A proof of the derivative formula is given complex function. 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.
ContentSee 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
.
) we have already repeatedly encountered partial derivatives of complex functions like 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 to specific 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, 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 this case can be performed 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 repeatedly urged not to “cling to letters” as a life preserver, 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’ll give you general formula for 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 anything 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 the examples are taken from real works, 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 form and carefully modify the 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: 
Differentiating complex functions
Let for the function n- variables' arguments are also functions of variables:
The following theorem on differentiation of a complex function is valid.
Theorem 8. If functions are differentiable at a point and a function is differentiable at the corresponding point, where , . Then the complex function is differentiable at the point, and the partial derivatives are determined by the formulas
where partial derivatives are calculated at the point and are calculated at the point .
Let us prove this theorem for a function of two variables. Let , a .
Let there be arbitrary increments of arguments at the point . They correspond to increments of functions and at the point . Increments and correspond to the increment of the function at the point. Since it is differentiable at the point, its increment can be written in the form
where and are calculated at the point , at and . Due to the differentiability of functions and at the point , we obtain
where is calculated at the point ; .
Let's substitute (14) into (13) and rearrange the terms
Note that at , since and tend to zero at . This follows from the fact that infinitesimals at and . But the functions are also differentiable, and, therefore, continuous at the point. Therefore, if and, then. Then and at .
Since partial derivatives are calculated at the point, we obtain
Let's denote
and this means that it is differentiable with respect to the variables and , and
Consequence. If , and , , i.e. , then the derivative with respect to the variable t calculated by the formula
If , then
The last expression is called the total derivative formula for a function of many variables.
Examples. 1) Find the total derivative of the function, where , .
Solution.
2) Find the complete derivative of the function if , .
Solution.
Using the rules of differentiation of a complex function, we obtain one important property of the differential of a function of several variables.
If independent variable functions, then the differential by definition is equal to:
Let now the arguments be differentiable functions at some point of the function with respect to the variables , and the function be differentiable with respect to the variables , . Then can be considered as a complex function of the variables , . According to the previous theorem, it is differentiable and the relation holds
where is determined by formulas (12). Let us substitute (12) into (17) and, collecting the coefficients for , we obtain
Since the coefficient of the derivative is equal to the differential of the function, we again obtained formula (16) for the differential of a complex function.
Thus, the first differential formula does not depend on whether its arguments are functions or whether they are independent. This property is called invariance of the form of the first differential.
Taylor formula (29) can also be written as
We will carry out the proof for a function of two variables or .
First let's look at a function of one variable. Let once be differentiable in a neighborhood of the point . Taylor's formula for a function of one variable with a remainder term in Lagrange's formula has
Since is an independent variable, then . By definition of the differential of a function of one variable
If we denote , then (31) can be written as
Let's consider some neighborhood of a point and an arbitrary point in it and connect the points with a straight line segment. It is clear that the coordinates and points of this line are linear functions of the parameter.
On a straight line segment, the function is a complex function of the parameter, because . Moreover, it is once differentiable with respect to and the Taylor formula (32) is valid for, where , i.e.
The differentials in formula (32) are differentials of a complex function, where , , , i.e.
Substituting (33) into (32) and taking into account that , we get
The last term in (34) is called the remainder term of the Taylor formula in Lagrange form
Without proof, we note that if, under the conditions of the theorem, the function is differentiable at the point m times, then the remainder term can be written as Peano form:
Chapter 7. Functions of Several Variables
7.1. Space Rn. Sets in linear space.
A set whose elements are all possible ordered sets of n real numbers, denoted and called n-dimensional arithmetic space, and the number n called dimension of space. An element of a set is called a point in space, or a vector, and the numbers coordinates this point. Point =(0, 0, …0) is called zero or origin.
Space is a set of real numbers, i.e. – number line; and – are a two-dimensional coordinate geometric plane and a three-dimensional coordinate geometric space, respectively. Vectors , , …, are called unit basis.
For two elements, a set, the concepts of the sum of elements and the product of an element by a real number are defined:
It is obvious that, due to this definition and the properties of real numbers, the equalities are true:
According to these properties, space is also called linear (vector) space.
In linear space it is defined dot product elements and as a real number, calculated according to the following rule:
The number is called vector length or the norm. Vectors are called orthogonal, If . Magnitude
, )= │ - │ =
called distance between elements And .
If and are nonzero vectors, then angle between them is called an angle such that
It is easy to verify that for any elements and a real number, the scalar product is satisfied:
A linear space with a scalar product defined in it by formula (1) is called Euclidean space.
Let the point and . The set of all points for which the inequalities hold
called n -measuring cube with an edge and center at point . For example, a two-dimensional cube is a square with a side centered at point .
The set of points satisfying the inequality is called n-dimensional ball radius centered at point , which is also called
- neighborhood of the point in and denote ,
Thus, a one-dimensional ball is an interval of length . 2D ball
there is a circle for which the inequality holds
Definition 1. The set is called limited, if exists
n- a dimensional ball containing this set.
Definition 2. A function defined on the set of natural numbers and taking values belonging to is called sequence in space and is denoted by where .
Definition 3. The point is called limit of the sequence, if for an arbitrary positive number there is a natural number such that the inequality holds for any number.
Symbolically, this definition is written as follows:
Designation:
From Definition 3 it follows that , for . This sequence is called convergent To .
If a sequence does not converge to any point, then it is called divergent.
Theorem 1. In order for the sequence to converge to a point, it is necessary and sufficient that for any number , i.e. to sequence i- x coordinates of the points converged to i-th coordinate of the point.
□ The proof follows from the inequalities
The sequence is called limited, if the set of its values is limited, i.e.
Like a number sequence, a convergent sequence of points is bounded and has a single limit.
Definition 4. The sequence is called fundamental(Cauchy sequence), if for any positive number it is possible to specify a natural number such that for arbitrary natural numbers and , large , it holds, i.e.
Theorem 2(Cauchy criterion). In order for a sequence to be convergent, it is necessary and sufficient that it be fundamental.
□ Necessity. Let it converge to the point . Then we get a sequence converging to . . . , ..., X is called region V . If X - region, then its closure is called closed area.
Sets X And Y called separable, if none of them contains touch points of the other.
Many X called related, if it cannot be represented as a union of two separable sets.
Many X called convex , if any two of its points can be connected by a segment entirely belonging to this set.
Example. Based on the definitions formulated above, it can be argued that
– a connected, linearly connected, open, non-convex set, is a region.
– connected, linearly connected, unopened, non-convex set, not a region.
– unconnected, not linearly connected, open, non-convex set, not a region.
– unconnected, not linearly connected, open set, not a region.
– connected, linearly connected, open set, is a region.
Let the function z - /(x, y) be defined in some domain D on the xOy plane. Let's take an internal point (x, y) from the area D and give x an increment Ax such that the point (x + Ax, y) 6 D (Fig. 9). Let's call the quantity the partial increment of the function z with respect to x. Let's make a relation: For a given point (x, y), this relation is a function of Definition. If for Ax -* 0 the relation ^ has a finite limit, then this limit is called the partial derivative of the function z = /(x, y) with respect to the independent variable x at the point (x, y) and is denoted by the symbol jfc (or /i(x, jj ), or z"x(x, In the same way, by definition, or, which is the same thing, Similarly, If u is a function of n independent variables, then Noticing that Arz is calculated with a constant value of the variable y, and Atz - with a constant value of the variable x, definitions of partial derivatives can be formulated as follows: Partial derivatives Geometric meaning of partial derivatives of a function of two variables Differentiability of a function of several variables Necessary conditions for the differentiability of a function Sufficient conditions for the differentiability of functions of several variables Total differential Partial differentials Derivatives of a complex function of the partial derivative with respect to x of the function z = /(x, y. ) is the ordinary derivative of this function with respect to x, calculated under the assumption that y is a constant; the partial derivative with respect to y of the function z - /(x, y) is its derivative with respect to y, calculated under the assumption that x is a constant. It follows that the rules for calculating partial derivatives coincide with the rules proven for a function of one variable. Example. Find the partial derivatives of the function 4 We have Substitutions*. The existence of the function r = f(x, y) at a given point of partial derivatives with respect to all arguments does not imply the continuity of the function at this point. Thus, the function is not continuous at the point 0(0,0). However, at this point the specified function has partial derivatives with respect to x and y. This follows from the fact that /(x, 0) = 0 and /(0, y) = 0 and therefore the Geometric meaning of the partial derivatives of a function of two variables Let the surface S in three-dimensional space be defined by the equation where f(x, y) is the function continuous in some domain D and having there partial derivatives with respect to x and y. Let us find out the geometric meaning of these derivatives at the point Mo(xo,yo) 6 D, which corresponds to the point f(x0)yo) on the surface z = f(x)y). When finding the partial derivative of the point M0, we assume that z is only a function of the argument x, while the argument y retains a constant value y = y0, i.e. The function fi(x) is geometrically represented by the curve L along which the surface S is intersected by the plane y = at o. Due to the geometric meaning of the derivative of a function of one variable, f\(xo) = tan a, where a is the angle formed by the tangent to the line L at the point JV0 with the Ox axis (Fig. 10). But so Thus, the partial derivative ($|) is equal to the tangent of the angle a between the Ox axis and the tangent at point N0 to the curve obtained in the section of the surface z = /(x, y) by the y plane. Similarly, we obtain that §6. Differentiability of a function of several variables Let the function z = /(x, y) be defined in some domain D on the xOy plane. Let's take a point (x, y) € D and give the selected values of x and y any increments Ax and Dy, but such that the point. Definition. A function r = /(x, y) is called differentiable * point (x, y) € 2E if the complete increment of this function, corresponding to the increments of Dx, Dy arguments, can be represented in the form where A and B do not depend on Dx and Dy ( but generally depend on x and y), and a(Dx, Dy) and /?(Dx, Dy) tend to zero as Dx and Dy tend to zero. . If the function z = /(x, y) is differentiable at the point (x, y), then the part A Dx 4- VDy of the increment of the function, linear with respect to Dx and Dy, is called the total differential of this function at the point (x, y) and is denoted by the symbol dz: In this way, Example. Let r = x2 + y2. At any point (r,y) and for any Dx and Du we have Here. now that a and /3 tend to zero as Dx and Dy tend to zero. According to the definition, this function is differentiable at any point in the xOy plane. At the same time, we note that in our reasoning we did not formally exclude the case when the increments of Dx, Du separately, or even both are equal to zero at once. Formula (1) can be written more compactly if we introduce the expression (distance between points (Using it, we can write Denoting the expression in parentheses by e, we have where c depends on J, Du and tends to zero if J 0 and DN 0, or, in short, if p 0. Formula (1), expressing the condition for the differentiability of the function z = f(xt y) at the point (x, y), can now be written in the form So, in the above example 6.1. Necessary conditions. differentiable™ function Theorem 4. If a function r = /(x, y) is differentiable at some point, then it is continuous at this point. 4 If at the point (x, y) the function r = /(x, y) is differentiable, then complete. the increment of the function i at this point, corresponding to the increments J and Dy of the arguments, can be represented in the form (the quantities A, B for a given point are constant; , from which it follows that the latter means that at the point (x, y) the function r /(x, y) is continuous. Theorem! b. If the function r = /(x, y) is differentiable at a given point, there are partial derivatives $§ and at this point. Let the function z = /(x, y) be differentiable at the point (x, y). Then the increment Dg of this function, corresponding to the increments Dx, Ay of the arguments, can be represented in the form (1). Taking in equality (1) Dx Φ 0, Dy = 0, we obtain whence Since on the right side of the last equality the value A does not depend on, This means that at the point (x, y) there is a partial derivative of the function r = /(x, y) in x, and by similar reasoning we are convinced (x, there is a partial derivative of the function zy, and from the theorem it follows that We emphasize that Theorem 5 states the existence of partial derivatives only at the point (x, y), but says nothing about their continuity at this point, as well as their behavior in the neighborhood of the point (x, y). derivative /"(x) at the point x0. In the case when the function depends on several variables, the situation is much more complicated: there are no necessary and sufficient conditions for differentiability for the function z = /(x, y) of two independent variables x, y; there are only separately necessary conditions (see above) and separately - sufficient. These sufficient conditions for differentiability of functions of several variables are expressed by the following theorem. Theorem c. If a function has partial derivatives /ε and f"v in some neighborhood of thin (xo, V0) and if these derivatives are continuous at the point (xo, V0), then the function z = f(x, y) is differentiable at the point (x- Example. Let's consider the function Partial derivatives Geometric meaning of partial derivatives of a function of two variables Differentiability of a function of several variables Sufficient conditions for the differentiability of functions of several variables Total differential Derivatives of a complex function It is defined everywhere Based on the definition of partial derivatives, we have For oschdrlm* differentiable. ™ of this function at the point 0(0,0) we find and the increment of this point For the differentiability of the function /(x,y) = at the point 0(0,0) it is necessary that the function e(Dx, Dy) be completely small at Dx 0 and Ду 0. Let us set D0. Then from formula (1) we have Therefore the function f(x,y) = is not differentiable at the point 0(0,0), although it has fa and f"r at this point. The result obtained is explained by the fact that the derivatives f"z and f"t are discontinuous at the point §7. Full differential. Partial differentials If the function z - f(z> y) is differentiable, then its total differential dz is equal to Noticing that A = B = u, we write formula (1) in the following form. We extend the concept of the differential of a function to independent variables, setting the differentials of independent variables equal their increments: After this, the formula for the total differential of the function is taken as Example. Let i - 1l(x + y2). Then Similarly, if u =) is a differentiable function of n independent variables, then the Expression is called the post differential of the function z = f(x, y) with respect to the variable x; the expression is called the partial differential of the function z = /(x, y) of the variable y. From formulas (3), (4) and (5) it follows that the total differential of a function is the sum of its partial differentials: Note that the total increment Az of the function z = /(x, y), generally speaking, is not equal to the sum of the partial increments. If at the point (i, y) the function z = /(x, y) is differentiable and the differential dz Φ 0 at this point, then its total increment differs from its linear part only by the sum of the last terms aAx 4- /?DE, which at Ax 0 and Ау -» О are infinitesimals of a higher order than the terms of the linear part. Therefore, when dz Ф 0, the linear part of the increment of the differentiable function is called the main part of the increment of the function and an approximate formula is used, which will be the more accurate, the smaller in absolute value the increments of the arguments are. §8. Derivatives of a complex function 1. Let the function be defined in some domain D on the xOy plane, and each of the variables x, y in turn is a function of the argument t: We will assume that when t changes in the interval (the corresponding points (x, y) do not leave outside the region D. If we substitute values into the function z = / (x, y), we obtain a complex function of one variable t and for appropriate values the function / (x, y) is differentiable, then the complex function has a derivative at the point t. M Let us give t an increment Dt. Then x and y will receive some increments Ax and Dy. As a result of this, for (J)2 + (Dy)2 Ф 0, the function z will also receive some increment Dt, which, due to the differentiability of the function z = /(x). , y) at the point (x, y) can be represented in the form where a) tend to zero as Ax and Du tend to zero. Let us define a and /3 for Ax = Ay = 0 by setting a Then a(will be continuous for J = Dn = 0. Consider the relation We have In each term^ in the right side of (2) both factors have limits at indeed, partial derivatives and ^ for a given are constant, by condition there are limits from the existence of derivatives ^ and at the point £ the functions x = y(t) and y = are continuous at this point; therefore, as At 0, both J and Dy tend to zero, which in turn entails is a tendency to zero a(Dx, Dy) and P(Ax, Ay). Thus, the right-hand side of equality (2) at 0 has a limit equal to. Hence, at At 0 there is also a limit of the left-hand side of (2), i.e. e. there is an equal one. Passing in equality (2) to the limit as At -» 0, we obtain the required formula. In the special case, when, therefore, z is a complex function of x, we obtain In formula (5) there is a partial derivative funadiig = /(x , y) by x, when calculating which in the expression /(x, y) the argument y is taken as a constant. And there is a complete derivative of the function z with respect to the independent variable x, when calculating which y in the expression /(x, y) is no longer taken as a constant, but is in turn considered a function of x: y = tp(x)t and therefore the dependence of z on is taken into account completely. Example. Find and jg if 2. Let us now consider the differentiation of a complex function of several variables. Let where in turn so that Assume that at the point (() there are continuous partial derivatives u, 3? and at the corresponding point (x, y), where the Function f(x, y) is differentiable. Let us show that under these conditions the complex function z = z(() y) at point t7) has derivatives and π, and we will find expressions for these derivatives. Note that this case does not differ significantly from the one already studied. Indeed, when differentiating z with respect to £, the second independent variable rj is taken as a constant, as a result of which x and y in this operation become functions of one variable x" = c), y = c) and the question of the derivative ζ is solved in exactly the same way as the question of derivative when deriving formula (3). Using formula (3) and formally replacing the derivatives § and ^ in it with the derivatives u and, respectively, we obtain Similarly, we find Example. Find the partial derivatives ^ and ^ of the function r = x2 y - x - y = If a complex function “ is given by formulas so that then, when the appropriate conditions are met, we have In the special case when And = where Partial derivatives Geometric meaning of partial derivatives of a function of two variables Differentiability of a function of several variables Necessary conditions for the differentiability of a function Sufficient conditions for the differentiability of functions of several variables Total differential. Derivatives of a complex function we have Here m is the total partial derivative of the function and with respect to the independent variable x, taking into account the complete dependence of and on x, including through z = z(x,y),a ^ -partial derivative of the function u = /(r, y, d) by x, when calculating k
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 the 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.
