to the axis, that is, the height difference at its ends is zero (does not seek, namely equal to). So derivative

It is possible to understand this: when we stand on the top of the top, the little displacement to the left or right changes our height is negligible.

There is a purely algebraic explanation: the left of the top is the function increases, and to the right - decreases. As we have already found out earlier, with an increase in the function, the derivative is positive, and as descending, is negative. But it changes smoothly, without jumps (because the road does not change the slope anywhere). Therefore, between negative and positive values \u200b\u200bmust be. He will be where the function neither increases, nor decreases - at the point of the vertex.

The same is true for the depression (the area where the function on the left decreases, and on the right - increases):

A little more about increments.

So, we change the argument by magnitude. Change from what value? What is he (argument) now? We can choose any point, and now we will dance from it.

Consider a point with the coordinate. The value of the function in it is equal. Then make something increment: increase the coordinate on. What is the argument now? Very easy: . And what is the value of the function now? Where the argument, there and the function :. And what about the increment of the function? Nothing new: it is still the magnitude of which the function has changed:

Practice to find increments:

1. Find the increment of the function at the point when the argument is increasing.
2. The same for the function at the point.

Solutions:

At different points at one and the same increment of the argument, the increment of the function will be different. It means that the derivative at every point is its own (we discussed at the very beginning - the steepness of the road at different points is different). Therefore, when we write a derivative, you must specify at what point:

Power function.

The power is called the function where the argument is to some extent (logical, yes?).

Moreover, to either :.

The simplest case is when the degree indicator:

We find its derivative at the point. We remember the definition of the derivative:

So, the argument changes from before. What is the increment of the function?

Increment is. But the function at any point is equal to its argument. Therefore:

The derivative is equal to:

Derived from equal:

b) now consider quadratic function (): .

And now remember that. This means that the value of increment can be neglected, since it is infinitely small, and therefore insignificantly against the background of another term:

So, we were born next rule:

c) We continue the logical range :.

This expression can be simplified in different ways: to reveal the first bracket by the formula of the abbreviated multiplication of the cube amount, or decompose the entire expression on the factors by the Cube difference formula. Try to do it yourself by any of the proposed ways.

So, I got the following:

And again remember that. This means that you can neglect by all the terms containing:

We get :.

d) similar rules can be obtained for large degrees:

e) it turns out that this rule can be generalized for power function With an arbitrary indicator, not even:

(2)

You can formulate the rule with words: "The degree is taken forward as a coefficient, and then decreases by".

Let us prove this rule later (almost at the very end). And now consider a few examples. Find derived functions:

1. (in two ways: by the formula and using the derivative determination - considering the increment of the function);

1. . You will not believe, but this is a power function. If you have any questions like "How is it? And where is the degree? ", Remember the topic" "!
   Yes, the root is also the degree, only fractional :.
   So our square root is just a degree with an indicator:
   .
   We are looking for a recently learned formula:

   If in this place it became incomprehensible again, repeat the topic "" !!! (about the degree with a negative indicator)

2. . Now the indicator of the degree:

   And now through the definition (I have not forgotten yet?):
   ;
   .
   Now, as usual, neglect the terms containing:
   .

3. . Combination of previous cases :.

Trigonometric functions.

Here we will use one fact of the highest mathematics:

When expressing.

Proof You will know in the first year of the Institute (and to be there, you need to pass it well). Now just show it graphically:

We see that when the function does not exist - the point on the graph of the population. But the closer to the value, the closer the function to. This is the most "striving."

You can additionally check this rule using the calculator. Yes, yes, do not be shy, take a calculator, we are not on the exam yet.

So try :;

Do not forget to transfer the calculator to "Radian" mode!

etc. We see that the smaller, the closer the value of the relationship to.

a) Consider the function. As usual, we will find its increment:

Turn the difference in sines into the work. To do this, we use the formula (remember the topic "") :.

Now the derivative:

We will replace :. Then, with infinitely small, it is also infinitely small :. The expression for takes the form:

And now you remember that when expressing. And also that if the infinitely low value can be neglected in the amount (that is, when).

So, we get the following rule: sinus derivative equal to cosine:

This is basic ("tabular") derivatives. Here they are one list:

Later we add to them a few more, but these are the most important, as they are most often used.

Practice:

1. Find derived function at point;
2. Find derived function.

Solutions:

1. At first we will find the derivative in general form, and then substitute instead of its value:
   ;
   .
2. Here we have something similar to the power function. Let's try to bring it to
   Normal form:
   .
   Excellent, now you can use the formula:
   .
   .
3. . Eeeeee ... .. What is it ????

Okay, you are right, we still do not know how to find such derivatives. Here we have a combination of several types of functions. To work with them, you need to learn a few more rules:

Exhibitor and natural logarithm.

There is such a function in mathematics, the derivative of which with any equal value of the function itself at the same way. It is called "Exhibitor", and is an indicative function

The basis of this function is a constant - it is an infinite decimalThat is, the number is irrational (such as). It is called "the number of Euler", therefore and denote the letter.

So, the rule:

Remember very easy.

Well, let's not go far, immediately consider reverse function. What function is the reverse for an indicative function? Logarithm:

In our case, the basis is the number:

Such a logarithm (that is, a logarithm with a base) is called "natural", and for it we use a special designation: instead of writing.

What is equal to? Of course, .

The derivative of the natural logarithm is also very simple:

Examples:

1. Find derived function.
2. What is the derived function equal?

Answers: Exhibitor I. natural logarithm - Functions are uniquely simple from the point of view of the derivative. Exchange and logarithmic functions with any other base will have another derivative, which we will analyze later with you, after passing the differentiation rules.

Differentiation rules

Rules What? Again the new term, again?! ...

Differentiation - This is the process of finding a derivative.

Only and everything. And how else to name this process in one word? Not a production of ... The differential of mathematics is called the most increment of the function at. This term is happening from Latin Differentia - a difference. Here.

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

Total there are 5 rules.

The constant is made out of the sign of the derivative.

If - some kind of constant number (constant), then.

Obviously, this rule works for difference :.

We prove. Let, or easier.

Examples.

Find derived functions:

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

Solutions:

1. (the derivative is the same in all points, as it is linear function, remember?);

Derived work

Here everything is similar: we introduce a new function and find its increment:

Derivative:

Examples:

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

Solutions:

Derivative indicative function

Now your knowledge is enough to learn how to find a derivative of any indicative function, and not just exhibitors (not forgotten what it is?).

So, where is some number.

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

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

Well, it turned out. Now try to find a derivative, and do not forget that this feature is complex.

Happened?

Here, check yourself:

The formula turned out to be very similar to the derivative exhibit: as it was, it remained, only a multiplier appeared, which is just a number, but not a variable.

Examples:
Find derived functions:

Answers:

This is just a number that cannot be counted without a calculator, that is, not to record in a simpler form. Therefore, in response in this form and leave.

Derivative logarithmic function

Here is similar: you already know the derivative from the natural logarithm:

Therefore, to find an arbitrary from logarithm with another reason, for example:

You need to bring this logarithm to the base. And how to change the basis of the logarithm? I hope you remember this formula:

Only now instead we will write:

In the denominator, it turned out just a constant (constant number, without a variable). The derivative is very simple:

Derivatives indicative I. logarithmic functions Almost not found in the exam, but it will not be superfluous to know them.

Derivative complex function.

What is a "complex function"? No, it is not a logarithm, and not Arcthangence. These functions can be complex for understanding (although if the logarithm seems to you difficult, read the topic "Logarithms" and everything will pass), but from the point of view of mathematics the word "complex" does not mean "difficult".

Imagine a small conveyor: two people are sitting and have some kind of actions with some objects. For example, the first wraps a chocolate in the wrapper, and the second implies it with a ribbon. It turns out such an integral object: a chocolate, wrapped and lined with a ribbon. To eat a chocolate, you need to do reverse action in reverse order.

Let's create a similar mathematical conveyor: first we will find a cosine of the number, and then the resulting number to be erected into a square. So, we give a number (chocolate), I find his cosine (wrap), and then you will be erected by what I did, in a square (tie to the ribbon). What happened? Function. This is an example of a complex function: when to find its meanings we do the first action directly with the variable, and then another action with what happened as a result of the first one.

We can completely do the same actions and in the reverse order: first you will be erected into a square, and then I'm looking for a cosine of the resulting number :. It is easy to guess that the result will be almost always different. Important feature Complex functions: when the procedure change, the function changes.

In other words, a complex function is a function, the argument of which is another feature.: .

For the first example,.

The second example: (the same). .

Action that we do the latter will call "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 myself what function is external, and which is internal:

Answers:The separation of internal and external functions is very similar to replacement of variables: for example, in function

1. First we will perform what action? First, consider sinus, but only then erected into the cube. So, the internal function, and the external one.
   And the initial function is their composition :.
2. Inner:; External :.
   Check :.
3. Inner:; External :.
   Check :.
4. Inner:; External :.
   Check :.
5. Inner:; External :.
   Check :.

we produce a replacement of variables and get a function.

Well, now we will extract our chocolate chocolate - search for a derivative. The procedure is always reverse: first we are looking for an external function derivative, then multiply the result on the derivative of the internal function. With regard to the original example, it looks like this:

Another example:

So, we finally formulate the official rule:

The algorithm for finding a derivative complex function:

It seems to be simple, yes?

Check on the examples:

Solutions:

1) internal :;

External:;

2) internal :;

(Only do not think now to cut on! From under the cosine, nothing is done, remember?)

3) internal :;

External:;

It is immediately seen that here is a three-level complex function: after all, it is already a complex function itself, and it is still removing the root from it, that is, we perform the third action (chocolate in the wrapper and with a ribbon put into the portfolio). But there are no reason to be afraid: all the same "unpack" this function will be in the same order as usual: from the end.

That is, first use the root, then cosine, and only then expression in brackets. And then all this variables.

In such cases, it is convenient to numbered actions. That is, imagine that we are known. What order are we going to perform actions to calculate the value of this expression? We will examine on the example:

The later the action takes place, the more the "external" will be the corresponding function. Sequence of actions - as before:

Here the nesting is generally 4-level. Let's determine the procedure.

1. Forced expression. .

2. Root. .

3. Sinus. .

4. Square. .

5. We collect everything in a bunch:

DERIVATIVE. Briefly about the main thing

Derived function - The ratio of the increment of the function to the increment of the argument with an infinitely small increment of the argument:

Basic derivatives:

Differentiation Rules:

The constant is made for the sign of the derivative:

Derived amount:

Production work:

Private derivative:

Derivative complex function:

Algorithm for finding a derivative of complex function:

1. We define the "internal" function, we find its derivative.
2. We define the "external" function, we find its derivative.
3. Multiply the results of the first and second items.

This video, I start a long series of lessons dedicated to the derivative. This lesson consists of several parts.

First of all, I will tell you that in general such derivatives and how to count them, but not a wisdom academic language, but as I myself understand and as I explain to my students. Secondly, we will consider the simplest rule to solve problems in which we will look for derivative sums, derivative differences and derivatives of the power function.

We will look at more complex combined examples, of which you, in particular, learn that such problems containing roots and even fractions can be solved using the formula of the derivative of the power function. In addition, of course, there will be many tasks and examples of solutions of the most different level of complexity.

In general, initially I was going to write a short 5-minute roller, but see what happened from it. Therefore, the lyrics are enough - proceed to business.

What is a derivative?

So let's start from afar. Many years ago, when the trees were greine, and life was more fun, mathematics thought about what: Consider a simple function specified by your schedule, we call it $ Y \u003d F \\ Left (x \\ right) $. Of course, the schedule exists in itself, so you need to spend the axis of $ x $, as well as the axis $ y $. And now let's choose any point on this chart, absolutely any. The abscissa is called $ ((x) _ (1)) $, ordinate, as it is not difficult to guess, there will be $ F \\ left (((x) _ (1)) \\ Right) $.

Consider on the same schedule another point. It doesn't matter what, most importantly, it differs from initial. In her, again, there is an abscissa, we call it $ ((x) _ (2)) $, as well as the ordinate - $ F \\ left (((x) _ (2)) \\ Right) $.

So, we received two points: they have different abscissa and, therefore, different values Functions, although the latter is optional. But what is really important, so this is that from the course of the planimeria we know: you can spend directly in two points and, and only one. Here let's spend it and spend.

And now I will spend through the very first of them direct, parallel axis of the abscissa. Receive right triangle. Let's give it $ ABC $, a direct angle of $ C $. This triangle has one very interesting property: the fact is that the angle $ \\ alpha $ is actually equal to the corner under which the direct $ AB $ is intersect with the continuation of the abscissa axis. Judge for yourself:

1. direct $ AC $ parallel to the axis of $ ox $ by construction,
2. direct $ AB $ crosses $ AC $ under $ \\ alpha $
3. consequently, $ AB $ crosses $ Ox $ under the same $ \\ alpha $.

What we can say about $ \\ text () \\! \\! \\ Alpha \\! \\! \\ Text () $? Nothing concrete, except that in the $ ABC $ triangle ratio of the $ BC $ rattu to the $ AC $ cathelet is equal to the tangent of this very corner. So write:

Of course, $ AC $ in this case is easily considered:

Similarly, $ BC $:

In other words, we can record the following:

\\ [\\ Operatorname (TG) \\ Text () \\! \\! \\ alpha \\! \\! \\ Text () \u003d \\ FRAC (F \\ Left (((x) _ (2)) \\ Right) -f \\ left ( ((x) _ (1)) \\ RIGHT)) (((x) _ (2)) - ((x) _ (1))) \\]

Now that we all found out, let's go back to our schedule and consider a new point $ B $. Stretch old values \u200b\u200band take and take $ b $ somewhere closer to $ ((x) _ (1)) $. Again, we denote it to abscissue for $ ((x) _ (2)) $, and the ordinate is $ f \\ left (((x) _ (2)) \\ Right) $.

We will consider our little triangle $ ABC $ and $ \\ text () \\! \\! \\ Alpha \\! \\! \\ Text () $ inside it. It will be quite obvious that it will be a completely different angle, the tangent will also be different because the lengths of the divisions of $ AC $ and $ BC $ have changed significantly, and the formula for the tangent of the angle did not change at all - this is still the ratio between the change of function and the change of the argument .

Finally, we continue to move $ b $ getting closer to the original point $ a $, as a result, the triangle will still decrease, and the direct containing the $ AB segment will increasingly be like a function tangent.

As a result, if you continue the rapprochement of points, i.e., reduce the distance to zero, then the direct $ AB $, indeed, will turn into a tangent to the schedule at this point, and $ \\ text () \\! \\! \\ Alpha \\! \\! \\! \\! \\ Alpha And here we smoothly go to the definition of $ F $, namely, the derivative function at $ (((x) _ (1)) $ is called the $ \\ alpha $ tangent between the tangent to the graph at $ ((x) _ ((x) _ ( 1)) $ and the positive direction of the axis of $ OX $:

\\ [(f) "\\ left (((x) _ (1)) \\ Right) \u003d \\ Operatorname (TG) \\ Text () \\! \\! \\ Alpha \\! \\! \\ Text () \\]

Returning to our schedule, it should be noted that as $ ((x) _ (1)) $ you can choose any point on the chart. For example, with the same success, we could remove the bar at the point shown in the picture.

The angle between the tangent and positive direction of the axis will call $ \\ beta $. Accordingly, $ F $ per $ ((x) _ (2)) $ will be equal to the tangent of this angle of $ \\ beta $.

\\ [(f) \\ left (((x) _ (2)) \\ Right) \u003d TG \\ TEXT () \\! \\! \\ BETA \\! \\! \\ Text () \\]

At each point of the graph there will be its own tangent, and, therefore, its value of the function. In each of these cases, in addition to the point in which we are looking for a differential derivative or amount, or a derivative of a power function, you need to take another point that is at some distance from it, and then rush this point to the original and, of course, to find out how in the process This movement will change the tangent angle of inclination.

The derivative of the power function

Unfortunately, this definition does not suit us. All these formulas, pictures, corners do not give us the slightest idea of \u200b\u200bhow to consider a real derivative in real tasks. Therefore, let's take a little bit from formal definition and consider more efficient formulas and techniques, with which these tasks can already be solved.

Let's start with the most simple structures, namely, the functions of the form $ y \u003d ((x) ^ (n)) $, i.e. power functions. In this case, we can write the following: $ (y) "\u003d n \\ cdot ((x) ^ (n-1)) $. In other words, the degree that stood in the indicator is shown in the multiplier in front, and the indicator itself decreases Unit. For example:

\\ [\\ begin (align) & y \u003d ((x) ^ (2)) \\\\ & (y) "\u003d 2 \\ Cdot ((x) ^ (2-1)) \u003d 2x \\\\\\ End (Align) \\]

But another option:

\\ [\\ begin (align) & y \u003d ((x) ^ (1)) \\\\ & (y) "\u003d ((\\ left (x \\ right)) ^ (\\ Prime)) \u003d 1 \\ CDOT ((x ) ^ (0)) \u003d 1 \\ Cdot 1 \u003d 1 \\\\ \\ ((\\ left (x \\ right)) ^ (\\ Prime)) \u003d 1 \\\\\\ End (Align) \\]

Using these simple rules, let's try to remove the barcode of the following examples:

So we get:

\\ [((\\ left (((x) ^ (6)) \\ Right)) ^ (\\ Prime)) \u003d 6 \\ Cdot ((x) ^ (5)) \u003d 6 ((x) ^ (5)) \\]

Now we solve the second expression:

\\ [\\ begin (align) & f \\ left (x \\ right) \u003d ((x) ^ (100)) \\\\ \\ ((\\ left (((x) ^ (100)) \\ Right)) ^ (\\ Of course, it was very

Simple tasks . butReal tasks more complex and they are not limited to the only degrees of the function. So, Rule No. 1 - if the function is represented as other two, the derivative of this amount is equal to the sum of the derivatives:

\\ [((\\ left (f + g \\ right)) ^ (\\ Prime)) \u003d (f) "+ (g)" \\]

Similarly, the derivative of the difference of two functions is equal to the difference of derivatives:

\\ [((\\ left (F-G \\ Right)) ^ (\\ Prime)) \u003d (f) "- (g)" \\]

\\ [((\\ left (((x) ^ (2)) + x \\ right)) ^ (\\ Prime)) \u003d ((\\ left (((x) ^ (2)) \\ Right)) ^ (\\ In addition, there is another important rule: if there is a $ C $ constant before some $ F $, to which this function is multiplied, then $ F $ all this design is considered as:

\\ [((\\ left)) ^ (\\ Prime)) \u003d C \\ CDot (F) "\\]

\\ [((\\ left (3 ((x) ^ (3)) \\ RIGHT)) ^ (\\ Prime)) \u003d 3 ((\\ left (((x) ^ (3)) \\ Right)) ^ (\\ Finally, another very important rule: In tasks, a separate term is often found, which does not contain $ x $. For example, we can observe it in our current expressions. Derivative constant, i.e., the numbers, in no way dependent on $ x $, is always equal to zero, and it is completely no matter what the $ C constant is equal to:

\\ [((\\ left (C \\ RIGHT)) ^ (\\ Prime)) \u003d 0 \\]

Example Solution:

\\ [((\\ left (1001 \\ right)) ^ (\\ Prime)) \u003d ((\\ left (\\ FRAC (1) (1000) \\ RIGHT)) ^ (\\ Prime)) \u003d 0 \\]

Once again key points:

The derivative of the two functions is always equal to the sum of the derivatives: $ ((\\ left (f + g \\ right)) ^ (\\ Prime)) \u003d (f) "+ (g)" $;

1. For similar reasons, the derivative of the difference of two functions is equal to the difference of two derivatives: $ ((\\ left (F-G \\ Right)) ^ (\\ Prime)) \u003d (f) "- (g)" $;
2. If the function has a constant multiplier, then this constant can be made for a derivative sign: $ ((\\ left (C \\ CDOT F \\ RIGHT)) ^ (\\ Prime)) \u003d C \\ CDot (F) "$;
3. If the entire function is a constant, then its derivative is always zero: $ ((\\ left (C \\ Right)) ^ (\\ Prime)) \u003d 0 $.
4. Let's see how it all works on real examples. So:

We write:

\\ [\\ begin (align) & ((\\ left (((x) ^ (5)) - 3 ((x) ^ (2)) + 7 \\ right)) ^ (\\ Prime)) \u003d ((\\ left (((x) ^ (5)) \\ RIGHT)) ^ (\\ Prime)) - ((\\ left (3 ((x) ^ (2)) \\ Right)) ^ (\\ Prime)) + (7) "\u003d \\\\ \\ \u003d 5 ((x) ^ (4)) - 3 ((\\ left (((x) ^ (2)) \\ Right)) ^ (\\ Prime)) + 0 \u003d 5 ((x) ^ (4)) - 6x \\\\\\ End (Align) \\]

In this example, we see the derivative sum, and the difference derivative. Total, the derivative is $ 5 ((x) ^ (4)) - 6x $.

Go to the second function:

We write down the solution:

\\ [\\ begin (align) & ((\\ left (3 ((x) ^ (2)) - 2x + 2 \\ Right)) ^ (\\ Prime)) \u003d ((\\ left (3 ((x) ^ ( 2)) \\ Right)) ^ (\\ Prime)) - ((\\ left (2x \\ right)) ^ (\\ Prime)) + (2) "\u003d \\\\ · 3 (((\\ left (((x) ^ (2)) \\ Right)) ^ (\\ Prime)) - 2 (x) "+ 0 \u003d 3 \\ CDOT 2x-2 \\ Cdot 1 \u003d 6x-2 \\\\\\ End (Align) \\]

So we found the answer.

Go to the third function - it is already trying:

\\ [\\ begin (align) & ((\\ left (2 ((x) ^ (3)) - 3 (((x) ^ (2)) + \\ FRAC (1) (2) X-5 \\ RIGHT)) ^ (\\ Prime)) \u003d ((\\ left (2 ((x) ^ (3)) \\ right)) ^ (\\ Prime)) - ((\\ left (3 ((x) ^ (2)) \\ RIGHT )) ^ (\\ Prime)) + ((\\ left (\\ FRAC (1) (2) X \\ RIGHT)) ^ (\\ Prime)) - (5) "\u003d \\\\ & \u003d 2 ((\\ left (( (x) ^ (3)) \\ RIGHT)) ^ (\\ Prime)) - 3 ((\\ left (((x) ^ (2)) \\ Right)) ^ (\\ Prime)) + \\ FRAC (1) (2) \\ Cdot (x) "\u003d 2 \\ Cdot 3 ((x) ^ (2)) - 3 \\ CDOT 2X + \\ FRAC (1) (2) \\ Cdot 1 \u003d 6 ((x) ^ (2)) -6x + \\ FRAC (1) (2) \\\\\\ End (Align) \\]

We found the answer.

Go to the last expression - the most complex and most long:

So, we believe:

\\ [\\ begin (align) & ((\\ left (6 ((x) ^ (7)) - 14 (((x) ^ (3)) + 4x + 5 \\ right)) ^ (\\ Prime)) \u003d ( (\\ left (6 ((x) ^ (7)) \\ RIGHT)) ^ (\\ Prime)) - ((\\ left (14 ((x) ^ (3)) \\ Right)) ^ (\\ Prime)) + ((\\ left (4x \\ Right)) ^ (\\ Prime)) + (5) "\u003d \\\\ \\ \u003d 6 \\ CDOT 7 \\ CDOT ((X) ^ (6)) - 14 \\ CDOT 3 ((x ) ^ (2)) + 4 \\ Cdot 1 + 0 \u003d 42 (((x) ^ (6)) - 42 ((x) ^ (2)) + 4 \\\\\\ End (Align) \\]

But this decision does not end, because we are asked not to just remove the touch, but to calculate its value at a specific point, so we substitute in the expression -1 instead of $ x $:

\\ [(y) "\\ left (-1 \\ right) \u003d 42 \\ CDOT 1-42 \\ CDOT 1 + 4 \u003d 4 \\]

We are following and go to even more complex and interesting examples. The fact is that the formula for solving a power derivative $ ((\\ left (((x) ^ (n)) \\ Right)) ^ (\\ Prime)) \u003d n \\ cdot ((x) ^ (n-1)) $ It has an even wider area of \u200b\u200bapplication than it is usually customary. With it, it is possible to solve examples with fractions, roots, etc. It is this that we will now go.

To begin with, write down once again, which will help us find a derivative of the power function:

And now attention: so far we have considered as $ n $ only integersHowever, do not interfere with considering the fractions and even negative numbers. For example, we can record the following:

\\ [\\ begin (align) \\ sqrt (x) \u003d ((x) ^ (\\ FRAC (1) (2))) \\\\ & ((\\ left (\\ sqrt (x) \\ right)) ^ (\\ ^ (- \\ FRAC (1) (2))) \u003d \\ FRAC (1) (2) \\ CDOT \\ FRAC (1) (\\ SQRT (X)) \u003d \\ FRAC (1) (2 \\ SQRT (X)) \\\\\\ End (Align) \\]

Nothing complicated, so let's see how this formula will help us when solving more complex tasks. So example:

\\ [\\ begin (align) & ((\\ left (3 ((x) ^ (2)) - 2x + 2 \\ Right)) ^ (\\ Prime)) \u003d ((\\ left (3 ((x) ^ ( 2)) \\ Right)) ^ (\\ Prime)) - ((\\ left (2x \\ right)) ^ (\\ Prime)) + (2) "\u003d \\\\ · 3 (((\\ left (((x) ^ (2)) \\ Right)) ^ (\\ Prime)) - 2 (x) "+ 0 \u003d 3 \\ CDOT 2x-2 \\ Cdot 1 \u003d 6x-2 \\\\\\ End (Align) \\]

\\ [\\ begin (align) \\ left (\\ sqrt (x) + \\ sqrt (x) + \\ sqrt (x) \\ right) \u003d ((\\ left (\\ sqrt (x) \\ right)) ^ (\\ Prime )) + ((\\ left (\\ sqrt (x) \\ right)) ^ (\\ Prime)) + ((\\ left (\\ sqrt (x) \\ right)) ^ (\\ Prime)) \\\\ & ((\\ \\ Prime)) \u003d ((\\ left (((x) ^ (\\ FRAC (1) (3))) \\ Right)) ^ (\\ Prime)) \u003d \\ FRAC (1) (3) \\ CDOT ((X ) ^ (- \\ FRAC (2) (3))) \u003d \\ FRAC (1) (3) \\ CDOT \\ FRAC (1) (\\ sqrt (((x) ^ (2)))) \\\\ & (( \\ left (\\ sqrt (x) \\ right)) ^ (\\ Prime)) \u003d ((\\ left ((((x) ^ (\\ FRAC (1) (4))) \\ Right)) ^ (\\ Prime)) \u003d \\ FRAC (1) (4) ((x) ^ (- \\ FRAC (3) (4))) \u003d \\ FRAC (1) (4) \\ CDOT \\ FRAC (1) (\\ SQRT (((X) ^ (3)))) \\\\\\ End (Align) \\]

Returning to our example and write:

\\ [(y) "\u003d \\ FRAC (1) (2 \\ SQRT (X)) + \\ FRAC (1) (3 \\ SQRT (((x) ^ (2)))) + \\ FRAC (1) (4 \\ SQRT (((x) ^ (3)))) \\]

Here is a difficult decision.

Go to the second example, there are only two terms here, but each of them contains both a classic degree and roots.

Now we learn how to find a derivative of a power function, which, in addition, also contains the root:

\\ [\\ begin (align) & ((\\ left (((x) ^ (3)) \\ sqrt ((((x) ^ (2))) + ((x) ^ (7)) \\ SQRT (X) \\ Right)) ^ (\\ Prime)) \u003d ((\\ left (((x) ^ (3)) \\ Cdot \\ SQRT (((x) ^ (2))) \\ Right)) ^ (\\ Prime)) \u003d ((\\ left (((x) ^ (3)) \\ Cdot ((x) ^ (\\ FRAC (2) (3))) \\ Right)) ^ (\\ Prime)) \u003d \\\\ & \u003d (( \\ left (((x) ^ (3+ \\ FRAC (2) (3))) \\ RIGHT)) ^ (\\ Prime)) \u003d ((\\ left (((x) ^ (\\ FRAC (11) (3 ))) \\ Right)) ^ (\\ Prime)) \u003d \\ FRAC (11) (3) \\ Cdot ((x) ^ (\\ FRAC (8) (3))) \u003d \\ FRAC (11) (3) \\ ))) \\\\ \\ ((\\ left (((x) ^ (7)) \\ cdot \\ sqrt (x) \\ right)) ^ (\\ Prime)) \u003d ((\\ left (((x) ^ (7 )) \\ Cdot ((x) ^ (\\ FRAC (1) (3))) \\ Right)) ^ (\\ Prime)) \u003d ((\\ left (((x) ^ (7 \\ FRAC (1) (3 ))) \\ Right)) ^ (\\ Prime)) \u003d 7 \\ FRAC (1) (3) \\ Cdot ((x) ^ (6 \\ FRAC (1) (3)) \u003d \\ FRAC (22) (3 ) \\ Cdot ((x) ^ (6)) \\ CDOT \\ SQRT (X) \\\\\\ End (Align) \\]

Both allegations are considered, it remains to write down the final answer:

\\ [(y) "\u003d \\ FRAC (11) (3) \\ Cdot ((x) ^ (2)) \\ CDOT \\ SQRT (((x) ^ (2))) + \\ FRAC (22) (3) \\ Cdot ((x) ^ (6)) \\ CDOT \\ SQRT (X) \\]

We found the answer.

Fraction derivative through power function

But on this possibility of the formula for the solution of the derivative of the power function does not end. The fact is that with its help, not only examples with roots can be considered, but also with fractions. This is just a rare possibility that greatly simplifies the solution of such examples, but at the same time it is often ignored not only by students, but also by teachers.

So, now we will try to combine two formulas at once. On the one hand, the classical derivative of the power function

\\ [((\\ left (((x) ^ (n)) \\ right)) ^ (\\ Prime)) \u003d n \\ Cdot ((x) ^ (n-1)) \\]

On the other hand, we know that the expression of the type $ \\ FRAC (1) (((x) ^ (n))) $ is represented as $ ((x) ^ (- n)) $. Hence,

\\ [\\ left (\\ FRAC (1) (((x) ^ (n))) \\ Right) "\u003d ((\\ left (((x) ^ (- n)) \\ Right)) ^ (\\ Prime) ) \u003d - n \\ cdot ((x) ^ (- n - 1)) \u003d - \\ FRAC (n) (((x) ^ (n + 1))) \\]

\\ [((\\ left (\\ FRAC (1) (X) \\ RIGHT)) ^ (\\ Prime)) \u003d \\ left (((x) ^ (- 1)) \\ Right) \u003d - 1 \\ CDot ((x ) ^ (- 2)) \u003d - \\ FRAC (1) (((x) ^ (2))) \\]

Thus, derivatives of simple fractions, where there is a constant in the numener, and in the denominator - the degree is also considered using the classical formula. Let's see how it works in practice.

So the first function:

\\ [((\\ left (\\ FRAC (1) (((x) ^ (2))) \\ Right)) ^ (\\ Prime)) \u003d ((\\ left (((x) ^ (- 2)) \\ The first example is resolved, go to the second:

\\ [\\ Begin (Align) & ((\\ Left (\\ FRAC (7) (4 ((x) ^ (4))) - \\ FRAC (2) (3 ((x) ^ (3))) + \\ \\ & \u003d ((\\ left (\\ FRAC (7) (4 ((x) ^ (4))) \\ Right)) ^ (\\ Prime)) - ((\\ left (\\ FRAC (2) (3 (( x) ^ (3))) \\ right)) ^ (\\ prime)) + ((\\ left (2 ((x) ^ (3)) \\ Right)) ^ (\\ Prime)) - ((\\ left ( 3 ((x) ^ (4)) \\ right)) ^ (\\ Prime)) \\\\ \\ ((\\ left (\\ FRAC (7) (4 ((x) ^ (4))) \\ Right)) ^ (\\ Prime)) \u003d \\ FRAC (7) (4) ((\\ Left (\\ FRAC (1) (((x) ^ (4))) \\ Right)) ^ (\\ Prime)) \u003d \\ FRAC (7 ) (4) \\ CDOT ((\\ left (((x) ^ (- 4)) \\ Right)) ^ (\\ Prime)) \u003d \\ FRAC (7) (4) \\ CDOT \\ LEFT (-4 \\ RIGHT) \\ Cdot ((x) ^ (- 5)) \u003d \\ FRAC (-7) (((x) ^ (5))) \\\\ & ((\\ left (\\ FRAC (2) (3 ((x) ^ (3))) \\ Right)) ^ (\\ Prime)) \u003d \\ FRAC (2) (3) \\ CDOT ((\\ left (\\ FRAC (1) (((x) ^ (3))) \\ Right) ) ^ (\\ Prime)) \u003d \\ FRAC (2) (3) \\ CDOT ((\\ left (((x) ^ (- 3)) \\ right)) ^ (\\ Prime)) \u003d \\ FRAC (2) ( 3) \\ cdot \\ left (-3 \\ right) \\ Cdot ((x) ^ (- 4)) \u003d \\ FRAC (-2) (((x) ^ (4))) \\\\ & ((\\ left ( \\ FRAC (5) (2) ((x) ^ (2)) \\ Right)) ^ (\\ Prime)) \u003d \\ FRAC (5) (2) \\ Cdot 2x \u003d 5x \\\\ ≤ (\\ left (2 ((x) ^ (3)) \\ RIGHT)) ^ (\\ Prime)) \u003d 2 \\ Cdot 3 ((x) ^ (2)) \u003d 6 ((x) ^ (2)) \\\\ & ((\\ (3)) \u003d 12 ((x) ^ (3)) \\\\\\ End (Align) \\] ...

Now we collect all these components in a single formula:

\\ [(y) "\u003d - \\ FRAC (7) (((x) ^ (5))) + \\ FRAC (2) (((x) ^ (4))) + 5x + 6 ((x) ^ (2)) - 12 ((x) ^ (3)) \\]

We got the answer.

However, before moving on, I would like to draw your attention to the form of records of the original expressions: In the first expression, we recorded $ F \\ Left (X \\ Right) \u003d ... $, in the second: $ y \u003d ... $ Many students Lose when they see different recording forms. What is the difference between $ f \\ left (x \\ right) $ and $ y $? In fact, nothing. These are just different records with the same meaning. Just when we talk $ f \\ left (x \\ right) $ then we are talkingFirst of all, about the function, and when it comes to $ y $, it is most often meant a function schedule. Otherwise, this is the same, that is, the derivative in both cases is considered the same.

Difficult tasks with derivatives

In conclusion, I would like to consider a pair of complex combined tasks, which are used at once all that we have considered today. They are waiting for both roots, and fractions, and amounts. However, these examples will be complex only within the framework of today's video tutorial, because truly complex functions of derivatives will be waiting for you ahead.

So, the final part of today's video tutorial consisting of two combined tasks. Let's start with the first one:

\\ [\\ begin (align) & ((\\ left (((x) ^ (3)) - \\ FRAC (1) (((x) ^ (3))) + \\ sqrt (x) \\ right)) ^ (\\ Prime)) \u003d ((\\ left (((x) ^ (3)) \\ Right)) ^ (\\ Prime)) - ((\\ left (\\ FRAC (1) ((((x) ^ (3) )) \\ Right)) ^ (\\ Prime)) + \\ left (\\ sqrt (x) \\ right) \\\\ \\ ((\\ left (((x) ^ (3)) \\ Right)) ^ (\\ Prime) ) \u003d 3 ((x) ^ (2)) \\\\ Δ (\\ left (\\ FRAC (((((x) ^ (3))) \\ Right)) ^ (\\ Prime)) \u003d ((\\ (4))) \\\\ \\ ((\\ left (\\ sqrt (x) \\ right)) ^ (\\ Prime)) \u003d ((\\ left (((x) ^ (\\ FRAC (1) (3))) \\ Right)) ^ (\\ Prime)) \u003d \\ FRAC (1) (3) \\ CDOT \\ FRAC (1) (((x) ^ (\\ FRAC (2) (3)))) \u003d \\ FRAC (1) (3 \\ SQRT (((x) ^ (2)))) \\\\\\ End (Align) \\]

The derivative function is:

\\ [(y) "\u003d 3 ((x) ^ (2)) - \\ FRAC (3) (((x) ^ (4))) + \\ FRAC (1) (3 \\ sqrt (((x) ^ (2)))) \\]

The first example is resolved. Consider the second task:

In the second example, act in the same way:

\\ [((\\ left (- \\ FRAC (2) (((x) ^ (4))) + \\ sqrt (x) + \\ FRAC (4) (X \\ SQRT (((x) ^ (3)) )) \\ Right)) ^ (\\ Prime)) \u003d ((\\ left (- \\ FRAC (2) (((x) ^ (4))) \\ Right)) ^ (\\ Prime)) + ((\\ left (\\ SQRT (X) \\ Right)) ^ (\\ Prime)) + ((\\ left (\\ FRAC (4) (X \\ CDOT \\ SQRT (((x) ^ (3)))) \\ Right)) ^ (\\ Prime)) \\]

Calculate every term separately:

\\ [\\ begin (align) & ((\\ left (- \\ FRAC (2) (((x) ^ (4))) \\ Right)) ^ (\\ Prime)) \u003d - 2 \\ CDOT ((\\ left ( ((x) ^ (- 4)) \\ RIGHT)) ^ (\\ Prime)) \u003d - 2 \\ Cdot \\ left (-4 \\ right) \\ Cdot ((x) ^ (- 5)) \u003d \\ FRAC (8 ) (((x) ^ (5))) \\\\ & ((\\ left (\\ sqrt (x) \\ right)) ^ (\\ Prime)) \u003d ((\\ left (((x) ^ (\\ FRAC ( 1) (4))) \\ RIGHT)) ^ (\\ Prime)) \u003d \\ FRAC (1) (4) \\ Cdot ((x) ^ (- \\ FRAC (3) (4))) \u003d \\ FRAC (1 ) (4 \\ Cdot ((x) ^ (\\ FRAC (3) (4)))) \u003d \\ FRAC (1) (4 \\ sqrt (((x) ^ (3)))) \\\\ & ((\\ ((x) ^ (\\ FRAC (3) (4)))) \\ Right)) ^ (\\ Prime)) \u003d ((\\ left (\\ FRAC (4) (((x) ^ (1 \\ FRAC (3 ) (4)))) \\ Right)) ^ (\\ Prime)) \u003d 4 \\ Cdot ((\\ left (((x) ^ (- 1 \\ FRAC (3) (4))) \\ Right)) ^ ( \\ Prime)) \u003d \\\\ & \u003d 4 \\ Cdot \\ left (-1 \\ FRAC (3) (4) \\ RIGHT) \\ Cdot ((x) ^ (- 2 \\ FRAC (3) (4)) \u003d 4 \\ Cdot \\ Left (- \\ FRAC (7) (4) \\ RIGHT) \\ CDOT \\ FRAC (1) (((x) ^ (2 \\ FRAC (3) (4)))) \u003d \\ FRAC (-7) (((x) ^ (2)) \\ cdot ((x) ^ (\\ FRAC (3) (4)))) \u003d - \\ FRAC (7) (((x) ^ (2)) \\ CDOT \\ SQRT (((x) ^ (3)))) \\\\\\ End (Align) \\]

All the terms are counted. Now we return to the initial formula and fold together all three terms. We get that the final answer will be like this:

\\ [(y) "\u003d \\ FRAC (8) (((x) ^ (5))) + \\ FRAC (1) (4 \\ sqrt ((((x) ^ (3)))) - \\ FRAC (7 ) (((x) ^ (2)) \\ CDOT \\ SQRT ((((x) ^ (3)))) \\]

And that is all. It was our first lesson. In the following lessons, we will look at more complex designs, as well as find out why the derivatives are generally needed.

On which we disassemble the simplest derivatives, and also got acquainted with the rules of differentiation and some technical techniques finding derivatives. Thus, if you are not very clear with derivatives of functions, you will not be completely clear, then first read the above lesson. Please set up to a serious way - the material is not simple, but I still try to set it out simply and accessible.

In practice, a derivative of a complex function has to face very often, I would even say, almost always when you tasks to find derivatives.

We look at the table for a rule (No. 5) of differentiation of a complex function:

We understand. First of all, pay attention to the record. Here we have two functions - and, moreover, the function, figuratively speaking, is invested in the function. The function of this type (when one function is embedded in another) and is called a complex function.

I will call the function external function, and function - internal (or nested) function.

! These definitions are not theoretical and should not appear in the piston design of tasks. I use informal expressions "External Function", "Internal" function only to make it easier for you to understand the material.

In order to clarify the situation, consider:

Example 1.

Find a derivative function

Under the sinus, we are not just the letter "X", but an integer expression, so it will not be possible to find a derivative immediately on the table. We also notice that here it is impossible to apply the first four rules, it seems there is a difference, but the fact is that the sinus is not "separated into parts":

In this example, from my explanations, it is intuitive that the function is a complex function, and the polynomial is an internal function (attachment), and is an external function.

First stepto perform when finding a derivative complex function is to figure out what function is internal and what is the external.

In the case of simple examples, it seems it seems that a polynomial is invested under sine. But what if everything is not obvious? How to determine exactly what function is external, and what is the inner? To do this, I propose to use the next reception, which can be carried out mentally or on the draft.

Imagine that we need to calculate the value of an expression value on the calculator (instead of a unit there may be any number).

What do we calculate first? First of all You will need to perform the following:, Therefore, the polynomial and will be internal function:

Secondly It will be necessary to find, so sinus - it will be an external function:

After we Have figured out With internal and external functions, it's time to apply the differentiation rule of a complex function .

We start to solve. From the lesson How to find a derivative? We remember that the decoration of the solution of any derivative always begins so - we conclude an expression in the brackets and put on the right at the top of the barcode:

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

Note that the internal function did not change, we do not touch her.

Well, it is quite obvious that

The result of the application of formula In the piston design it looks like this:

A permanent multiplier usually endure expressions:

If any misunderstanding remains, rewrite the decision on paper and read the explanations again.

Example 2.

Find a derivative function

Example 3.

Find a derivative function

As always, write:

We understand where we have an external function, and where is the inner. To do this, try (mentally or on a draft) to calculate the value of the expression at. What needs to be performed first? First of all, it is necessary to count what is equal to the base:, it means that the polynomial is internal function:

And, only then the exercise is carried out into the extent, therefore, the power function is an external function:

According to the formula First you need to find the derivative from the external function, in this case, on the extent. We wanted the necessary formula in the table :. We repeat again: any tabular formula is valid not only for "X", but also for complex expression. Thus, the result of applying the range of differentiation of a complex function following:

I emphasize again that when we take a derivative of an external function, the internal function does not change with us:

Now it remains to find a completely simple derivative from the internal function and a little "combing" the result:

Example 4.

Find a derivative function

This is an example for self-decide (response at the end of the lesson).

To secure an understanding of the derivative complex function, I will give an example without comment, try to figure it out yourself, paint, where external and where is the internal function, why are the tasks solved this way?

Example 5.

a) find a derivative function

b) find a derivative function

Example 6.

Find a derivative function

Here we have a root, and in order to indifferentiate the root, it must be represented in the form of a degree. Thus, first give the function to the proper form:

Analyzing the function, we conclude that the sum of the three terms is an internal function, and the external function is the external function. Apply the differentiation rule of a complex function :

The degree again represent in the form of a radical (root), and for the derivative of the internal function, use a simple rule of differentiation amount:

Ready. You can also lead the expression to common denominator And write down everything with one fraction. Beautiful, of course, but when bulky long derivatives are obtained - it is better not to do this (it's easy to get confused, to allow an unnecessary error, and the teacher will inconveniently check).

Example 7.

Find a derivative function

This is an example for an independent decision (answer at the end of the lesson).

It is interesting to note that sometimes instead of the procedure for differentiation of a complex function, you can use the proportion differentiation rule But this decision will look like a perversion unusual. Here is a characteristic example:

Example 8.

Find a derivative function

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

We prepare the function for differentiation - we take a minus per sign of the derivative, and the cosine raise into the numerator:

Cosine is an internal function, the external function is an external function.
We use our rule :

We find the derivative of the internal function, the cosine is discarding back down:

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

Example 9.

Find a derivative function

This is an example for an independent decision (answer at the end of the lesson).

So far, we have considered cases when only one investment was in our complex function. In the practical tasks, it is often possible to meet derivatives, where, as matryoshki, one to another, are embedded at once 3, or even 4-5 functions.

Example 10.

Find a derivative function

We understand in the investments of this function. We try to calculate the expression using the experimental value. How would we believe on the calculator?

First you need to find, it means, Arksinus is the deepest investment:

Then this arxinus units should be built into the square:

And finally, the seven is erected into a degree:

That is, in this example, we have three different functions and two attachments, while the inner function is arxinus, and the external function itself is an indicative function.

We begin to decide

According to the rule First you need to take a derivative from the external function. We look at the table of derivatives and find a derivative of the indicative function: the only difference is instead of "X" we have a difficult expression that does not cancel the validity of this formula. So, the result of applying the differentiation runt of a complex function following.