Derivative and differential of a complex function of several variables. Partial derivatives Partial derivatives of a complex function examples with solution

1°. The case of one independent variable. If z=f(x,y) is a differentiable function of the arguments x and y, which in turn are differentiable functions of the independent variable t: , then the derivative of the complex function can be calculated using the formula

Example. Find if , where .

Solution. According to formula (1) we have:

Example. Find the partial derivative and total derivative if .

Solution. .

Based on formula (2) we obtain .

2°. The case of several independent variables.

Let z =f (x ;y) - 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.

Theorem. If z == f(x ; y) - differentiable at a point M(x;y)D function and x =x (t) And at =y (t) - differentiable functions of the independent variable t, then the derivative of a complex function z (t) == f(x (t);y (t )) calculated by the formula

Special case:z = f (x ; y), where y = y(x), those. z = f (x ;y (x )) - complex function of one independent variable X. This case reduces to the previous one, and the role of the variable t plays X. According to formula (3) we have:

.

The last formula is called total derivative formulas.

General case:z = f (x ;y ), Where x =x (u ;v ),y=y (u ;v). Then z = f (x (u ;v);y (u ;v)) - complex function of independent variables And And v. Its partial derivatives can be found using formula (3) as follows. Having fixed v, we replace in it , the corresponding partial derivatives

Thus, the derivative of a complex function (z) with respect to each independent variable (And And v) is equal to the sum of the products of partial derivatives of this function (z) with respect to its intermediate variables (x and y) to their derivatives with respect to the corresponding independent variable (u and v).

In all cases considered, the formula is valid

(invariance property of a total differential).

Example. Find and if z = f(x ,y ), where x =uv , .

Solution. Applying formulas (4) and (5), we obtain:

Example. Show that the function satisfies the equation .

Solution. The function depends on x and y through an intermediate argument, so

Substituting partial derivatives into the left side of the equation, we have:

That is, the function z satisfies this equation.

Derivative in a given direction and gradient of the function

1°. Derivative of a function in a given direction. Derivative functions z= f(x,y) in this direction called , where and are the values ​​of the function at points and . If the function z is differentiable, then the formula is valid

where are the angles between the directions l and the corresponding coordinate axes. The derivative in a given direction characterizes the rate of change of a function in that direction.

Example. Find the derivative of the function z = 2x 2 - 3 2 at point P (1; 0) in the direction making an angle of 120° with the OX axis.

Solution. Let's find the partial derivatives of this function and their values ​​at point P.

A proof of the formula for the derivative of a complex function is given. Cases when a complex function depends on one or two variables are considered in detail. A generalization is made to the case of an arbitrary number of variables.

See also: Examples of using the formula for the derivative of a complex function

Basic formulas

Here we provide the derivation of the following formulas for the derivative of a complex function.
If , then
.
If , then
.
If , then
.

Derivative of a complex function from one variable

Let a function of variable x be represented as a complex function in the following form:
,
where there are some functions. The function is differentiable for some value of the variable x. The function is differentiable at the value of the variable.
Then the complex (composite) function is differentiable at point x and its derivative is determined by the formula:
(1) .

Formula (1) can also be written as follows:
;
.

Proof

Let us introduce the following notation.
;
.
Here there is a function of the variables and , there is a function of the variables and . But we will omit the arguments of these functions so as not to clutter the calculations.

Since the functions and are differentiable at points x and , respectively, then at these points there are derivatives of these functions, which are the following limits:
;
.

Consider the following function:
.
For a fixed value of the variable u, is a function of . It's obvious that
.
Then
.

Since the function is a differentiable function at the point, it is continuous at that point. That's why
.
Then
.

Now we find the derivative.

.

The formula is proven.

Consequence

If a function of a variable x can be represented as a complex function of a complex function
,
then its derivative is determined by the formula
.
Here , and there are some differentiable functions.

To prove this formula, we sequentially calculate the derivative using the rule for differentiating a complex function.
Consider the complex function
.
Its derivative
.
Consider the original function
.
Its derivative
.

Derivative of a complex function from two variables

Now let the complex function depend on several variables. First let's look at case of a complex function of two variables.

Let a function depending on the variable x be represented as a complex function of two variables in the following form:
,
Where
and there are differentiable functions for some value of the variable x;
- a function of two variables, differentiable at the point , . Then the complex function is defined in a certain neighborhood of the point and has a derivative, which is determined by the formula:
(2) .

Proof

Since the functions and are differentiable at the point, they are defined in a certain neighborhood of this point, are continuous at the point, and their derivatives exist at the point, which are the following limits:
;
.
Here
;
.
Due to the continuity of these functions at a point, we have:
;
.

Since the function is differentiable at the point, it is defined in a certain neighborhood of this point, is continuous at this point, and its increment can be written in the following form:
(3) .
Here

- increment of a function when its arguments are incremented by values ​​and ;
;

- partial derivatives of the function with respect to the variables and .
For fixed values ​​of and , and are functions of the variables and . They tend to zero at and:
;
.
Since and , then
;
.

Function increment:

. :
.
Let's substitute (3):



.

The formula is proven.

Derivative of a complex function from several variables

The above conclusion can easily be generalized to the case when the number of variables of a complex function is more than two.

For example, if f is function of three variables, That
,
Where
, and there are differentiable functions for some value of the variable x;
- differentiable function of three variables at point , , .
Then, from the definition of differentiability of the function, we have:
(4)
.
Because, due to continuity,
; ; ,
That
;
;
.

Dividing (4) by and passing to the limit, we obtain:
.

And finally, let's consider the most general case.
Let a function of variable x be represented as a complex function of n variables in the following form:
,
Where
there are differentiable functions for some value of the variable x;
- differentiable function of n variables at a point
, , ... , .
Then
.

§ 5. Partial derivatives of complex functions. differentials of complex functions

1. Partial derivatives of a complex function.

Let be a function of two variables whose arguments And , are themselves functions of two or more variables. For example, let
,
.

Then will complex function independent variables And , the variables will be for her intermediate variables. In this case, how to find the partial derivatives of a function with respect to And ?

You can, of course, express it directly in terms of and:

and look for partial derivatives of the resulting function. But the expression can be very complex, and finding partial derivatives , then it will require a lot of effort.

If the functions
,
,
are differentiable, then find and it is possible without resorting to direct expression through and . In this case, the formulas will be valid

(5.1)

Indeed, let's give the argument increment
, – const. Then the functions
And will receive increments

and the function will be incremented

Where , – infinitesimal at
,
. Let's divide all terms of the last equality by . We get:

Since by condition the functions and are differentiable, they are continuous. Therefore, if
, then and . This means that passing to the limit at in the last equality we obtain:

(since , are infinitesimal for , ).

The second equality from (5.1) is proved in a similar way.

EXAMPLE. Let
, Where
,
. Then is a complex function of the independent variables and . To find its partial derivatives, we use formula (5.1). We have

Substituting into (5.1), we obtain

,

Formulas (5.1) are naturally generalized to the case of a function with a larger number of independent and intermediate arguments. Namely, if

………………………

and all the functions under consideration are differentiable, then for any
there is equality

It is also possible that the function arguments are functions of only one variable, i.e.

,
.

Then it will be a complex function of only one variable and we can raise the question of finding the derivative . If the functions
,
are differentiable, then it can be found by the formula
(5.2)

EXAMPLE. Let
, Where
,
. Here is a complex function of one independent variable. Using formula (5.2) we obtain

.

And finally, it is possible that the role of the independent variable is played by , i.e. ,

Where
.

From formula (5.2) we then obtain

(5.3)

(because
). Derivative , standing in formula (5.3) on the right is the partial derivative of the function with respect to . It is calculated with a fixed value. Derivative on the left side of formula (5.3) is called complete derivative of the function . When calculating it, it was taken into account that it depends on in two ways: directly and through the second argument.

EXAMPLE. Find and for function
, Where
.

We have
.

To find, we use formula (5.3). We get

.

And in conclusion of this paragraph, we note that formulas (5.2) and (5.3) are easy to generalize to the case of functions with a large number of intermediate arguments.

2. Differential of a complex function.

Let us recall that if

is a differentiable function of two independent variables, then by definition

, (5.4)

or in another form
. (5.5)

The advantage of formula (5.5) is that it remains true even when is a complex function.

Indeed, let , where , . Let us assume that the functions , , are differentiable. Then the complex function will also be differentiable and its total differential according to formula (5.5) will be equal to

.

Applying formula (5.1) to calculate the partial derivatives of a complex function, we obtain

Since the complete differentials of the functions and are in parentheses, we finally have

So, we are convinced that both in the case when and are independent variables, and in the case when and are dependent variables, the differential of the function can be written in the form (5.5). In this regard, this form of recording the total differential is called invariant . The form of writing the differential proposed in (5.4) will not be invariant; it can only be used in the case when and are independent variables. The form of writing the differential will not be invariant either -th order. Recall that we showed earlier that a differential of order function of two variables can be found by the formula

. (4.12)

But if they are not independent variables, then formula (4.12) with
ceases to be true.

Obviously, all the reasoning carried out in this section for a function of two variables can be repeated in the case of a function with a larger number of arguments. Therefore, for a function the differential can also be written in two forms:

and the second form of notation will be invariant, i.e. fair even in the case when
are not independent variables, but intermediate arguments.

§ 6. Differentiation of implicit functions

Speaking about ways to define a function of one or several variables, we noted that the analytical definition of a function can be explicit or implicit. In the first case, the value of the function is found from the known values ​​of the arguments; in the second, the value of the function and its arguments are related by some equation. However, we did not specify when the equations

And

define implicitly specified functions and respectively. Easy-to-use sufficient conditions for the existence of an implicit function variables (
) are contained in the following theorem.

THEOREM6.1 . (existence of an implicit function) Let the function
and its partial derivatives
are defined and continuous in some neighborhood of the point. If
And
, then there is such a neighborhood point at which the equation

defines a continuous function and

1) Consider the equation
. The conditions of the theorem are satisfied, for example, in any neighborhood of the point
. Therefore, in some neighborhood of the point
this equation defines as an implicit function of two variables and . An explicit expression of this function can be easily obtained by solving the equation for:

2) Consider the equation
. It defines two functions of two variables and . Indeed, the conditions of the theorem are satisfied, for example, in any neighborhood of the point

, in which the given equation defines a continuous function taking on the value
.

On the other hand, the conditions of the theorem are satisfied in any neighborhood of the point
. Consequently, in a certain neighborhood of the point the equation defines a continuous function that takes the value at the point
.

Since a function cannot take on two values ​​at one point, this means we are talking about two different functions
and correspondingly. Let us find their explicit expressions. To do this, let's solve the original equation for . We get

3) Consider the equation
. It is obvious that the conditions of the theorem are satisfied in any neighborhood of the point
. Consequently, there is such a neighborhood of the point
, in which the equation defines as an implicit function of the variable . It is impossible to obtain an explicit expression for this function, since the equation cannot be resolved with respect to .

4) Equation
does not define any implicit function, since there are no pairs of real numbers and that satisfy it.

Function
, given by the equation
, according to Theorem 6.1, has continuous partial derivatives with respect to all arguments in the neighborhood of the point. Let's find out how to find them without explicitly specifying the function.

Let the function
satisfies the conditions of Theorem 6.1. Then the equation
continuous function
. Consider the complex function
, Where . The function is a complex function of one variable, and if
, That

(6.1)

On the other hand, according to formula (5.3) to calculate the total derivative
(6.2)

From (6.1) and (6.2) we obtain that if , then

(6.3)

Comment. Divide by is possible, since according to Theorem 6.1
anywhere in the vicinity.

EXAMPLE. Find the derivative of the implicit function given by the equation and calculate its value at
.

,
.

Substituting partial derivatives into formula (6.3), we obtain

.

Next, substituting into the original equation, we find two values:
And
.

Consequently, in the neighborhood of the point the equation defines two functions:
And
, Where
,
. Their derivatives at will be equal

And
.

Let now the equation
defines in some neighborhood of a point
function Let's find it. Let us recall that in fact this is the ordinary derivative of a function considered as a function of a variable at a constant value. Therefore, we can apply formula (6.3) to find it, considering it a function, an argument, a constant. We get

. (6.4)

Similarly, considering a function, an argument, a constant, using formula (6.3) we find

. (6.5)

EXAMPLE. Find the partial derivatives of the function given by the equation
.

,
,
.

Using formulas (6.4) and (6.5), we obtain

,
.

Finally, consider the general case when the equation

defines a function of variables in a certain neighborhood of a point. Repeating the arguments carried out for an implicitly given function of two variables, we obtain

,
, …,
.

§ 7. Directional derivative

1. Directional derivative.

Let a function of two variables be defined in some domain
plane
, – point of the region, –vector of any direction. Let's move from the point
to a point in the direction of the vector. The function will receive an increment

Let's divide the function increment
by the length of the offset segment
. Resulting ratio
gives the average rate of change of the function in the area
. Then the limit of this ratio at
(if it exists and is finite) will be the rate of change of the function at the point
in the direction of the vector. He is called derivative of a function at a point in the direction of the vector and denote
or
.

In addition to the rate of change of the function, it also allows you to determine the nature of the change in the function at a point in the direction of the vector (increasing or decreasing):

These statements are proven in the same way as similar ones for a function of one variable.

Note that partial derivatives of a function are a special case of directional derivative. Namely,
this is the derivative of the function in the direction of the vector (axis direction
), is the derivative of the function in the direction of the vector (axis direction
).

Let us assume that the function is differentiable at the point. Then

Where – infinitesimal at
.