Examples of complex and implicit functions. Derivation of a function specified implicitly: guidance, examples. Extrema of a function of two variables

A 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 the function f(x;y) into the equation instead of Z, we obtain the identity F(x,y, f(x,y))=0. Partial derivatives of a function, identically equal to zero, are also equal to zero.

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

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

Where
And

Example: Find the partial derivatives of the function Z given by the equation
.

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

And

1. Directional derivative

Let a function of two variables Z= f(x; y) be given in a certain neighborhood of the point M (x,y). Consider some direction defined by the unit vector
, Where
(see picture).

On a straight line passing in this direction through point M, we take point M 1 (
) so that the length
segmentMM 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 the 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= f(x; y) in the direction:

1. Gradient

Consider a function of three variables
differentiable at some point
.

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

.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 formula for the scalar product of vectors And
. Let's rewrite the last formula as follows:

, Where - angle between vector And
. Because the
, then it follows that the derivative of the function in direction takes the max value at =0, i.e. when the direction of the vectors And
match up. Wherein
That is, in fact, the gradient of a 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 the max point of the function Z= f(x; y) if there is such a δ-neighborhood of the point
, that for each point from this neighborhood the inequality
. The point min is determined in a similar way, only the inequality sign will change
. The value of the function at the point max(min) is called maximum (minimum). The maximum and minimum of a function are called extrema.

1. Necessary and sufficient conditions for an extremum

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

Proof: Having fixed one of the variables x or y, we transform Z = f(x; y) into a function of one variable, for the extremum of which the above conditions must be met. Geometrically equalities
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 exampleZ=|-
| has max at the point O(0,0), but has no derivatives at this 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. The equality of partial derivatives to zero 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 the I and III quarters Z>0, and in the 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 in a certain neighborhood the function f(x; y) has continuous partial derivatives up to the 2nd order inclusive. Let's calculate at the point
values
,
And
. Let's denote

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

As is known, an implicitly given function of one variable is defined as follows: the function y of the independent variable x is called implicit if it is given by an equation that is not resolved with respect to y:

Example 1.11.

The equation

implicitly specifies two functions:

And the equation

does not specify any function.

Theorem 1.2 (existence of an implicit function).

Let the function z =f(x,y) and its partial derivatives f"x and f"y be defined and continuous in some neighborhood UM0 of the point M0(x0y0). In addition, f(x0,y0)=0 and f"(x0,y0)≠0, then equation (1.33) defines in the neighborhood of UM0 an implicit function y= y(x), continuous and differentiable in some interval D with center at point x0, and y(x0)=y0.

No proof.

From Theorem 1.2 it follows that on this interval D:

that is, there is an identity in

where the “total” derivative is found according to (1.31)

That is, (1.35) gives a formula for finding the derivative of an implicitly given function of one variable x.

An implicit function of two or more variables is defined similarly.

For example, if in some region V of Oxyz space the equation holds:

then under certain conditions on the function F it implicitly defines the function

Moreover, by analogy with (1.35), its partial derivatives are found as follows:

Example 1.12. Assuming that the equation

implicitly defines a function

find z"x, z"y.

therefore, according to (1.37), we get the answer.

11.Use of partial derivatives in geometry.

12.Extrema of a function of two variables.

The concepts of maximum, minimum, and extremum of a function of two variables are similar to the corresponding concepts of a function of one independent variable (see section 25.4).

Let the function z = ƒ(x;y) be defined in some domain D, point N(x0;y0) О D.

A point (x0;y0) is called a maximum point of the function z=ƒ(x;y) if there is a d-neighborhood of the point (x0;y0) such that for each point (x;y) different from (xo;yo), from this neighborhood the inequality ƒ(x;y) holds<ƒ(хо;уо).

A The minimum point of the function is determined in a similar way: for all points (x; y) other than (x0; y0), from the d-neighborhood of the point (xo; yo) the following inequality holds: ƒ(x; y)>ƒ(x0; y0).

In Figure 210: N1 is the maximum point, and N2 is the minimum point of the function z=ƒ(x;y).

The value of the function at the point of maximum (minimum) is called the maximum (minimum) of the function. The maximum and minimum of a function are called its extrema.

Note that, by definition, the extremum point of the function lies inside the domain of definition of the function; maximum and minimum have a local (local) character: the value of the function at the point (x0; y0) is compared with its values ​​at points sufficiently close to (x0; y0). In region D, a function may have several extrema or none.

46.2. Necessary and sufficient conditions for an extremum

Let us consider the conditions for the existence of an extremum of a function.

Theorem 46.1 (necessary conditions for an extremum). If at the point N(x0;y0) the differentiable function z=ƒ(x;y) has an extremum, then its partial derivatives at this point are equal to zero: ƒ"x(x0;y0)=0, ƒ"y(x0;y0 )=0.

Let's fix one of the variables. Let us put, for example, y=y0. Then we obtain a function ƒ(x;y0)=φ(x) of one variable, which has an extremum at x = x0. Therefore, according to the necessary condition for the extremum of a function of one variable (see section 25.4), φ"(x0) = 0, i.e. ƒ"x(x0;y0)=0.

Similarly, it can be shown that ƒ"y(x0;y0) = 0.

Geometrically, the equalities ƒ"x(x0;y0)=0 and ƒ"y(x0;y0)=0 mean that at the extremum point of the function z=ƒ(x;y) the tangent plane to the surface representing the function ƒ(x;y) ), is parallel to the Oxy plane, since the equation of the tangent plane is z=z0 (see formula (45.2)).

Z note. A function can have an extremum at points where at least one of the partial derivatives does not exist. For example, the function has a maximum at the point O(0;0) (see Fig. 211), but does not have partial derivatives at this point.

The point at which the first order partial derivatives of the function z ≈ ƒ(x; y) are equal to zero, i.e. f"x=0, f"y=0, is called a stationary point of the function z.

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. The equality of partial derivatives to zero is a necessary but not sufficient condition for the existence of an extremum. Consider, for example, the function z = xy. For it, the point O(0; 0) is critical (at it z"x=y and z"y - x vanish). However, the function z=xy does not have an extremum in it, since in a sufficiently small neighborhood of the point O(0; 0) there are points for which z>0 (points of the first and third quarters) and z< 0 (точки II и IV четвертей).

Thus, to find the extrema of a function in a given area, it is necessary to subject each critical point of the function to additional research.

Theorem 46.2 (sufficient condition for an extremum). Let the function ƒ(x;y) at a stationary point (xo; y) and some of its neighborhood have continuous partial derivatives up to the second order inclusive. Let us calculate at the point (x0;y0) the values ​​A=f""xx(x0;y0), B=ƒ""xy(x0;y0), C=ƒ""yy(x0;y0). Let's denote

1. if Δ > 0, then the function ƒ(x;y) at the point (x0;y0) has an extremum: maximum if A< 0; минимум, если А > 0;

2. if Δ< 0, то функция ƒ(х;у) в точке (х0;у0) экстремума не имеет.

In the case of Δ = 0, there may or may not be an extremum at the point (x0;y0). More research is needed.

TASKS

1.

Example. Find the intervals of increasing and decreasing functions. Solution. The first step is finding the domain of definition of a function. In our example, the expression in the denominator should not go to zero, therefore, . Let's move on to the derivative function: To determine the intervals of increase and decrease of a function based on a sufficient criterion, we solve inequalities on the domain of definition. Let's use a generalization of the interval method. The only real root of the numerator is x = 2, and the denominator goes to zero at x = 0. These points divide the domain of definition into intervals in which the derivative of the function retains its sign. Let's mark these points on the number line. We conventionally denote by pluses and minuses the intervals at which the derivative is positive or negative. The arrows below schematically show the increase or decrease of the function on the corresponding interval. Thus, And . At the point x = 2 the function is defined and continuous, so it should be added to both the increasing and decreasing intervals. At the point x = 0 the function is not defined, so we do not include this point in the required intervals. We present a graph of the function to compare the results obtained with it. Answer: the function increases with , decreases on the interval (0; 2] .

2.

Examples.

Set the intervals of convexity and concavity of a curve y = 2 – x 2 .

We'll find y"" and determine where the second derivative is positive and where it is negative. y" = –2x, y"" = –2 < 0 на (–∞; +∞), следовательно, функция всюду выпукла.

y = e x. y"" = e Because x x > 0 for any

y = x 3 . y"" = 6x, then the curve is concave everywhere. y"" < 0 при x < 0 и y, That x"" > 0 at x < 0 кривая выпукла, а при x> 0. Therefore, when

3.

4. Given the function z=x^2-y^2+5x+4y, vector l=3i-4j and point A(3,2). Find dz/dl (as I understand it, the derivative of the function in the direction of the vector), gradz(A), |gradz(A)|. Let's find the partial derivatives: z(with respect to x)=2x+5 z(with respect to y)=-2y+4 Let's find the values ​​of the derivatives at point A(3,2): z(with respect to x)(3,2)=2*3+ 5=11 z(by y)(3,2)=-2*2+4=0 From where, gradz(A)=(11,0)=11i |gradz(A)|=sqrt(11^2+0 ^2)=11 Derivative of the function z in the direction of the vector l: dz/dl=z(in x)*cosa+z(in y)*cosb, a, b-angles of the vector l with the coordinate axes. cosa=lx/|l|, cosb=ly/|l|, |l|=sqrt(lx^2+ly^2) lx=3, ly=-4, |l|=5. cosa=3/5, cosb=(-4)/5. dz/dl=11*3/5+0*(-4)/5=6.6.

Higher order derivatives 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 given 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].

From here, applying formula (1), we obtain:

.

To find the second derivative, differentiate with respect to X the first derivative found, taking into account that at there is a function x:

.

2°. The case of several independent variables. Likewise, if the equation F(x, y, z)=0, Where F(x, y, z) - 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 using the formulas

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

.

From here we can determine dz, and therefore .

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

1st method. Denoting the left side of this equation by F(x, y, z), let's find the 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 method. 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 formula , we see that

.

3°. Implicit Function System. If a 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 therefore 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 functions X And at; find .

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

.

In a similar way we find:

.

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

Solving this system for differentials du And dv, we get:

4°. Parametric function specification. If the function of r variables X And at is 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, we find the partial derivatives and .

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

Find and .

Solution. 1st method. By differentiation we find three equations connecting the differentials of all five variables:

From the first two equations we determine du And dv:

.

Let's substitute the found values ​​into the third equation du And dv:

.

2nd method. From the third given equation we can find:

Let us first differentiate the first two equations with respect to X, then by at:

From the first system we find: .

From the second system we find: .

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

Replacing variables

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

1°. Replacing variables in expressions containing ordinary derivatives.

,

believing .

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

,

.

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

Example. Convert Equation

,

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

Solution. Let us express the derivatives of at By X through derivatives of X By u.

.

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

,

or, finally,

.

Example. Convert Equation

going to polar coordinates

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

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

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 an 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; the variables x and y are intermediate variables.

If z = ƒ(x;y) is a function differentiable at the point M(x;y) є D and x = x(t) and y = 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 using 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 Δх and Δу, respectively. They, in turn, will cause the function z to increment Az.

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

where а→0, β→0 at Δх→0, Δу→0 (see paragraph 44.3). Let's divide the expression Δz by Δt and go to the limit at 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 conditions 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)) - complex function one independent variable x. This case reduces to the previous one, and the role of the variable t is played by x. 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