The derivative of a function given implicitly.
Derivative of a parametrically defined function

In this article, we will consider two more typical tasks that are often found in tests in higher mathematics. In order to successfully master the material, it is necessary to be able to find derivatives at least at an average level. You can learn how to find derivatives practically from scratch in two basic lessons and Derivative of a compound function. If everything is in order with differentiation skills, then let's go.

Derivative of a function defined implicitly

Or, in short, the derivative implicit function. What is an implicit function? Let's first recall the very definition of a function of one variable:

Function of one variable is the rule that each value of the independent variable corresponds to one and only one value of the function.

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

So far, we have considered functions defined in explicit form. What does it mean? Let's arrange a debriefing on specific examples.

Consider the function

We see that on the left we have a lone “y”, and on the right - only x's. That is, the function explicitly expressed in terms of the independent variable .

Let's consider another function:

Here the variables and are located "mixed". And impossible in any way express "Y" only through "X". What are these methods? Transferring terms from part to part with a change of sign, bracketing, throwing factors according to the rule of proportion, etc. Rewrite the equality and try to express “y” explicitly:. You can twist and turn the equation for hours, but you will not succeed.

Allow me to introduce: - an example implicit function.

In the course of mathematical analysis, it was proved that the implicit function exists(but not always), it has a graph (just like a "normal" function). It's the same for an implicit function. exists first derivative, second derivative, etc. As they say, all the rights of sexual minorities are respected.

And in this lesson we will learn how to find the derivative of a function given implicitly. It's not that hard! All differentiation rules, the table of derivatives of elementary functions remain in force. The difference is in one peculiar point, which we will consider right now.

Yes, and I will tell you the good news - the tasks discussed below are performed according to a rather rigid and clear algorithm without a stone in front of three tracks.

Example 1

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

2) We use the rules of linearity of the derivative (the first two rules of the lesson How to find the derivative? Solution examples):

3) Direct differentiation.
How to differentiate and completely understandable. What to do where there are “games” under the strokes?

- just to disgrace, the derivative of a function is equal to its derivative: .

How to differentiate
Here we have complex function. Why? It seems that under the sine there is only one letter "Y". But, the fact is that only one letter "y" - IS A FUNCTION IN ITSELF(see the definition at the beginning of the lesson). Thus, the sine is an external function, is an internal function. We use the differentiation rule complex function :

The product is differentiable according to the usual rule :

Note that is also a complex function, any “twist toy” is a complex function:

The design of the solution itself should look something like this:


If there are brackets, then open them:

4) On the left side, we collect the terms in which there is a “y” with a stroke. On the right side - we transfer everything else:

5) On the left side, we take the derivative out of brackets:

6) And according to the rule of proportion, we drop these brackets into the denominator of the right side:

The derivative has been found. Ready.

It is interesting to note that any function can be rewritten implicitly. For example, the function can be rewritten like this: . And differentiate it according to the algorithm just considered. In fact, the phrases "implicit function" and "implicit function" differ in one semantic nuance. The phrase "implicitly defined function" is more general and correct, - this function is given implicitly, but here you can express "y" and present the function explicitly. The words “implicit function” are more often understood as a “classical” implicit function, when “y” cannot be expressed.

It should also be noted that the “implicit equation” can implicitly define two or even more functions at once, for example, the equation of a circle implicitly defines the functions , , which define semicircles. But, within the framework of this article, we will not make a special distinction between the terms and nuances, it was just information for general development.

The second way to solve

Attention! You can familiarize yourself with the second method only if you know how to confidently find partial derivatives. Calculus Beginners and Dummies Please do not read and skip this paragraph, otherwise the head will be a complete mess.

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

We move all the terms to the left side:

And consider a function of two variables:

Then our derivative can be found by the formula
Let's find partial derivatives:

Thus:

The second solution allows you to perform a check. But it is undesirable to draw up a final version of the task for him, since partial derivatives are mastered later, and a student studying the topic “Derivative of a function of one variable” should not know partial derivatives.

Let's look at a few more examples.

Example 2

Find the derivative of a function given implicitly

We hang strokes on both parts:

We use the rules of linearity:

Finding derivatives:

Expanding all parentheses:

We transfer all the terms with to the left side, the rest - to the right side:

Final answer:

Example 3

Find the derivative of a function given implicitly

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

It is not uncommon for fractions to appear after differentiation. In such cases, fractions must be discarded. Let's look at two more examples.

Example 4

Find the derivative of a function given implicitly

We conclude both parts under strokes and use the linearity rule:

We differentiate using the rule of differentiation of a complex function and the rule of differentiation of the quotient :

Expanding the brackets:

Now we need to get rid of the fraction. This can be done later, but it is more rational to do it right away. The denominator of the fraction is . Multiply on . In detail, it will look like this:

Sometimes after differentiation, 2-3 fractions appear. If we had one more fraction, for example, then the operation would have to be repeated - multiply each term of each part on

On the left side, we put it out of brackets:

Final answer:

Example 5

Find the derivative of a function given implicitly

This is a do-it-yourself example. The only thing in it, before getting rid of the fraction, you will first need to get rid of the three-story structure of the fraction itself. Full solution and answer at the end of the lesson.

Derivative of a parametrically defined function

Do not strain, in this paragraph, too, everything is quite simple. You can write down the general formula of a parametrically given function, but, in order to make it clear, I will immediately write down a specific example. In parametric form, the function is given by two equations: . Often, equations are written not under curly braces, but sequentially:,.

The variable is called a parameter and can take values ​​from "minus infinity" to "plus infinity". Consider, for example, the value and substitute it into both equations: . Or humanly: "if x is equal to four, then y is equal to one." You can mark a point on the coordinate plane, and this point will correspond to the value of the parameter. Similarly, you can find a point for any value of the parameter "te". As for the "ordinary" function, for the American Indians of a parametrically given function, all rights are also respected: you can plot a graph, find derivatives, and so on. By the way, if there is a need to build a graph of a parametrically given function, you can use my program.

In the simplest cases, it is possible to represent the function explicitly. We express the parameter from the first equation: and substitute it into the second equation: . The result is an ordinary cubic function.

In more "severe" cases, such a trick does not work. But this is not a problem, because in order to find the derivative parametric function there is a formula:

We find the derivative of "the player with respect to the variable te":

All the rules of differentiation and the table of derivatives are valid, of course, for the letter , thus, there is no novelty in the process of finding derivatives. Just mentally replace all the "x"s in the table with the letter "te".

We find the derivative of "x with respect to the variable te":

Now it only remains to substitute the found derivatives into our formula:

Ready. The derivative, like the function itself, also depends on the parameter .

As for the notation, instead of writing in the formula, one could simply write it without a subscript, since this is the “ordinary” derivative “by x”. But there is always a variant in the literature, so I will not deviate from the standard.

Example 6

We use the formula

In this case:

Thus:

A feature of finding the derivative of a parametric function is the fact that at each step, it is advantageous to simplify the result as much as possible. So, in the considered example, when finding, I opened the brackets under the root (although I might not have done this). There is a great chance that when substituting and into the formula, many things will be well reduced. Although there are, of course, examples with clumsy answers.

Example 7

Find the derivative of a function given parametrically

This is a do-it-yourself example.

In the article The simplest typical problems with a derivative we considered examples in which it was required to find the second derivative of a function. For a parametrically given function, you can also find the second derivative, and it is found by the following formula: . It is quite obvious that in order to find the second derivative, one must first find the first derivative.

Example 8

Find the first and second derivatives of a function given parametrically

Let's find the first derivative first.
We use the formula

In this case:

Derivatives of higher orders are found by successive differentiation of formula (1).

Example. Find and if (x ²+y ²)³-3(x ²+y ²)+1=0.

Solution. Denoting the left side of this equation through f(x, y) find the partial derivatives

f"x(x,y)=3(x²+y²)²∙2x-3∙2x=6x[(x²+y²)-1],

f"y(x,y)=3(x²+y²)²∙2y-3∙2y=6y[(x²+y²)-1].

Hence, applying formula (1), we obtain:

.

To find the second derivative, we differentiate with respect to X found first derivative, keep in mind that at there is a function x:

.

2°. Case of several independent variables. Similarly, if the equation F(x, y, z)=0, Where F(x, y, z) is a differentiable function of variables x, y And z, defines z as a function of independent variables X And at And Fz(x, y, z)≠ 0, then the partial derivatives of this implicitly given function, generally speaking, can be found by the formulas

.

Another way to find the derivatives of the function z is as follows: differentiating the equation F(x, y, z) = 0, we get:

.

From here it is possible to determine dz, and hence also .

Example. Find and if x ² - 2y²+3z² -yz +y=0.

1st way. Denoting the left side of this equation through F(x, y, z), find partial derivatives F"x(x,y,z)=2x, F"y(x,y,z)=-4y-z+1, F"z(x,y,z)=6z-y.

Applying formulas (2), we obtain:

2nd way. Differentiating this equation, we get:

2xdx-4ydy+6zdz-ydz-zdy +dy=0

From here we determine dz, i.e. the total differential of the implicit function:

.

Comparing with the formula , we see that

.

3°. System of implicit functions. If the system of two equations

defines u And v as functions of the variables x and y and the Jacobian

,

then the differentials of these functions (and, consequently, their partial derivatives) can be found from the system of equations

Example: Equations u+v=x+y, xu+yv=1 determine u And v as a function X And at; find .

Solution. 1st way. Differentiating both equations with respect to x, we get:

.

Similarly, we find:

.

2nd way. By differentiation, we find two equations relating the differentials of all four variables: du +dv=dx +dy ,xdu +udx +ydv +vdy=0.

Having solved this system with respect to the differentials du And dv, we get:

4°. Parametric function definition. If the function of r variables X And at given parametrically by the equations x=x(u,v), y=y(u,v), z=z(u,v) And

,

then the differential of this function can be found from the system of equations

Knowing the differential dz=p dx+q dy, find the partial derivatives and .

Example. Function z arguments X And at given by the equations x=u+v, y=u²+v², z=u²+v² (u≠v).

Find and .

Solution. 1st way. By differentiation, we find three equations relating the differentials of all five variables:

From the first two equations, we determine du And dv:

.

Substitute the found values ​​into the third equation du And dv:

.

2nd way. From the third given equation, one can find:

Differentiate the first two equations first with respect to X, then by at:

From the first system we find: .

From the second system we find: .

Substituting the expressions and into formula (5), we obtain:

Change of variables

When changing variables in differential expressions, the derivatives included in them should be expressed in terms of other derivatives according to the rules of differentiation of a complex function.

1°. Change of variables in expressions containing ordinary derivatives.

,

assuming .

at By X through derivatives of at By t. We have:

,

.

Substituting the found expressions of derivatives into this equation and replacing X through , we get:

Example. Convert Equation

,

taking it as an argument at, and for the function x.

Solution. We express the derivatives of at By X through derivatives of X By y.

.

Substituting these derivative expressions into this equation, we will have:

,

or, finally,

.

Example. Convert Equation

going to polar coordinates

x=r cos φ, y=r cos φ.

Solution. Considering r as a function φ , from formulas (1) we obtain:

dх = сosφ dr – r sinφ d φ, dy=sinφ+r cosφ dφ,

Let a continuous function at from X is set implicitly F(x, y) = 0, where F(x, y), F" x(x, y), F"y(x, y) There is continuous functions in some domain D containing the point ( X, at), whose coordinates satisfy the relations F (x, y) = 0, F"y(x, y) ≠ 0. Then the function at from X has a derivative

Proof (see figure). Let F"y(x, y) > 0. Since the derivative F"y(x, y) is continuous, then we can construct a square [ X 0 - δ" , X 0 + δ" , at 0 - δ" , at 0 + δ" ], so that for all its points F"y (x, y) > 0, i.e. F(x, y) is monotonic in at at a fixed X. Thus, all the conditions of the implicit function existence theorem are satisfied at = f (x), such that F(x, f (x)) º 0.
Let us set the increment Δ X. new value X + Δ X will match at + Δ at = f (x + Δ x), such that these values ​​satisfy the equation F (x + Δ x, y + Δ y) = 0. Obviously,

Δ F = F(x + Δ x, y + Δ y) − F(x, y) = 0

and in this case

.

From (7) we have

.

Since the implicit function at = f (x) is continuous, then Δ at→ 0 as Δ X→ 0, hence α → 0 and β → 0. Whence we finally have

.

Q.E.D.

Partial derivatives and differentials of higher orders.

Let the partial derivatives of the functions z = f (x, y) defined in a neighborhood of the point M exist at every point in this neighborhood. In this case, partial derivatives are functions of two variables X And at defined in the indicated neighborhood of the point M. Let us call them partial derivatives of the first order. In turn, the partial derivatives with respect to the variables X And at from functions at the point M, if they exist, are called second-order partial derivatives of the function f (M) at this point and are denoted by the following symbols

Second-order partial derivatives of the form , , are called mixed partial derivatives.

Higher order differentials

We will consider dx in the expression for dy as a constant factor. Then the function dy is a function of only the argument x and its differential at a point x has the form (when considering the differential from dy we will use the new notation for differentials):

δ ( d y) = δ [ f " (x) d x] = [f " (x) d x] " δ x = f "" (x) d(x) δ x .

Differential δ ( d y) from differential dy at the point x, taken at δ x = dx, is called the second-order differential of the function f (x) at the point x and denoted d 2 y, i.e.

d 2 y = f ""(x)·( dx) 2 .

In turn, the differential δ( d 2 y) from differential d 2 y, taken at δ x = dx, is called the third-order differential of the function f(x) and denoted d 3 y etc. Differential δ( d n-1 y) from differential d n -1 f, taken at δ x = dx, is called the differential n-th order (or n- m differential) functions f(x) and denoted d n y.
Let's prove that for n-th differential of the function, the formula

d n y = y (n) ·( dx)n, n = 1, 2, … (3.1)

In the proof, we use the method of mathematical induction. For n= 1 and n= 2 formula (3.1) is proved. Let it be true for differentials of order n - 1

d n −1 y=y( n−1) ( dx)n −1 ,

and function y (n-1) (x) is differentiable at some point x. Then

Letting δ x = dx, we get

Q.E.D.
For anyone n fair equality

or

those. n- i derivative of the function y= f (x) at the point x is equal to the ratio n-th differential of this function at the point x To n-th degree of the differential of the argument.

Directional derivative of functions of several variables.

The function and the unit vector are considered. Direct l through t. M 0 with direction vector

Definition 1. Function derivative u = u(x, y, z) by variable t called derivative in direction l

Since on this line u is a complex function of one variable, then the derivative with respect to t is equal to the total derivative with respect to t(§ 12).

It is denoted and equal to

Derivative of a complex function. total derivative

Let z=ƒ(x;y) be a function of two variables x and y, each of which is a function of the independent variable t: x = x(t), y = y(t). In this case, the function z = f(x(t);y(t)) is a complex function of one independent variable t; variables x and y are intermediate variables.

If z \u003d ƒ (x; y) is a function differentiable at the point M (x; y) є D and x \u003d x (t) and y \u003d y (t) are differentiable functions of the independent variable t, then the derivative of the complex function z (t ) = f(x(t);y(t)) is calculated by the formula

Let's give the independent variable t an increment Δt. Then the functions x = x(t) and y = y(t) will receive increments Δx and Δy, respectively. They, in turn, will cause the function z to increment Az.

Since, by the condition, the function z - ƒ(x; y) is differentiable at the point M(x; y), its total increment can be represented as

where a→0, β→0 as Δх→0, Δу→0 (see item 44.3). We divide the expression Δz by Δt and pass to the limit as Δt→0. Then Δх→0 and Δу→0 due to the continuity of the functions x = x(t) and y = y(t) (according to the condition of the theorem, they are differentiable). We get:

Special case: z=ƒ(x;y), where y=y(x), i.e. z=ƒ(x;y(x)) is a complex function of one independent variable x. This case reduces to the previous one, with x playing the role of the variable t. According to formula (44.8) we have:

Formula (44.9) is called the total derivative formula.

General case: z=ƒ(x;y), where x=x(u;v), y=y(u;v). Then z= f(x(u;v);y(u;v)) is a complex function of the independent variables u and v. Its partial derivatives can be found using formula (44.8) as follows. Having fixed v, we replace it with the corresponding partial derivatives

The formula for the derivative of a function given implicitly. Proof and examples of application of this formula. Examples of calculating derivatives of the first, second and third order.

Content

First order derivative

Let the function be given implicitly using the equation
(1) .
And let this equation, for some value , has a unique solution . Let the function be a differentiable function at the point , and
.
Then, for this value , there is a derivative , which is determined by the formula:
(2) .

Proof

For proof, consider the function as a complex function of the variable :
.
We apply the rule of differentiation of a complex function and find the derivative with respect to the variable of the left and right sides of the equation
(3) :
.
Since the derivative of the constant is equal to zero and , then
(4) ;
.

The formula has been proven.

Derivatives of higher orders

Let us rewrite equation (4) using other notation:
(4) .
Moreover, and are complex functions of the variable :
;
.
Dependence defines the equation (1):
(1) .

We find the derivative with respect to the variable from the left and right sides of equation (4).
According to the formula for the derivative of a complex function, we have:
;
.
According to the derivative product formula:

.
According to the derivative sum formula:


.

Since the derivative of the right side of equation (4) is equal to zero, then
(5) .
Substituting the derivative here, we obtain the value of the second-order derivative in implicit form.

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

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

Examples

Example 1

Find the first derivative of the function given implicitly by the equation:
(P1) .

Formula 2 Solution

We find the derivative by formula (2):
(2) .

Let's move all the variables to the left side so that the equation takes the form .
.
From here.

We find the derivative with respect to , assuming that it is constant.
;
;
;
.

We find the derivative with respect to the variable, assuming the variable is constant.
;
;
;
.

By formula (2) we find:
.

We can simplify the result if we note that according to the original equation (A.1), . Substitute :
.
Multiply the numerator and denominator by:
.

Solution in the second way

Let's solve this example in the second way. To do this, we find the derivative with respect to the variable of the left and right parts of the original equation (P1).

We apply:
.
We apply the formula for the derivative of a fraction:
;
.
We apply the formula for the derivative of a complex function:
.
We differentiate the original equation (P1).
(P1) ;
;
.
Multiply by and group the terms.
;
.

Substitute (from equation (P1)):
.
Let's multiply by:
.

Example 2

Find the second order derivative of the function given implicitly using the equation:
(P2.1) .

Differentiate the original equation with respect to the variable , assuming that it is a function of :
;
.
We apply the formula for the derivative of a complex function.
.

We differentiate the original equation (A2.1):
;
.
It follows from the original equation (A2.1) that . Substitute :
.
Expand the brackets and group the members:
;
(P2.2) .
We find the derivative of the first order:
(P2.3) .

To find the second order derivative, we differentiate equation (A2.2).
;
;
;
.
We substitute the expression for the first order derivative (A2.3):
.
Let's multiply by:

;
.
From here we find the derivative of the second order.

Example 3

Find the third order derivative for of the function given implicitly using the equation:
(P3.1) .

Differentiate the original equation with respect to the variable, assuming that is a function of .
;
;
;
;
;
;
(P3.2) ;

We differentiate equation (A3.2) with respect to the variable .
;
;
;
;
;
(P3.3) .

We differentiate equation (A3.3).
;
;
;
;
;
(P3.4) .

From equations (A3.2), (A3.3) and (A3.4) we find the values ​​of derivatives at .
;
;
.