Formulas for differentiating implicitly given functions. Derivative of a function defined implicitly
Very often, when solving practical problems (for example, in higher geodesy or analytical photogrammetry), complex functions of several variables appear, i.e., arguments x, y, z one function f(x,y,z) ) are themselves functions of the new variables U, V, W ).
So, for example, it happens when moving from a fixed coordinate system Oxyz to the mobile system O 0 UVW and back. In this case, it is important to know all the partial derivatives with respect to the "fixed" - "old" and "moving" - "new" variables, since these partial derivatives usually characterize the position of an object in these coordinate systems, and, in particular, affect the correspondence of aerial photographs to a real object . In such cases, the following formulas apply:
That is, given a complex function T three "new" variables U, V, W through three "old" variables x, y, z then:
Comment. Variations in the number of variables are possible. For example: if
In particular, if z = f(xy), y = y(x) , then we get the so-called "total derivative" formula:
The same formula for the "total derivative" in the case of:
will take the form:
Other variations of formulas (1.27) - (1.32) are also possible.
Note: the "total derivative" formula is used in the course of physics, section "Hydrodynamics" when deriving the fundamental system of equations of fluid motion.
Example 1.10. Given:
According to (1.31):
§7 Partial derivatives of an implicitly given function of several variables
As you know, an implicitly defined function of one variable is defined as follows: the function 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 defines two functions:
And the equation
does not define any function.
Theorem 1.2 (existence of an implicit function).
Let the function z \u003d f (x, y) and its partial derivatives f" x and f" y defined and continuous in some neighborhood U M0 points M 0 (x 0 y 0 ) . Besides, f(x 0 ,y 0 )=0 and f"(x 0 ,y 0 )≠0 , then equation (1.33) determines in the neighborhood U M0 implicit function y= y(x) , continuous and differentiable in some interval D centered on a point x 0 , and y(x 0 )=y 0 .
Without 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 implicitly given function one variable x .
An implicit function of two or more variables is defined similarly.
For example, if in some area V space Oxyz the equation is fulfilled:
then under certain conditions on the function F it implicitly defines a function
At the same time, 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 obtain the answer.
§8 Partial derivatives of the second and higher orders
Definition 1.9 Partial derivatives of the second order of a function z=z(x,y) are defined like this:
There were four of them. Moreover, under certain conditions on the functions z(x,y) equality holds:
Comment. Partial derivatives of the second order can also be denoted as follows:
Definition 1.10 Partial derivatives of the third order - eight (2 3).
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
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:
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
. Because the
, 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.
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.
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.
We will learn to find derivatives of functions that are given implicitly, that is, given by some equations that relate variables to each other x and y. Examples of functions defined implicitly:
,
Derivatives of implicit functions, or derivatives of implicit functions, are fairly easy to find. Now let's analyze the corresponding rule and example, and then find out why this is needed at all.
In order to find the derivative of a function given implicitly, it is necessary to differentiate both sides of the equation with respect to x. Those terms in which only x is present will turn into the usual derivative of a function of x. And the terms with y must be differentiated using the rule of differentiation of a complex function, since y is a function of x. If it’s quite simple, then in the resulting derivative of the term with x it should turn out: the derivative of the function from the y, multiplied by the derivative from the y. For example, the derivative of the term will be written as , the derivative of the term will be written as . Further, from all this it is necessary to express this "y stroke" and the desired derivative of the function given implicitly will be obtained. Let's look at this with an example.
Example 1
Solution. We differentiate both sides of the equation with respect to x, assuming that y is a function of x:
From here we get the derivative that is required in the task:
Now something about the ambiguous property of implicitly defined functions, and why special rules for their differentiation are needed. In some cases, you can make sure that substitution in a given equation (see examples above) instead of the y of its expression through x leads to the fact that this equation turns into an identity. So. the above equation implicitly defines the following functions:
After substituting the expression y squared through x into the original equation, we get the identity:
.
The expressions that we substituted were obtained by solving the equation for the y.
If we were to differentiate the corresponding explicit function
then we would get a response as in example 1 - from a function specified implicitly:
But not every function given implicitly can be represented in the form y = f(x) . So, for example, the implicitly defined functions
are not expressed in terms of elementary functions, that is, these equations cannot be solved with respect to the player. Therefore, there is a rule for differentiating a function given implicitly, which we have already studied and will be consistently applied in other examples.
Example 2 Find the derivative of a function given implicitly:
.
We express the y prime and - at the output - the derivative of the function given implicitly:
Example 3 Find the derivative of a function given implicitly:
.
Solution. Differentiate both sides of the equation with respect to x:
.
Example 4 Find the derivative of a function given implicitly:
.
Solution. Differentiate both sides of the equation with respect to x:
.
We express and get the derivative:
.
Example 5 Find the derivative of a function given implicitly:
Solution. We transfer the terms on the right side of the equation to the left side and leave zero on the right. Differentiate both sides of the equation with respect to x.
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