Consider the function y (x), which is recorded in an implicit way in general $ F (x, y (x)) \u003d 0 $. The derivative of the implicit function is in two ways:

1. Differentiation of both parts of the equation
2. Using the use of the finished formula $ Y "\u003d - \\ FRAC (f" _x) (f "_y) $

How to find?

Method 1.

It is not necessary to provide a function to a clear mind. It is necessary to immediately begin to differentiate the left and right of the equation of $ x $. It is worth noting that the $ y "$ derivative" $ is calculated according to the range of differentiation of a complex function. For example, $ (y ^ 2) "_ x \u003d 2yy" $. After finding a derivative, it is necessary to express $ Y "$ from the obtained equation and place $ y" $ in the left side.

Method 2

You can use the formula in which the private derivatives of the implicit function $ F (x, y (x)) \u003d 0 $ are used in the numerator and denominator. To find the numerator, we take a $ x $ derivative, and for the denominator a derivative of $ y $.

The second derivative of the implicit function can be found by repeated differentiation of the first derivative of the implicit function.

Examples of solutions

Consider practical examples of solutions to calculate the derivative implicitly specified function.

Example 1.

Find the derivative of the implicit function $ 3x ^ 2y ^ 2 -5x \u003d 3Y - $ 1

Decision

We use the method number 1. Namely, the left and right-hand part of the equation are preferred: $$ (3x ^ 2y ^ 2 -5x) "_ x \u003d (3y - 1)" _ x $$ Do not forget when differentiation, use the formula of the derivative product of functions: $$ (3x ^ 2) "_ x y ^ 2 + 3x ^ 2 (y ^ 2)" _ x - (5x) "_ x \u003d (3y)" _ x - (1) "_ x $$ $$ 6x y ^ 2 + 3x ^ 2 2yy "- 5 \u003d 3Y" $$ $$ 6x y ^ 2 - 5 \u003d 3Y "- 6x ^ 2 yy" $$ $$ 6x y ^ 2 - 5 \u003d y "(3-6x ^ 2 y) $$ $$ y "\u003d \\ FRAC (6x y ^ 2 - 5) (3 - 6x ^ 2y) $$ If it is impossible to solve your task, then send it to us. We will provide detailed solution. You can familiarize yourself with the course of calculation and learn information. This will help in a timely manner at the teacher!

Answer

$$ y "\u003d \\ FRAC (6x y ^ 2 - 5) (3 - 6x ^ 2y) $$

Example 2.

The function is defined implicitly, to find the derivative of $ 3x ^ 4 y ^ 5 + E ^ (7x-4y) -4x ^ 5 -2y ^ 4 \u003d 0 $

Decision

We use the method number 2. We find private derivatives of the functions $ f (x, y) \u003d 0 $ We put $ y $ constant and indifferentiate $ x $: $$ f "_x \u003d 12x ^ 3 y ^ 5 + E ^ (7x-4y) \\ CDOT 7 - 20X ^ 4 $$ $$ F "_x \u003d 12x ^ 3 y ^ 5 + 7e ^ (7x-4y) - 20x ^ 4 $$ We now consider $ x $ constant and differentiate on $ y $: $$ F "_Y \u003d 15x ^ 4 y ^ 4 + E ^ (7x-4y) \\ CDOT (-4) - 8Y ^ 3 $$ $$ f "_Y \u003d 15x ^ 4 y ^ 4 - 4e ^ (7x-4y) - 8Y ^ 3 $$ Now we substitute in the $ y formula "\u003d - \\ FRAC (f" _y) (f "_x) $ and get: $$ y "\u003d - \\ FRAC (12x ^ 3 y ^ 5 + 7e ^ (7x-4y) - 20x ^ 4) (15x ^ 4 y ^ 4 - 4e ^ (7x-4y) - 8Y ^ 3) $$

Answer

$$ y "\u003d - \\ FRAC (12x ^ 3 y ^ 5 + 7e ^ (7x-4y) - 20x ^ 4) (15x ^ 4 y ^ 4 - 4e ^ (7x-4y) - 8Y ^ 3) $$

Or shorter - derivative of an implicit function. What is an implicit function? Since my lessons are practical, I try to avoid definitions, wording theorems, but it will be appropriate here. And what is a function in general?

The function of one variable is a rule by which each value of an independent variable corresponds to one and only one function value.

The variable is called independent variable or argument.
The variable is called dependent variable or function.

Roughly speaking, Beakok "Igarek" in this case - and there is a function.

So far, we have considered the functions specified in applied form. What does it mean? We arrange the parsing of flights on specific examples.

Consider a function

We see that on the left we have a lonely "igrek" (function), and on the right - only "Ikers". That is, a function explicitly expressed through an independent variable.

Consider another feature:

Here are variables and arranged "intention". Moreover no way impossible Express "Ix" only through "X". What kind of ways is it? The transfer of the components from the part to the part with the change of the sign, to the brackets, transferring multipliers to the rule of proportion and others. Rewrite the equality and try to express "Igarek" explicitly :. You can twist-thrust the hour equation, but you will not succeed.

Allow me to introduce: - Example implicit function.

In the course of mathematical analysis it is proved that an implicit function exists (However, not always), she has a schedule (just like the "normal" function). The implicit function in the same way exists The first derivative, the second derivative, etc. As they say, all the rights of sex minorities are observed.

And in this lesson, we will learn to find a derivative of the function specified implicitly. It's not so difficult! All differentiation rules, the table of derivative elementary functions remain in force. The difference in one kind of moment, which we consider right now.

Yes, and I will inform good news - the tasks discussed below are performed on a pretty hard and clear algorithm without a stone in front of three tracks.

Example 1.

1) At the first stage, hanging strokes on both parts:

2) We use the linearity rules derivative (the first two rules of the lesson How to find a derivative? Examples of solutions):

3) direct differentiation.
How to differentiate and completely understandable. What to do where under strokes is "igraki"?

Just before disgrace the derivative of the function is equal to its derivative: .

How to differentiate

Here we have complex function. Why? It seems to be under the sinus only one letter "Igarek". But, the fact is that only one letter "Igarek" - In itself is a function(See definition at the beginning of the lesson). Thus, the sinus is an external function - an internal function. Use the differentiation rule of a complex function :

Work differentiating by the usual rule :

Please note that - also a complex function, any "Cheerk with Founds" - a complex function:

The decision itself should look something like this:

If there are brackets, then reveal them:

4) In the left side we collect the terms, in which there is "igrek" with a touche. On the right side - tolerate everything else:

5) In the left part, we carry out the derivative of the brackets:

6) And according to the rule of proportion, we discard these brackets to the denominator of the right part:

The derivative was found. Ready.

It is interesting to note that in an implicit form you can rewrite any function. For example, a function You can rewrite: . And differentiate it according to the algorithm just discussed. In fact, the phrase "function specified in implicit form" and "implicit function" are distinguished by one semantic nuance. Phrase "The function specified in implicit form" is more general and correct, - This function is set in an implicit form, but here you can express "igrek" and present the function explicitly. Under the phrase, the "implicit function" understand the "classic" implicit function, when "igrek" cannot be expressed.

The second way of solving

Attention!With the second way, it is possible to familiarize yourself in the event that you can confidently find private derivatives. Beginners Study Mathematical Analysis and Teapots, please do not read and skip this item, otherwise there will be a full porridge in my head.

Find a derivative of an implicit function in the second way.

We carry all the components to the left:

And we consider the function of two variables:

Then our derivative can be found by the formula

We find private derivatives:

In this way:

The second solution method allows you to check. But it is undesirable to make it a finishing version of the task, since the private derivatives are mastered later, and the student learns the topic "derived function of one variable", to know the private derivatives as it should not yet.

Consider a few more examples.

Example 2.

Find a derivative from the function specified implicitly

Turn the touches on both parts:

We use linearity rules:

Find derivatives:

Reveal all brackets:

We transfer all the components with the left part, the rest - the right side:

In the left part, we endure the bracket:

Final answer:

Example 3.

Find the derivative of the function specified

Complete solution and sample design at the end of the lesson.

Not uncommon, when fractions arise after differentiation. In such cases, fractions need to get rid. Consider two more examples.

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:

$$ C "\u003d 0 $$$$ x" \u003d 1 $$$$$ (f + g) "\u003d f" + g "$$$$ (FG)" \u003d F "G + FG" $$$$ ( CF) "\u003d CF" $$$$ \\ left (\\ FRAC (F) (G) \\ Right) "\u003d \\ FRAC (F" G-FG ") (G ^ 2) $$$$ \\ left (\\ FRAC (C) (G) \\ Right) "\u003d - \\ FRAC (CG") (G ^ 2) $$ Derivative complex function:
$$ F "_X (G (x)) \u003d f" _g \\ cdot g "_x $$

Table of derivatives of some functions

$$ \\ left (\\ FRAC (1) (X) \\ RIGHT) "\u003d - \\ FRAC (1) (x ^ 2) $$$$ (\\ SQRT (X))" \u003d \\ FRAC (1) (2 \\ $$$$ \\ left (E ^ X \\ RIGHT) "\u003d E ^ x $$$$ (\\ ln x)" \u003d \\ FRAC (1) (X) $$$$ (\\ log_a x) "\u003d \\ FRAC (1) (x \\ ln a) $$$$ (\\ sin x) "\u003d \\ cos x $$$$ (\\ cos x)" \u003d - \\ sin x $$$$ (\\ Text (TG) X) "\u003d \\ FRAC (1) (\\ cos ^ 2 x) $$$$ (\\ Text (CTG) X)" \u003d - \\ FRAC (1) (\\ sin ^ 2 x) $$$$ (\\ arcsin x) "\u003d \\ FRAC (1) (\\ sqrt (1-x ^ 2)) $$$$ (\\ arccos x)" \u003d \\ FRAC (-1) (\\ SQRT (1-x ^ 2)) $$$$ (\\ Text (arctg) x) "\u003d \\ FRAC (1) (1 + x ^ 2) $$$$ (\\ Text (ArcCTG) X)" \u003d \\ FRAC (-1) (1 + x ^ 2) $ $

Suppose the function is defined implicitly with the equation
(1) .
And let this equation, with some meaning, has a single solution. Let the function be differentiated by the function at the point, and
.
Then, with this meaning, there is a derivative, which is determined by the formula:
(2) .

Evidence

To prove, consider the function as a complex function from the variable:
.
Apply differentiation rule of complex function and find a derivative by variable from the left and right parts of the equation
(3) :
.
Since the constant derivative is zero and, then
(4) ;
.

The formula is proved.

Derivatives of higher orders

We rewrite equation (4) using other notation:
(4) .
At the same time, they are complex functions from the variable:
;
.
Dependence Determines equation (1):
(1) .

We find the derivative of the variable from the left and right of the equation (4).
By the formula of the derivative complex function We have:
;
.
By the formula of the derivative of the work :

.
By the formula of the derivative of the amount :


.

Since the derivative of the right part of the equation (4) is zero,
(5) .
Substituting a derivative here, we obtain the value of the second order derivative in an implicit form.

Differentiating, in a similar way, equation (5), we obtain an equation containing a third-order derivative:
.
Substituting the found values \u200b\u200bof the derivatives of the first and second orders here, we find the value of the third order derivative.

Continuing differentiation, you can find a derivative of any order.

Examples

Example 1.

Find the first order derivative from the function specified implicitly by the equation:
(P1) .

Decision by formula 2

Find a derivative according to formula (2):
(2) .

We transfer all the variables into the left part so that the equation takes the form.
.
From here.

Find a derivative of software, counting constant.
;
;
;
.

Find a variable derivative, considering the variable constant.
;
;
;
.

By formula (2) we find:
.

We can simplify the result if we note that according to the initial equation (clause 1) ,. Substitute:
.
Multiply the numerator and denominator on:
.

Solution in the second way

I solve this example in the second way. To do this, find a derivative along the variable left and right parts of the original equation (P1).

We use:
.
Apply formula derivative fraci :
;
.
Apply formula derivative complex function :
.
Differentiating the initial equation (P1).
(P1) ;
;
.
We multiply on and group members.
;
.

Substitute (from the equation (P1)):
.
Multiply on:
.

Answer

Example 2.

Find a second order derivative from the function specified implicitly using the equation:
(P2.1) .

Decision

Differentiating the original equation, by variable, counting that is a function from:
;
.
Apply the formula of the derivative complex function.
.

Differentiating the initial equation (P2.1):
;
.
From the initial equation (P2.1) it follows that. Substitute:
.
Reveal brackets and group members:
;
(P2.2) .
Find a first-order derivative:
(P2.3) .

To find a second-order derivative, differentiate equation (p2.2).
;
;
;
.
Substitute the expression of the first-order derivative (P2.3):
.
Multiply on:

;
.
From here we find a second-order derivative.

Answer

Example 3.

Find the third order derivative with the function specified implicitly using the equation:
(P3.1) .

Decision

Differentiate the initial equation on the variable considering what is a function from.
;
;
;
;
;
;
(P3.2) ;

Differentiating the equation (p3.2) by variable.
;
;
;
;
;
(P3.3) .

Differentiating equation (p3.3).
;
;
;
;
;
(P3.4) .

From equations (p3.2), (p3.3) and (p3.4) we find the values \u200b\u200bof derivatives at.
;
;
.