Question: Derivatives of complex functions of several variables (consider cases of one and several independent variables). Examples. Derivative and differential of a complex function of several variables Complex function of two variables

) we have already repeatedly encountered partial derivatives complex functions similar and more difficult examples. So what else can you talk about?! ...And everything is like in life - there is no complexity that cannot be complicated =) But mathematics is what mathematics is for, to fit the diversity of our world into a strict framework. And sometimes this can be done with one single sentence:

In general, the complex function has the form , Where, at least one of letters represents function, which may depend on arbitrary number of variables.

The minimum and simplest option is the long-familiar complex function of one variable, whose derivative we learned how to find last semester. You also have the skills to differentiate functions (take a look at the same functions ) .

Thus, now we will be interested in just the case. Due to the great variety of complex functions, the general formulas for their derivatives are very cumbersome and difficult to digest. In this regard, I will limit myself concrete examples, from which you can understand the general principle of finding these derivatives:

Example 1

Given a complex function where . Required:
1) find its derivative and write down the 1st order total differential;
2) calculate the value of the derivative at .

Solution: First, let's look at the function itself. We are offered a function depending on and , which in turn are functions one variable:

Secondly, let’s pay close attention to the task itself - we are required to find derivative, that is, we are not talking about partial derivatives at all, which we are used to finding! Since the function actually depends on only one variable, then the word “derivative” means total derivative. How to find her?

The first thing that comes to mind is direct substitution and further differentiation. Let's substitute to function:
, after which there are no problems with the desired derivative:

And, accordingly, the total differential:

This solution is mathematically correct, but a small nuance is that when the problem is formulated the way it is formulated, no one expects such barbarism from you =) But seriously, you can really find fault here. Imagine that a function describes the flight of a bumblebee, and the nested functions change depending on the temperature. Performing a direct substitution , we only get private information, which characterizes flight, say, only in hot weather. Moreover, if a person who is not knowledgeable about bumblebees is presented with the finished result and even told what this function is, then he will never learn anything about the fundamental law of flight!

So, completely unexpectedly, our buzzing brother helped us understand the meaning and importance of the universal formula:

Get used to the “two-story” notation for derivatives - in the task under consideration, they are the ones in use. In this case, one should be very neat in the entry: derivatives with direct symbols “de” are complete derivatives, and derivatives with rounded icons are partial derivatives. Let's start with the last ones:

Well, with the “tails” everything is generally elementary:

Let's substitute the found derivatives into our formula:

When a function is initially proposed in an intricate way, it will be logical (and this is explained above!) leave the results as they are:

At the same time, in “sophisticated” answers it is better to refrain from even minimal simplifications (here, for example, it begs to be removed 3 minuses)- and you have less work, and your furry friend is happy to review the task easier.

However, a rough check will not be superfluous. Let's substitute into the found derivative and carry out simplifications:

(at the last step we used trigonometric formulas , )

As a result, the same result was obtained as with the “barbaric” solution method.

Let's calculate the derivative at the point. First it is convenient to find out the “transit” values (function values ) :

Now we draw up the final calculations, which in in this case can be done in different ways. I use an interesting technique in which the 3rd and 4th “floors” are simplified not according to the usual rules, but are transformed as the quotient of two numbers:

And, of course, it’s a sin not to check using a more compact notation :

Answer:

It happens that the problem is proposed in a “semi-general” form:

"Find the derivative of the function where »

That is, the “main” function is not given, but its “inserts” are quite specific. The answer should be given in the same style:

Moreover, the condition can be slightly encrypted:

"Find the derivative of the function »

In this case you need on one's own designate nested functions with some suitable letters, for example, through and use the same formula:

By the way, about letter designations. I have already repeatedly urged not to “cling to letters” as to lifebuoy, and now this is especially relevant! Analyzing various sources on the topic, I generally got the impression that the authors “went crazy” and began to mercilessly throw students into the stormy abyss of mathematics =) So forgive me :))

Example 2

Find the derivative of a function , If

Other designations should not be confusing! Every time you come across a task like this, you need to answer two simple questions:

1) What does the “main” function depend on? In this case, the function “zet” depends on two functions (“y” and “ve”).

2) What variables do nested functions depend on? In this case, both “inserts” depend only on the “X”.

So you shouldn't have any difficulty adapting the formula to this task!

A short solution and answer at the end of the lesson.

Additional examples of the first type can be found in Ryabushko's problem book (IDZ 10.1), well, we are heading for function of three variables:

Example 3

Given a function where .
Calculate derivative at point

The formula for the derivative of a complex function, as many guess, has a related form:

Decide once you guessed it =)

Just in case, I will give a general formula for the function:
, although in practice you are unlikely to see anything longer than Example 3.

In addition, sometimes it is necessary to differentiate a “truncated” version - as a rule, a function of the form or. I leave this question for you to study on your own - come up with some simple examples, think, experiment and derive shortened formulas for derivatives.

If anything is still unclear, please slowly re-read and comprehend the first part of the lesson, because now the task will become more complicated:

Example 4

Find the partial derivatives of a complex function, where

Solution: this function has the form , and after direct substitution and we get the usual function of two variables:

But such fear is not only not accepted, but one no longer wants to differentiate =) Therefore, we will use ready-made formulas. To help you quickly grasp the pattern, I will make some notes:

Look carefully at the picture from top to bottom and left to right….

First, let's find the partial derivatives of the “main” function:

Now we find the “X” derivatives of the “liners”:

and write down the final “X” derivative:

Similarly with the “game”:

And

You can stick to another style - find all the “tails” at once and then write down both derivatives.

Answer:

About substitution somehow I don’t think about it at all =) =), but you can tweak the results a little. Although, again, why? – only make it more difficult for the teacher to check.

If necessary, then full differential here it is written according to the usual formula, and, by the way, it is at this step that light cosmetics become appropriate:

This is... ...a coffin on wheels.

Due to the popularity of the type of complex function under consideration, there are a couple of tasks for independent solution. A simpler example in a “semi-general” form is for understanding the formula itself;-):

Example 5

Find the partial derivatives of the function, where

And more complicated - with the inclusion of differentiation techniques:

Example 6

Find the complete differential of a function , Where

No, I’m not trying to “send you to the bottom” at all - all examples are taken from real work, and “on the high seas” you can come across any letters. In any case, you will need to analyze the function (answering 2 questions – see above), present it in general view and carefully modify partial derivative formulas. You may be a little confused now, but you will understand the very principle of their construction! Because the real challenges are just beginning :)))

Example 7

Find partial derivatives and create the complete differential of a complex function
, Where

Solution: the “main” function has the form and still depends on two variables – “x” and “y”. But compared to Example 4, another nested function has been added, and therefore the partial derivative formulas are also lengthened. As in that example, for a better visualization of the pattern, I will highlight the “main” partial derivatives in different colors:

And again, carefully study the record from top to bottom and from left to right.

Since the problem is formulated in a “semi-general” form, all our work is essentially limited to finding partial derivatives of embedded functions:

A first grader can handle:

And even the full differential turned out quite nice:

I deliberately did not offer you any specific function - so that unnecessary clutter would not interfere with a good understanding of schematic diagram tasks.

Answer:

Quite often you can find “mixed-sized” investments, for example:

Here the “main” function, although it has the form , still depends on both “x” and “y”. Therefore, the same formulas work - just some partial derivatives will be equal to zero. Moreover, this is also true for functions like , in which each “liner” depends on one variable.

A similar situation occurs in the final two examples of the lesson:

Example 8

Find the total differential of a complex function at a point

Solution: the condition is formulated in a “budgetary” way, and we must label the nested functions ourselves. I think this is a good option:

The “inserts” contain ( ATTENTION!) THREE letters are the good old “X-Y-Z”, which means that the “main” function actually depends on three variables. It can be formally rewritten as , and the partial derivatives in this case are determined by the following formulas:

We scan, we delve into, we capture….

In our task:

Theorem.Let u = f (x, y) is given in domain D and let x = x(t) And y = y(t) identified in the area , and when , then x and y belong to region D. Let the function u be differentiable at the point M 0 (x 0 ,y 0 ,z 0), and functions x(t) and at(t) differentiable at the corresponding point t 0 , then the complex function u = f[x(t),y(t)]=F (t)differentiable at point t 0 and the equality holds:

Proof. Since u is differentiable by condition at the point ( x 0 , y 0), then its total increment is represented as

Dividing this ratio by , we get:

Let's go to the limit at and get the formula

Note 1. If u= u(x, y) And x= x, y= y(x), then the total derivative of the function u by variable X

or .

The last equality can be used to prove the rule for differentiating a function of one variable, given implicitly in the form F(x, y) = 0, where y= y(x) (see topic No. 3 and example 14).

We have: . From here . (6.1)

Let's return to example 14 of topic No. 3:

;

As you can see, the answers coincided.

Note 2. Let u = f (x, y), Where X= X(t ,v), at= at(t ,v). Then u is ultimately a complex function of two variables t And v. If now the function u is differentiable at the point M 0 (x 0 , y 0), and the functions X And at are differentiable at the corresponding point ( t 0 , v 0), then we can talk about partial derivatives with respect to t And v from a complex function at the point ( t 0 , v 0). But if we are talking about the partial derivative with respect to t at a specified point, then the second variable v is considered constant and equal to v 0 . Consequently, we are talking about the derivative only of a complex function with respect to t and, therefore, we can use the derived formula. Thus, we get.

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.

Content

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
.