The derivative of the function y 2 root x. The derivative of the power function (degree and roots)
In this lesson, we will learn to apply formulas and differentiation rules.
Examples. Find derived functions.
1. Y \u003d x 7 + x 5 -X 4 + x 3 -X 2 + X-9. Apply Rule I., Formulas 4, 2 and 1. We get:
y '\u003d 7x 6 + 5x 4 -4x 3 + 3x 2 -2x + 1.
2. y \u003d 3x 6 -2x + 5. We decide similarly using the same formulas and formula 3.
y '\u003d 3 ∙ 6x 5 -2 \u003d 18x 5 -2.
Apply Rule I., Formulas 3, 5
and 6
and 1.
Apply Rule IV, Formulas 5 and 1 .
In the fifth example of the rule I. The derivative of the amount is equal to the sum of the derivatives, and the derivative of the 1st terms we just found (an example 4 ), therefore, we will find derivatives 2ndand 3rd Slictic, A. for the 1st The foundation can immediately write the result.
Differential 2nd and 3-E. Formulas 4
. To do this, we transform the roots of the third and fourth degrees in denominators to degrees with negative indicators, and then 4
Formula, we find derivatives of degrees.
Look at this example and the result. Caught regularity? Okay. This means that we have received a new formula and can add it to our derivatives table.
I solve the sixth example and withdraw another formula.
We use the rule IV and formula 4
. The resulting fractions will cut.
We look at this feature and on its derivative. Of course, you understood the pattern and are ready to call the formula:
We learn new formulas!
Examples.
1. Find the increment of the argument and the increment of the function y \u003d x 2if the initial value of the argument was equal 4 , and new - 4,01 .
Decision.
New value of argument x \u003d x 0 + Δx. Substitut the data: 4.01 \u003d 4 + Δh, hence the increment of the argument ΔХ.\u003d 4.01-4 \u003d 0.01. The increment of the function, by definition, is equal to the difference between the new and former values \u200b\u200bof the function, i.e. ΔY \u003d F (x 0 + Δh) - f (x 0). Since we have a function y \u003d x 2T. Δu\u003d (x 0 + Δx) 2 - (x 0) 2 \u003d (x 0) 2 + 2x 0 · Δx + (Δx) 2 - (x 0) 2 \u003d 2x 0 · Δx + (Δx) 2 \u003d
2 · 4 · 0,01+(0,01) 2 =0,08+0,0001=0,0801.
Answer: Argument increment ΔХ.\u003d 0.01; Protecting function Δu=0,0801.
It was possible to increment the function differently: ΔY.\u003d y (x 0 + Δx) -y (x 0) \u003d y (4.01) -U (4) \u003d 4.01 2 -4 2 \u003d 16.0801-16 \u003d 0.0801.
2. Find an angle of inclination tangent to graphics function y \u003d f (x) At point x 0, if a f "(x 0) \u003d 1.
Decision.
The value of the derivative at the touch point x 0 And there is a tangent tilt angle of tangent (geometric meaning derivative). We have: f "(x 0) \u003d Tgα \u003d 1 → α \u003d 45 °,as TG45 ° \u003d 1.
Answer: Tangent of graphics of this function forms with a positive axis direction Oh an angle equal 45 °.
3. Derive a derivative formula y \u003d x n.
Differentiation - This is the action of finding a derived function.
When the derivatives are used, the formulas that were derived on the basis of the determination of the derivative, as well as we derived a derivative formula: (x n) "\u003d NX N-1.
Here are these formulas.
Table derivatives It will be easier to memorize, saying verbal wording:
1. The constant value derivative is zero.
2. Xo barcode is equal to one.
3. A permanent multiplier can be reached for a sign of the derivative.
4. The derivative of the degree is equal to the product of this degree to the degree with the same base, but the indicator per unit is smaller.
5. The root derivative is equal to one divided into two of the same root.
6. The derivative of the unit divided by the X is equal to minus unit divided into X square.
7. Sine derivative is equal to cosine.
8. The cosine derivative is minus sinus.
9. The tangent derivative is equal to a unit divided into a cosine square.
10. The Kotannce derivative is minus a unit divided into square sinus.
Learning differentiation rules.
1.
The derivative of the algebraic amount is equal to the algebraic amount of the derivatives of the terms.
2. The derivative of the work is equal to the product of the derivative of the first factor on the second plus the product of the first factor on the derivative of the second.
3. The derivative of "y", divided into "VE" is equal to the fraction, in the numerator of which "in the barcode multiplied to the" WE "minus" y, multiplied by the Bather ", and in the denominator -" WE in a square ".
4. Private case Formulas 3.
Learn together!
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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 the increment of the argument, a table of derivatives appeared and accurately certain rules Differentiation. 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 the 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. Derivative square root | |
| 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. it typical errorwhich meets on initial stage Studying derivatives, but as several single-stage examples have already solved, the average student of this error is no longer done.
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).
Other frequent error - mechanical solution of the 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 simple features.
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". Receive the following values 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 sinus derivatives, 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 is a square root of an independent variable, with the derivative of which we have read 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.
Hello, dear readers. After reading the article, you will probably have a natural question: "Why, in fact, it is necessary?". By virtue of this, I first consider it necessary to inform you in advance that the desired method of solving square equations is represented rather from the moral and aesthetic side of mathematics than from the side of practical dry use. Also, I apologize in advance before the readers who consider my amateuric sayings unacceptable. So, let's start scoring the nails with a microscope.
We have an algebraic equation of the second degree (it is square) in general form:
Let's go wrong square equation to quadratic function:
Where, obviously, it is necessary to find such values \u200b\u200bof the function argument in which these would return zero.
It seems that it is necessary to simply solve the square equation using the Vieta theorem or through the discriminant. But we have not gathered here for this. Let's better take a derivative! 
Based on the definition of the physical meaning of the derivative of the first order, it is clear that substituting the argument in the resulting function we (in particular) we obtain speed Changes in the function in the point given by this argument.
This time we received a "speed speed" change of the function (then you mean acceleration) At a specific point. After analyzing the received, we can conclude that the "acceleration" is a constant that does not depend on the function argument - we will remember this.
Now recall some physics and equivalent movement (ore). What do we have in arsenal? It is true, there is a formula for determining the coordinate of moving along the axis with the desired movement:
Where - time, - initial speed, - acceleration.
It is easy to see that our original function is just an ore.
Did the displacement formula for ores is not a consequence of the solution of the square equation?
Not. The formula for ores is higher in fact is the result of taking an integral from the speed formula with a pond. Or from the chart you can find the figure of the figure. There will come out the trapeze.
The displacement formula for ore does not follow from the solution of any square equations. It is very important, otherwise there would be no sense of the article.
Now it remains to figure out what is what, and what we lack.
"Acceleration" we already have - they are the second order derivative, derived above. But to get the initial speed, we need to take in general anyone (we denote it as) and substitute it in a derivative now already first order - for it will be the desired.
In this case, the question arises, what needs to be taken? Obviously, such that the initial speed is equal to zero, so that the "movement of ore" formula will be viewed:
In this case, make a search equation:
[Substituted in the first order derivative]
The root of such an equation is regarding:
And the value of the initial function with this argument will be: 
Now it becomes obvious that:
Connect all "Puzzle Details" together:
So we got the final solution to the task. In general, we did not open America - we simply came to the formula for solving the square equation through the discriminant of the district. It does not make practical sense (this does not carry this (approximately the same way you can solve the equations of the first / second degree of any (not necessarily general).
The purpose of this article is, in particular, heated interest in analyzing the mat. functions and generally to mathematics.
I was Peter, thanks for your attention!
Definition. Let the function \\ (y \u003d f (x) \\) define in a certain interval containing within itself the point \\ (x_0 \\). We give the argument the increment \\ (\\ Delta X \\) is so as not to get out of this interval. Find the appropriate increment of the function \\ (\\ Delta Y \\) (when moving from point \\ (x_0 \\) to the point \\ (x_0 + \\ deelta x \\)) and amounted to the ratio \\ (\\ FRAC (\\ Delta Y) (\\ Delta X) \\). If there is a limit of this relationship with \\ (\\ delta x \\ rightarrow 0 \\), then the specified limit is called derived function \\ (y \u003d f (x) \\) at point \\ (x_0 \\) and denote \\ (F "(x_0) \\).
$$ \\ LIM _ (\\ Delta X \\ To 0) \\ FRAC (\\ Delta Y) (\\ Delta X) \u003d F "(x_0) $$
To designate the derivative, the Y symbol often use. Note that Y "\u003d F (x) is a new function, but naturally associated with the function y \u003d f (x) defined in all points x in which the above limit exists . This feature is called this: derivative function y \u003d f (x).
Geometric meaning of the derivative It consists next. If the function of the function y \u003d F (x) at the abscissa point x \u003d a can be carried out by a tangent, non-parallel axis y, then f (a) expresses the angular coefficient of tangent:
\\ (k \u003d f "(a) \\)
Since \\ (k \u003d tg (a) \\), then the equality \\ (F "(a) \u003d Tg (A) \\) is true.
And now we interpret the definition of the derivative from the point of view of approximate equalities. Let the function \\ (y \u003d f (x) \\) has a derivative at a specific point \\ (x \\):
$$ \\ LIM _ (\\ Delta X \\ To 0) \\ FRAC (\\ Delta Y) (\\ Delta X) \u003d F "(x) $$
This means that approximate equality \\ (\\ FRAC (\\ Delta Y) (\\ Delta X) \\ APPROX F "(X) \\), i.e. \\ (\\ Delta Y \\ Approx F" (x) \\ Cdot \\ Delta X \\). 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 proportionality coefficient is the value of the derivative at a given point x. For example, for the function \\ (y \u003d x ^ 2 \\), the approximate equality \\ (\\ Delta Y \\ Approx 2x \\ Cdot \\ Delta X \\) is true. If you carefully analyze the definition of the derivative, then we will find that it is put on it algorithm.
Word it.
How to find the derivative function y \u003d f (x)?
1. Fix the value \\ (x \\), to find \\ (F (x) \\)
2. Give the argument \\ (x \\) increment \\ (\\ Delta X \\), go to a new point \\ (X + \\ Delta X \\), to find \\ (F (x + \\ Delta x) \\)
3. Find the increment of the function: \\ (\\ Delta Y \u003d F (X + \\ Delta X) - F (x) \\)
4. Make a relation \\ (\\ FRAC (\\ Delta Y) (\\ Delta X) \\)
5. Calculate $$ \\ LIM _ (\\ Delta X \\ To 0) \\ FRAC (\\ Delta Y) (\\ Delta X) $$
This limit is derived from the point x.
If the function y \u003d f (x) has a derivative at point x, it is called differentiable at point x. The procedure for finding the derivative function y \u003d f (x) is called differentiation Functions y \u003d f (x).
Let us discuss such a question: how are the continuity of the continuity and differentiability of the function at the point.
Let the function y \u003d f (x) differentiate at the point x. Then to the graph of the function at the point M (x; f (x)), it is possible to carry out a tangent, and, we recall, the angular coefficient of tangent is f "(x). Such a chart cannot" break "at the point M, i.e. the function is obliged be continuous at point x.
These were reasoning "on the fingers." We give a more stringent reasoning. If the function y \u003d f (x) is differentiable at the point x, then approximate equality is performed \\ (\\ Delta Y \\ Approx F "(x) \\ Cdot \\ Delta X \\). If in this equality \\ (\\ Delta X \\) rushed to zero, then \\ (\\ delta y \\) will strive for zero, and this is the condition of the continuity of the function at the point.
So, if the function is differentiable at the point x, it is continuous at this point.
The opposite statement is incorrect. For example: function y \u003d | x | Continuous everywhere, in particular at point x \u003d 0, but tangent to the graphics of the function in the "point of the joint" (0; 0) does not exist. If at some point to the graphics of the function can not be tanged, then at this point there is no derivative.
One more example. The function \\ (y \u003d \\ sqrt (x) \\) is continuous on the entire numeric line, including at the point x \u003d 0. And the function to the graphic function exists at any point, including at the point x \u003d 0. But at this point The tangent coincides with the axis of y, i.e. perpendicular to the abscissa axis, its equation has the form x \u003d 0. There is no corner of the coefficient, it means that there is no and \\ (F "(0) \\)
So, we got acquainted with the new feature of the function - differentiability. And how can the function of the function be concluded about its differentiability?
The answer is actually obtained above. If at some point to the graph of the function you can spend a tangential, non-perpendicular abscissa axis, then at this point the function is differentiable. If at some point tangent to the graphics function does not exist or it is perpendicular to the abscissa axis, then at this point the function is not differentiated.
Differentiation rules
Operation finding a derivative called differentiation. When performing this operation, it often has to work with private, sums, works of functions, as well as with "functions functions", that is, complex functions. Based on the definition of the derivative, you can withdraw the differentiation rules that facilitate this work. If C is a constant number and f \u003d f (x), G \u003d G (x) - some differentiable functions, then the following are valid differentiation rules:
$$ F "_X (G (x)) \u003d f" _g \\ cdot g "_x $$
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