Derivative of an implicit function. Derivative of implicitly defined function: manual, examples. Derivatives of higher orders

Consider a function y(x), which is written in an implicit way in the general form $ F(x,y(x)) = 0 $. The derivative of an implicit function is found in two ways:

1. By differentiating both sides of the equation
2. By using the ready-made formula $ y" = - \frac(F"_x)(F"_y) $

How to find?

Method 1

It is not required to bring the function to an explicit form. We must immediately start differentiating the left and right sides of the equation with respect to $ x $. It is worth noting that the derivative of $ y" $ is calculated by the rule of differentiation of a complex function. For example, $ (y^2)"_x = 2yy" $. After finding the derivative, you need to express $ y" $ from the resulting equation and place $ y" $ on the left side.

Method 2

You can use a formula that uses the partial derivatives of the implicit function $ F(x,y(x)) = 0 $ in the numerator and denominator. To find the numerator, we take the derivative with respect to $ x $, and for the denominator, we take the derivative with respect to $ y $.

The second derivative of an implicit function can be found by re-differentiating the first derivative of an implicit function.

Solution examples

Consider practical examples of solutions for calculating the derivative of an implicitly given function.

The function Z= f(x; y) is called implicit if it is given by the equation F(x, y, z)=0 unresolved with respect to Z. Let us find the partial derivatives of the function Z given implicitly. To do this, substituting in the equation instead of Z the function f (x; y) we obtain the identity F (x, y, f (x, y)) \u003d 0. The partial derivatives with respect to x and y of the function, which is identically equal to zero, are also equal to zero.

F(x, y, f(x, y)) =
=0 (y is considered constant)

F(x, y, f(x, y)) =
=0 (xconsider constant)

Where
And

Example: Find partial derivatives of a function Z given an equation
.

Here F(x,y,z)=
;
;
;
. According to the formulas above, we have:

And

1. Directional derivative

Let a function of two variables Z = f(x; y) be given in some neighborhood of m. M (x, y). Consider some direction determined by the unit vector
, where
(see fig.).

On a straight line passing in this direction through point M, we take point M 1 (
) so that the length
segment MM 1 is equal to
. The increment of the function f(M) is determined by the relation, where
connected by relationships. ratio limit at
will be called the derivative of the function
at the point
towards and be designated .

=

If the function Z is differentiable at a point
, then its increment at this point, taking into account the relations for
can be written in the following form.

dividing both parts by

and passing to the limit at
we obtain a formula for the derivative of the function Z \u003d f (x; y) in the direction:

1. Gradient

Consider a function of three variables
differentiable at some point
.

The gradient of this function
at the point M is called a vector whose coordinates are equal, respectively, to partial derivatives
at this point. The symbol used to denote a gradient is
.
=
.

.The gradient indicates the direction of the fastest growth of the function at a given point.

Since the unit vector has coordinates (
), then the directional derivative for the case of a function of three variables is written in the form, i.e. has the dot product formula of vectors And
. Let's rewrite the last formula as follows:

, where - angle between vector And
. Insofar as
, then it follows that the directional derivative of the function takes the max value at =0, i.e. when the direction of the vectors And
match. Wherein
.Ie, in fact, the gradient of the function characterizes the direction and magnitude of the maximum rate of increase of this function at a point.

1. Extremum of a function of two variables

The concepts of max, min, extremum of a function of two variables are similar to the corresponding concepts of a function of one variable. Let the function Z = f(x; y) be defined in some domain D, etc. M
belongs to this area. Point M
is called a point max of the function Z= f(x; y) if there exists such a δ-neighborhood of the point
, that for each point from this neighborhood the inequality
. The point min is defined in a similar way, only the inequality sign will change in this case
. The value of the function at the point max(min) is called the maximum (minimum). The maximum and minimum of a function are called extrema.

1. Necessary and sufficient conditions for an extremum

Theorem:(Necessary extremum conditions). If at point M
differentiable function Z= f(x; y) has an extremum, then its partial derivatives at this point are equal to zero:
,
.

Proof: fixing one of the variables x or y, we convert Z= f(x; y) into a function of one variable, for the extremum of which the above conditions must be satisfied. Geometrically equal
And
mean that at the extremum point of the function Z= f(x; y), the tangent plane to the surface representing the function f(x, y)=Z is parallel to the OXY plane, because the equation of the tangent plane is Z=Z 0. The point at which the first-order partial derivatives of the function Z= f(x; y) are equal to zero, i.e.
,
, are called the stationary point of the function. A function can have an extremum at points where at least one of the partial derivatives does not exist. For example Z=|-
| has max at O(0,0) but no derivatives at that point.

Stationary points and points at which at least one partial derivative does not exist are called critical points. At critical points, the function may or may not have an extremum. Equality to zero of partial derivatives is a necessary but not sufficient condition for the existence of an extremum. For example, when Z=xy, point O(0,0) is critical. However, the function Z=xy does not have an extremum in it. (Because in quarters I and III Z>0, and in II and IV–Z<0). Таким образом для нахождения экстремумов функции в данной области необходимо подвергнуть каждую критическую точку функции дополнительному исследованию.

Theorem: (Sufficient condition for extrema). Let at a stationary point
and some neighborhood, the function f(x; y) has continuous partial derivatives up to the 2nd order inclusive. Calculate at a point
values
,
And
. Denote

If
, extremum at the point
may or may not be. More research is needed.

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.

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 .
;
;
.

Undoubtedly, in our minds, the image of a function is associated with equality and the line corresponding to it - the graph of the function. For example, - functional dependence, the graph of which is a quadratic parabola with a vertex at the origin and upward branches; is the sine function known for its waves.

In these examples, the left side of the equality is y , and the right side is an expression that depends on the argument x . In other words, we have an equation resolved with respect to y . The representation of a functional dependence in the form of such an expression is called by explicitly setting the function(or function explicitly). And this type of function assignment is the most familiar to us. In most examples and problems, we are presented with explicit functions. We have already discussed in detail about the differentiation of functions of one variable, given explicitly.

However, the function implies a correspondence between a set of x values ​​and a set of y values, and this correspondence is NOT necessarily established by any formula or analytical expression. That is, there are many ways to specify a function in addition to the usual .

In this article, we will look at implicit functions and ways to find their derivatives. Examples of implicit functions are or .

As you noticed, the implicit function is defined by the relation . But not all such relationships between x and y define a function. For example, no pair of real numbers x and y satisfies the equality , therefore, this relation does not define an implicit function.

It can implicitly determine the law of correspondence between the values ​​x and y , and each value of the argument x can correspond to either one (in this case we have a single-valued function) or several values ​​of the function (in this case, the function is called multi-valued). For example, the value x = 1 corresponds to two real values ​​y = 2 and y = -2 of the implicitly defined function .

It is far from always possible to reduce an implicit function to an explicit form, otherwise it would not be necessary to differentiate the implicit functions themselves. For example, - is not converted to an explicit form, but - is converted.

Now to business.

To find the derivative of an implicitly given function, it is necessary to differentiate both sides of the equality with respect to the argument x, considering y to be a function of x, and then express .

Differentiation of expressions containing x and y(x) is carried out using the rules of differentiation and the rule for finding the derivative of a complex function. Let's immediately analyze a few examples in detail so that there are no further questions.

Example.

Differentiate Expressions in x , assuming y is a function of x .

Solution.

Because y is a function of x , then is a complex function. It can be conventionally represented as f(g(x)) , where f is the cube function and g(x) = y . Then, according to the formula for the derivative of a complex function, we have: .

When differentiating the second expression, we take the constant out of the sign of the derivative and act as in the previous case (here f is the sine function, g(x) = y ):

For the third expression, we use the formula for the derivative of the product:

Sequentially applying the rules, we differentiate the last expression:

Now you can move on to finding the derivative of an implicitly given function, for this we have all the knowledge.

Example.

Find the derivative of an implicit function.

Solution.

The derivative of an implicit function is always represented as an expression containing x and y : . To arrive at this result, we differentiate both sides of the equality:

Let's solve the resulting equation with respect to the derivative:

Answer:

.

COMMENT.

To consolidate the material, let's solve another example.