The concept of a complex function of many variables. Function of two variables. Domain and level lines
) 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 case the complex function looks like 
, 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, 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: 
(on last step 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 let's formalize 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 would be 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, oh letter designations. I have repeatedly urged not to “cling to letters” as if they were 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!
Quick Solution and the 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, just on this step Light cosmetics become appropriate: 
This is... ...a coffin on wheels.
Due to the popularity of the type of complex function under consideration, a couple of tasks for independent decision. 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 construct 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 vision of the pattern, I will highlight the “main” partial derivatives 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: 
Partial derivatives of a function of three variables
Let's continue everyone's favorite topic mathematical analysis– derivatives. In this article we will learn how to find partial derivatives of a function of three variables: first derivatives and second derivatives. What do you need to know and be able to do to master the material? Believe it or not, firstly, you need to be able to find “ordinary” derivatives of a function of one variable - at a high or at least average level. If it’s really difficult with them, then start with a lesson How to find the derivative? Secondly, it is very important to read the article and comprehend and solve, if not all, then most of the examples. If this has already been done, then walk with me with a confident gait, it will be interesting, you will even enjoy it!
Methods and principles of finding partial derivatives of a function of three variables are actually very similar to partial derivatives of functions of two variables. A function of two variables, let me remind you, has the form , where “x” and “y” are independent variables. Geometrically, a function of two variables is usually some surface in our three-dimensional space.
A function of three variables has the form , and the variables are called independent variables or arguments, the variable is called dependent variable or function. For example: – function of three variables
And now a little about science fiction films and aliens. You can often hear about four-dimensional, five-dimensional, ten-dimensional, etc. spaces. Nonsense or not?
After all, a function of three variables implies a four-dimensional space
(and indeed, there are three variables + the function itself). The graph of a function of three variables is the so-called hypersurface. It is impossible to imagine it, since we live in three-dimensional space (length/width/height). So that you don't get bored with me, I offer a quiz. I will ask a few questions, and anyone interested can try to answer them:
– Is there a fourth, fifth, etc. in the world? measurements in the sense of the philistine understanding of space (length/width/height)?
– Is it possible to build a four-dimensional, five-dimensional, etc. space in the broad sense of the word? That is, give an example of such a space in our lives.
– Is it possible to travel into the past?
– Is it possible to travel to the future?
– Do aliens exist?
For any question you can choose one of four answers:
Yes / No (science prohibits this) / Science does not prohibit this / I don’t know
Whoever answers all the questions correctly is most likely to have some item ;-)
I will gradually give out answers to questions as the lesson progresses, don’t miss the examples!
Actually, they flew. And right away good news: for a function of three variables, the rules of differentiation and the table of derivatives are valid. That's why you need to be good at dealing with "ordinary" derivatives of functions one variable. There are very few differences!
Example 1
Solution: It’s not hard to guess - for a function there are three variables three first order partial derivatives, which are denoted as follows:
Or – partial derivative with respect to “x”;
or – partial derivative with respect to “y”;
or – partial derivative with respect to “zet”.
The symbol with a prime is more common, but the compilers of collections and training manuals really like to use cumbersome symbols for problems - so don’t get lost! Perhaps not everyone knows how to correctly read these “dreaded fractions” out loud. Example: should be read as follows: “de u po de x.”
Let's start with the derivative with respect to “X”: . When we find the partial derivative with respect to, then the variables andare considered constants (constant numbers). And the derivative of any constant, oh, grace, is equal to zero:
Immediately pay attention to the subscript - no one forbids you to mark that they are constants. It’s even more convenient; I recommend that beginners use just such a notation, there’s less risk of getting confused.
(1) We use the linearity properties of the derivative, in particular, we take all constants outside the sign of the derivative. Please note that in the second term there is no need to remove the constant: since “Y” is a constant, then it is also a constant. In the term, the “ordinary” constant 8 and the constant “zet” are taken out of the derivative sign.
(2) We find the simplest derivatives, not forgetting that they are constants. Next we comb the answer.
Partial derivative. When we find the partial derivative with respect to “y”, then the variables andare considered constants:
(1) We use the properties of linearity. And again, note that the terms , are constants, which means that nothing needs to be taken out of the derivative sign.
(2) Find derivatives, not forgetting that they are constants. Next we simplify the answer.
And finally, the partial derivative. When we find the partial derivative with respect to “zet”, then the variables andare considered constants:
General rule obvious and unpretentious: When we find the partial derivative for any reason independent variable, then two others independent variables are considered constants.
When completing these tasks, you should be extremely careful, in particular, You can't lose subscripts(which indicate which variable is used to differentiate). Losing the index would be a GROSS MISCONDITION. Hmmm... It’s funny if after such intimidation I miss them somewhere)
Example 2
Find first order partial derivatives of a function of three variables
This is an example for you to solve on your own. Complete solution and the answer at the end of the lesson.
The two examples considered are quite simple and, after solving several similar problems, even a teapot will get used to dealing with them orally.
To relieve the stress, let's return to the first question of the quiz: Is there a fourth, fifth, etc. in the world? measurements in the sense of the philistine understanding of space (length/width/height)?
Correct answer: Science does not prohibit this. All fundamental mathematical axiomatics, theorems, mathematical apparatus are beautiful and consistent work in space of any dimension. It is possible that somewhere in the Universe there exist hypersurfaces beyond the control of our minds, for example, a four-dimensional hypersurface, which is defined by a function of three variables. Or maybe the hypersurfaces are next to us or even we are right in them, it’s just that our vision, other senses, and consciousness are capable of perceiving and comprehending only three dimensions.
Let's return to the examples. Yes, if anyone is heavily loaded with the quiz, it’s better to read the answers to the following questions after you learn how to find the partial derivatives of a function of three variables, otherwise I’ll blow your mind during the course of the article =)
In addition to the simplest Examples 1 and 2, in practice there are tasks that can be called a small puzzle. Such examples, to my chagrin, fell out of sight when I created the lesson Partial derivatives of a function of two variables. Let's catch up:
Example 3
Solution: it seems that “everything is simple” here, but the first impression is deceptive. When finding partial derivatives, many will guess at the tea leaves and make mistakes.
Let's look at the example consistently, clearly and understandably.
Let's start with the partial derivative with respect to "x". When we find the partial derivative with respect to “x”, the variables are considered constants. Therefore, the exponent of our function is also a constant. For dummies, I recommend the following solution: in the draft, change the constant to a specific positive integer, for example, “five”. The result is a function of one variable:
or you can also write it like this:
This power function with a complex base (sine). By :
Now we remember that, thus:
At the final stage, of course, the solution should be written like this:
We find the partial derivative with respect to the “y”, they are considered constants. If “x” is a constant, then it is also a constant. On the draft we do the same trick: replace, for example, with 3, “Z” - replace with the same “five”. The result is again a function of one variable:
This indicative function with a complex exponent. By rule of differentiation of complex functions:
Now let's remember our replacement:
Thus:
On the final page, of course, the design should look nice:
And the mirror case with the partial derivative with respect to “zet” ( – constants):
With some experience, the analysis can be carried out mentally.
Let's complete the second part of the task - compose a first-order differential. It is very simple, by analogy with a function of two variables, a first order differential is written using the formula:
In this case:
And that’s business. I note that in practical problems A complete differential of the 1st order for a function of three variables requires much less frequent compilation than for a function of two variables.
A funny example for solving it yourself:
Example 4
Find first-order partial derivatives of a function of three variables and construct a first-order complete differential 
Full solution and answer at the end of the lesson. If you encounter any difficulties, use the discussed “Chaynikovsky” algorithm, it is guaranteed to help. And further helpful advice – do not hurry. Even I cannot solve such examples quickly.
Let's digress and look at the second question: Is it possible to build a four-dimensional, five-dimensional, etc. space in the broad sense of the word? That is, give an example of such a space in our lives.
Correct answer: Yes. Moreover, it is very easy. For example, we add a fourth dimension to the length/width/height - time. The popular four-dimensional space-time and the well-known theory of relativity, carefully compiled by Einstein based on the works of Lobachevsky, Poincaré, Lorentz and Minkowski. Not everyone knows either. Why did he get the Nobel Prize? There was a serious scandal in the scientific world, and the Nobel Committee formulated the merit of the C student Einstein approximately as follows: “For his overall contribution to the development of physics.” Further, as they say, promotion and PR.
It is easy to add a fifth dimension to the considered four-dimensional space, for example: atmospheric pressure. And so on, so on, so on, as many dimensions as you specify in your model - that’s how many there will be. IN in a broad sense words, we live in a multidimensional space.
Let's look at a couple more typical tasks:
Example 5
Solution: A task in this formulation is often found in practice and involves performing the following two actions:
– you need to find first-order partial derivatives;
– you need to calculate the values of the 1st order partial derivatives at the point.
We decide:
(1) Before us is a complex function, and in the first step we should take the derivative of the arctangent. In this case, we, in fact, calmly use the tabular formula for the derivative of the arctangent 
. By rule of differentiation of complex functions the result must be multiplied by the derivative internal function(attachments): .
(2) We use the properties of linearity.
(3) And we take the remaining derivatives, not forgetting that they are constants.
According to the task conditions, it is necessary to find the value of the found partial derivative 
at point . Let's substitute the coordinates of the point into the found derivative:
Advantage of this assignment is the fact that other partial derivatives follow a very similar pattern:
As you can see, the solution template is almost the same.
Let's calculate the value of the found partial derivative 
at point:
And finally, the derivative with respect to “zet”:
Ready. The solution could have been formulated in another way: first find all three partial derivatives, and then calculate their values at the point. But, it seems to me, the above method is more convenient - just find the partial derivative, and immediately, without leaving the cash register, calculate its value at the point.
It is interesting to note that geometrically, a point is a very real point in our three-dimensional space. The values of the function, derivatives 
– is already the fourth dimension, and no one knows where it is geometrically located. As they say, no one crawled around the Universe with a tape measure or checked.
Since the philosophical topic is on the rise again, let's consider the third question: Is it possible to travel into the past?
Correct answer: No. Traveling into the past contradicts the second law of thermodynamics about irreversibility physical processes(entropy). So please don’t dive into a pool without water, the event can only be replayed in a video =) It’s not for nothing that folk wisdom came up with the opposite everyday law: “Measure twice, cut once.” Although, in fact, the sad thing is that time is unidirectional and irreversible, none of us will be younger tomorrow. And various science fiction films like “The Terminator” are complete nonsense from a scientific point of view. It is also absurd from a philosophical point of view when the Effect, returning to the past, can destroy its own Cause.
Example 6
Find first order partial derivatives at a point
Example 7
Find first order partial derivatives at a point 
These are two simple examples for you to solve on your own. Full solution and answer at the end of the lesson.
But don’t be upset about the second law of thermodynamics, now I’ll encourage everyone more complex examples:
Example 8
Find first order partial derivatives of a function of three variables 
Solution: Let's find the first order partial derivatives: 
(1) When starting to find the derivative, you should follow the same approach as for a function of one variable. We use the properties of linearity, in this case we take it out of the sign of the derivative constant.
(2) Under the derivative sign we have work two functions, each of which depends from our “live” variable “x”. Therefore, it is necessary to use the product differentiation rule 
.
(3) There are no difficulties with the derivative, but the derivative 
is the derivative of a complex function: first you need to find, essentially, a tabular logarithm and multiply it by the derivative of the embedding.
(4) I think everyone has already gotten used to the simplest examples like - here we only have a “live” one, the derivative of which is equal to
The case with the derivative with respect to the “y” is almost a mirror image; I will write it down briefly and without comment: 
It’s more interesting with the “zet” derivative, although it’s still almost the same: 
(1) We take the constants out of the sign of the derivative.
(2) Here again is the product of two functions, each of which depends from the “live” variable “zet”. In principle, you can use the formula for the derivative of a quotient, but it’s easier to go the other way - find the derivative of the product.
(3) The derivative is a tabular derivative. The second term contains the already familiar derivative of a complex function.
Example 9
Find first order partial derivatives of a function of three variables 
This is an example for you to solve on your own. Think about how to more rationally find this or that partial derivative. Full solution and answer at the end of the lesson.
Before moving on to the final examples of the lesson and looking at second order partial derivatives functions of three variables, I’ll cheer everyone up again with the fourth question:
Is it possible to travel to the future?
Correct answer: Science does not prohibit this. Paradoxically, there is no mathematical, physical, chemical or other natural science law that would prohibit travel to the future! Seems like nonsense? But almost everyone in life has had a premonition (and not supported by any logical arguments) that this or that event will happen. And it happened! Where did the information come from? From the future? Thus, science fiction films about traveling into the future, and, by the way, the predictions of all kinds of fortune tellers and psychics cannot be called such nonsense. At least science has not refuted this. Everything is possible! So, when I was in school, CDs and flat-panel monitors from films seemed incredible to me.
The famous comedy “Ivan Vasilyevich Changes His Profession” is half fiction (at most). No scientific law prohibited Ivan the Terrible from being in the future, but it is impossible for two peppers to end up in the past and perform the duties of a king.
Second order partial derivatives of a function of three variables
General principle finding second-order partial derivatives of a function of three variables is similar to the principle of finding second-order partial derivatives of a function of two variables. Therefore, if you have worked through the lesson well Partial derivatives of a function of two variables, then everything will be very simple.
In order to find second-order partial derivatives, you first need to find first-order partial derivatives or in another notation: .
There are nine second-order partial derivatives.
The first group is the second derivatives with respect to the same variables:
or – the second derivative with respect to “x”;
or – the second derivative with respect to “y”;
or – the second derivative with respect to “zet”.
The second group is mixed 2nd order partial derivatives, there are six of them:
or - mixed derivative of “x by y”;
or - mixed derivative “yy by x”;
or - mixed derivative "x by z";
or - mixed derivative of “Z by X”;
or - mixed derivative of “Igrek by Z”;
or - mixed derivative of “zet by igrek”.
Functions of many variables
§1. The concept of a function of many variables.
Let there be n variable quantities. Each set 
denotes a point n-
dimensional set 
(P-dimensional vector).
Let given sets 
And 
.
ODA. If each point 
matches the singular number 
, then we say that a numerical function is given n variables:
.
is called the domain of definition, 
- a set of values of a given function.
When n=2 instead 
usually write x,
y,
z. Then the function of two variables has the form:
z= f(x, y).
For example, 
- function of two variables;
- function of three variables;
Linear function n variables.
ODA. Function graph n variables are called n-
dimensional hypersurface in space 
, each point of which is specified by coordinates
For example, a graph of a function of two variables z=
f(x,
y)
is a surface in three-dimensional space, each point of which is specified by coordinates ( x,
y,
z)
, Where 
, And 
.
Since it is not possible to depict a graph of a function of three or more variables, we will mainly (for clarity) consider functions of two variables.
Plotting functions of two variables is quite easy challenging task. The construction of so-called level lines can provide significant assistance in solving this problem.
ODA. Level line of a function of two variables z= f(x, y) is called the set of points on the plane HOU, which are the projection of the section of the graph of the function by a plane parallel HOU. At each point on the level line the function has the same value. Level lines are described by the equation f(x, y)=c, Where With– a certain number. There are infinitely many level lines, and one of them can be drawn through each point of the definition domain.
ODA. Surface level function n variables y=
f
(
) is called a hypersurface in space 
, at each point of which the value of the function is constant and equal to a certain value With. Level surface equation: f
(
)=s.
Example. Graph a function of two variables
.
.
When c=1: 
;
.
With c=4: 
;
.
At c=9: 
;
.
Level lines are concentric circles, the radius of which decreases with increasing z.
§2. Limit and continuity of a function of several variables.
For functions of many variables, the same concepts are defined as for functions of one variable. For example, one can give definitions of the limit and continuity of a function.
ODA. The number A is called the limit of a function of two variables z=
f(x,
y)
at 
,
and is designated 
, if for any positive number 
there is a positive number 
, such that if the point 
away from the point 
less distance 
, then the quantities f(x,
y)
and A differ by less than 
.
ODA. If the function z=
f(x,
y)
defined at point 
and has a limit at this point equal to the value of the function 
, then it is called continuous at a given point.
.
§3. Partial derivatives of functions of several variables.
Consider a function of two variables 
.
Let's fix the value of one of its arguments, for example 
, putting 
. Then the function 
there is a function of one variable 
. Let it have a derivative at the point 
:
.
This derivative is called the partial derivative (or first order partial derivative) of the function 
By 
at the point 
and is designated: 
;
;
;
.
The difference is called the partial increment 
and is designated 
:
Taking into account the above notations, we can write
.
Defined similarly
.
Partial derivative functions of several variables in one of these variables is called the limit of the ratio of the partial increment of a function to the increment of the corresponding independent variable, when this increment tends to zero.
When finding the partial derivative with respect to any argument, the other arguments are considered constant. All rules and formulas for differentiating functions of one variable are valid for partial derivatives of functions of many variables.
Note that the partial derivatives of a function are functions of the same variables. These functions, in turn, can have partial derivatives, which are called second partial derivatives(or second order partial derivatives) of the original function.
For example, the function 
has four second-order partial derivatives, which are denoted as follows:
;
;
;
.
And 
- mixed partial derivatives.
Example. Find second order partial derivatives for a function
.
Solution.
,
.
,
.
, 
.
Exercise.
1. Find second-order partial derivatives for functions
,
;
2. For function 
prove that 
.
Full differential functions of many variables.
With simultaneous changes in values X And at function 
will change by an amount called the total increment of the function z
at the point 
. Just as in the case of a function of one variable, the problem arises of approximate replacement of the increment 
on linear function from 
And 
. The role of linear approximation is performed by full differential Features:
Second order total differential:
=
.
=
.
In general, a total differential P-th order has the form:
Directional derivative. Gradient.
Let the function z=
f(x,
y)
is defined in some neighborhood of the point M( x,
y) And 
- some direction specified by the unit vector 
. The coordinates of a unit vector are expressed through the cosines of the angles formed by the vector and the coordinate axes and called direction cosines:
,
.
When moving point M( x,
y) in this direction l
exactly 
function z will receive an increment
called the increment of the function in a given direction l.
E 
if MM 1 =∆ l, That
T
ABOUT
functions z=
f(x,
y)
towards
is called the limit of the ratio of the increment of the function in this direction to the magnitude of the displacement ∆ l
as the latter tends to zero:
The directional derivative characterizes the rate of change of a function in a given direction. It is obvious that partial derivatives 
And 
represent derivatives in directions parallel to the axes Ox
And Oy. It is easy to show that
Example. Calculate the derivative of a function 
at point (1;1) in the direction 
.
ODA. Gradient functions z= f(x, y) is a vector with coordinates equal to partial derivatives:
.
Consider the scalar product of vectors 
And 
:
It's easy to see that 
, i.e. the directional derivative is equal to the scalar product of the gradient and the unit direction vector 
.
Because the 
, then the scalar product is maximum when the vectors have the same directions. Thus, the gradient of a function at a point specifies the direction of the fastest increase in the function at this point, and the modulus of the gradient is equal to the maximum growth rate of the function.
Knowing the gradient of a function, one can locally construct function level lines.
Theorem. Let a differentiable function be given z=
f(x,
y)
and at the point 
the gradient of the function is not zero: 
. Then the gradient is perpendicular to the level line passing through the given point.
Thus, if, starting from a certain point, we construct the gradient of the function and a small part of the level line perpendicular to it at nearby points, then we can (with some error) construct level lines.
Local extremum of a function of two variables
Let the function 
defined and continuous in some neighborhood of the point 
.
ODA. Dot 
is called the local maximum point of the function 
, if there is such a neighborhood of the point 
, in which for any point 
inequality holds:
.
The concept of a local minimum is introduced similarly.
Theorem (necessary condition for local extremum).
In order for a differentiable function 
had a local extremum at the point 
, it is necessary that all its first-order partial derivatives at this point be equal to zero:
So, the points of possible presence of an extremum are those points at which the function is differentiable and its gradient is equal to 0: 
. As in the case of a function of one variable, such points are called stationary.
Functions of several variables. Geometric representation of a function of two variables. Level lines and surfaces. Limit and continuity of functions of several variables, their properties. Partial derivatives, their properties and geometric meaning.
Definition 1.1. Variable z(with change area Z) is called function of two independent variables x,y in abundance M, if each pair ( x,y) from many M z from Z.
Definition 1.2. A bunch of M, in which the variables are specified x,y, called domain of the function, and themselves x,y- her arguments.
Designations: z = f(x,y), z = z(x,y).
Comment. Since a couple of numbers ( x,y) can be considered the coordinates of a certain point on the plane, we will subsequently use the term “point” for a pair of arguments to a function of two variables, as well as for an ordered set of numbers that are arguments to a function of several variables.
Definition 1.3. . Variable z(with change area Z) is called function of several independent variables in abundance M, if each set of numbers from the set M according to some rule or law one is put into correspondence specific value z from Z. The concepts of arguments and domain are introduced in the same way as for a function of two variables.
Designations: z = f , z = z .
Geometric representation of a function of two variables.
Consider the function z = f(x,y), (1.1)
Defined in some area M on the O plane xy. Then the set of points in three-dimensional space with coordinates ( x,y,z), where , is the graph of a function of two variables. Since equation (1.1) defines a certain surface in three-dimensional space, it will be geometric image the function in question.
z = f(x,y)
Examples include the plane equations studied in the previous semester
z = ax + by + c
and second order surfaces:
z = x² + y² (paraboloid of revolution),
(cone), etc.
Comment. For a function of three or more variables we will use the term “surface in n-dimensional space,” although it is impossible to depict such a surface.
Level lines and surfaces.
For a function of two variables given by equation (1.1), we can consider a set of points ( x,y) O plane xy, for which z takes on the same constant value, that is z= const. These points form a line on the plane called level line.
Find the level lines for the surface z = 4 – x² - y². Their equations look like x² + y² = 4 – c(c=const) – equations of concentric circles with a center at the origin and with radii . For example, when With=0 we get a circle x² + y² = 4.
For a function of three variables u = u (x, y, z) the equation u(x, y, z) = c defines a surface in three-dimensional space, which is called level surface.
For function u = 3x + 5y – 7z–12 level surfaces will be a family of parallel planes given by equations 3 x + 5y – 7z –12 + With = 0.
FUNCTIONS OF SEVERAL VARIABLES
1. BASIC CONCEPTS
Let: z - a variable value with a range of changes R; R - number line; D - area on the coordinate plane R2.
Any mapping D->R is called a function of two variables with domain D and written z = f(x;y).
In other words:
If each pair (x; y) of two independent variables from the domain D, according to some rule, is associated with one specific value z from R, then variable value z is called a function of two independent variables x and y with domain D and written
http://pandia.ru/text/78/481/images/image002_44.jpg" width="215" height="32 src=">
Example 1.
http://pandia.ru/text/78/481/images/image005_28.jpg" width="157" height="29 src=">
http://pandia.ru/text/78/481/images/image007_16.jpg" align="left" width="110" height="89">
The domain of definition is a part of the plane lying inside a circle of radius r = 3, with the center at the origin, see figure.
Example 3. Find and draw the domain of a function
http://pandia.ru/text/78/481/images/image009_11.jpg" width="86" height="32 src=">
http://pandia.ru/text/78/481/images/image011_10.jpg" width="147" height="30 src=">
2. GEOMETRICAL INTERPRETATION OF THE FUNCTION OF TWO
VARIABLES
2.1.Graph of a function of two variables
Let us consider a rectangular coordinate system in space and a region D on the xOy plane. At each point M(x;y) from this area we restore a perpendicular to the xOy plane and plot the value z = f(x;y) on it. Geometric location of the obtained points
http://pandia.ru/text/78/481/images/image013_10.jpg" width="106" height="23 src=">
http://pandia.ru/text/78/481/images/image015_6.jpg" width="159" height="23 src=">
These are circles centered at the origin, radius R = C1/2 and the equation
x2 + y2 = R2, see figure.
Level lines allow us to represent the surface under consideration, which gives concentric circles when sectioned by planes z = C.
http://pandia.ru/text/78/481/images/image017_16.gif" width="88" height="29"> and find .
Solution. Let's use the section method.
http://pandia.ru/text/78/481/images/image020_11.gif" width="184 height=60" height="60">– in the plane – a parabola.
– in the plane – parabola.
http://pandia.ru/text/78/481/images/image025_5.gif" width="43" height="24 src="> – circle.
The required surface is a paraboloid of revolution.
Distance between two arbitrary points and (Euclidean) space is called a number
http://pandia.ru/text/78/481/images/image030_5.gif" width="153 height=24" height="24"> is called open circle radius centered at point r.
An open circle of radius ε with center at point A is called - ε - surroundings point A.
3task
Find and graphically depict the domain of definition of the function:
Draw function level lines:
3. LIMIT OF A FUNCTION OF TWO VARIABLES
The basic concepts of mathematical analysis introduced for a function of one variable extend to functions of several variables.
Definition:
A constant number A is called the limit of a function of two variables z = f(x;y) for x -> x0, y -> y0, if for any
ε >0 there is δ >0 such that |f(x; y) - A|< ε , как только
|x - x0|< δ и |у – у0| < δ.
This fact is indicated as follows:
http://pandia.ru/text/78/481/images/image042_2.jpg" width="160" height="39 src=">
http://pandia.ru/text/78/481/images/image044_2.gif" width="20" height="25 src=">. For a function of two variables, the tendency to a limit point on the plane can occur according to infinite number directions (and not necessarily in a straight line), and therefore the requirement for the existence of a limit for a function of two (or several) variables is “tighter” compared to a function of one variable.
Example 1. Find .
Solution. Let the desire to reach the limiting point http://pandia.ru/text/78/481/images/image048_2.gif" width="55 height=24" height="24">. Then
http://pandia.ru/text/78/481/images/image050_2.gif" width="72 height=48" height="48"> depends on.
Example 2. Find .
Solution. For any straight line the limit is the same:
http://pandia.ru/text/78/481/images/image054_2.gif" width="57" height="29">. Then
http://pandia.ru/text/78/481/images/image056_1.gif" width="64" height="21">, (the rest is by analogy).
Definition. The number is called limit functions for and , if for such that the inequalities and imply the inequality 
. This fact is briefly written as follows:
http://pandia.ru/text/78/481/images/image065_1.gif" width="124" height="48">.gif" width="236" height="48 src=">;
http://pandia.ru/text/78/481/images/image069_1.gif" width="247" height="60 src=">,
where is the limit point http://pandia.ru/text/78/481/images/image070_1.gif" width="85" height="24 src="> with the domain of definition and let – limit point of the set, i.e. the point to which the arguments tend X And at.
Definition 1. They say that the function is continuous at a point if:
1) ;
2) 
, i.e. 
.
Let us formulate the definition of continuity in an equivalent form..gif" width="89" height="25 src=">.gif" width="85 height=24" height="24"> is continuous at a point if the equality holds
http://pandia.ru/text/78/481/images/image079_0.gif" width="16" height="20 src=">.gif" width="15 height=16" height="16"> let's give an arbitrary increment. The function will receive a partial increment by X
http://pandia.ru/text/78/481/images/image084_0.gif" width="35" height="25 src="> is a function of one variable. Similarly,
http://pandia.ru/text/78/481/images/image058_1.gif" width="85" height="24"> is called continuous at a point over a variable (over a variable) if
http://pandia.ru/text/78/481/images/image087.gif" width="101" height="36">).
Theorem.If the functionis defined in a certain neighborhood of a point and is continuous at this point, then it is continuous at this point in each of the variables.
The reverse statement is not true.
EXAMPLE Let us prove that the function
continuous at the point http://pandia.ru/text/78/481/images/image081_0.gif" width="15 height=16" height="16">.gif" width="57" height="24" > at point corresponding to the increment http://pandia.ru/text/78/481/images/image081_0.gif" width="15" height="16 src=">:
http://pandia.ru/text/78/481/images/image092_0.gif" width="99" height="36 src=">, which means that it is continuous at a point in the variable.
Similarly, one can prove continuity at a point with respect to a variable.
Let us show that there is no limit. Let a point approach a point along a straight line passing through the point. Then we get
.
Thus, approaching the point http://pandia.ru/text/78/481/images/image051_1.gif" width="15" height="20">, we obtain different limit values. It follows that the limit of this function does not exist at the point, which means the function http://pandia.ru/text/78/481/images/image097.jpg" width="351" height="48 src=">
Other designations
http://pandia.ru/text/78/481/images/image099.jpg" width="389" height="55 src=">
Other designations
http://pandia.ru/text/78/481/images/image101_0.gif" width="60" height="28 src=">.
Solution. We have:
, 
Example 2.
http://pandia.ru/text/78/481/images/image105.jpg" width="411" height="51 src=">
Example 3. Find partial derivatives of a function
http://pandia.ru/text/78/481/images/image107.jpg" width="477" height="58 src=">
Example 4. Find partial derivatives of a function
http://pandia.ru/text/78/481/images/image109.jpg" width="321" height="54 src=">
5.2. First order differentials of a function of two variables
The partial differentials of the function z = f(x, y) with respect to the variables x and y are determined, respectively, by the formulas x(x;y) and f"y(x;y) exist at the point (x0;y0) and in some of its neighborhood and are continuous at this point, then, by analogy with a function of one variable, a formula is established for the complete increment of a function of two variables
http://pandia.ru/text/78/481/images/image112_0.gif" width="364" height="57 src=">
where http://pandia.ru/text/78/481/images/image114_0.gif" width="154" height="39 src=">
In other words, the function z = f(x, y) is differentiable at the point (x, y) if its increment Δz is equivalent to the function: 
Expression
http://pandia.ru/text/78/481/images/image116.jpg" width="192" height="57 src=">
Taking into account the fact that Δx = dx, Δy=dy:
http://pandia.ru/text/78/481/images/image090_0.gif" width="57" height="24 src="> is differentiable at the point, then it is continuous at this point.
The converse statement is false, i.e., continuity is only a necessary, but not a sufficient condition for the differentiability of a function. Let's show it.
EXAMPLE Let's find the partial derivatives of the function http://pandia.ru/text/78/481/images/image120.gif" width="253" height="57 src=">.
The resulting formulas lose their meaning at the point http://pandia.ru/text/78/481/images/image121.gif" width="147" height="33 src="> has no partial derivatives at the point. In fact, 
. This function of one variable, as is known, does not have a derivative at the point http://pandia.ru/text/78/481/images/image124.gif" width="25" height="48"> does not exist at the point. Similarly , there is no partial derivative. 
, is obviously continuous at the point .
So, we have shown that a continuous function may not have partial derivatives. It remains to establish the connection between differentiability and the existence of partial derivatives.
5.4. Relationship between differentiability and the existence of partial derivatives.
Theorem 1. A necessary condition for differentiability.
If the function z = f(x, y) is differentiable at the point M(x, y), then it has partial derivatives with respect to each variable and at the point M.
The converse theorem is not true, i.e. the existence of partial derivatives is necessary, but not a sufficient condition for the differentiability of a function.
Theorem 2. Sufficient condition differentiability. If the function z = f(x, y) has continuous partial derivatives at the point , then it is differentiable at the point (and its total differential at this point is expressed by the formula http://pandia.ru/text/78/481/images/image130 .gif" width="101 height=29" height="29">
Example 2. Calculate 3,021.97
3task
Calculate approximately using differential:
5.6. Rules for differentiating complex and implicit functions. Full derivative.
Case 1.
z=f(u, v); u=φ(x, y), v=ψ(x, y)
The functions u and v are continuous functions of the arguments x, y.
Thus, the function z is a complex function of the arguments x and y: z=f(φ(x, y),ψ(x, y))
Let us assume that the functions f(u, v), φ(x, y), ψ(x, y) have continuous partial derivatives with respect to all their arguments.
Let's set the task to calculate http://pandia.ru/text/78/481/images/image140.gif" width="23" height="44 src=">.
Let's give the argument x an increment Δx, fixing the value of the argument y. Then functions of two variables u= φ(x, y) and
v= φ(x, y) will receive partial increments Δxu and Δxv. Consequently, z=f(u, v) will receive the full increment defined in paragraph 5.2 (first-order differentials of a function of two variables):
http://pandia.ru/text/78/481/images/image142.gif" width="293" height="43 src=">
If xu→ 0, then Δxu → 0 and Δxv → 0 (due to the continuity of the functions u and v). Passing to the limit at Δx→ 0, we obtain:
http://pandia.ru/text/78/481/images/image144.gif" width="147" height="44 src="> (*)
EXAMPLE
Z=ln(u2+v), u=ex+y² , v=x2 + y;
http://pandia.ru/text/78/481/images/image146.gif" width="81" height="41 src=">.
http://pandia.ru/text/78/481/images/image148.gif" width="97" height="44 src=">.gif" width="45" height="44 src=">.
Then using formula (*) we get:
http://pandia.ru/text/78/481/images/image152.gif" width="219" height="44 src=">.
To obtain the final result, in the last two formulas, instead of u and v, it is necessary to substitute еx+y² and x2+y, respectively.
Case 2.
The functions x and y are continuous functions.
Thus, the function z=f(x, y) depends through x and y on one independent variable t, i.e. let’s assume that x and y are not independent variables, but functions of the independent variable t, and define the derivative http: //pandia.ru/text/78/481/images/image155.gif" width="235" height="44 src=">
Let's divide both sides of this equality by Δt:
http://pandia.ru/text/78/481/images/image157.gif" width="145" height="44 src="> (**)
Case 3.
Let us now assume that the role of the independent variable t is played by the variable x, that is, that the function z = f(x, y) depends on the independent variable x both directly and through the variable y, which is a continuous function of x.
Taking into account that http://pandia.ru/text/78/481/images/image160.gif" width="120" height="44 src="> (***)
Derivative x(x, y)=http://pandia.ru/text/78/481/images/image162.gif" width="27" height="27 src=">, y=sin x.
Finding partial derivatives
http://pandia.ru/text/78/481/images/image164.gif" width="72" height="48 src=">.gif" width="383" height="48 src=">
The proven rule for differentiating complex functions is applied to find the derivative of an implicit function.
Derivative of a function specified implicitly.
Let us assume that the equation
defines y as an implicit function of x having derivative
y' = φ'(x)_
Substituting y = φ(x) into the equation F(x, y) = 0, we would have to obtain the identity 0 = 0, since y = φ(x) is a solution to this equation. We see, therefore, that the constant zero can be considered as a complex function of x, which depends on x both directly and through y =φ(x).
The derivative with respect to x of this constant must be zero; applying rule (***), we get
F’x(x, y) + F’y(x, y) y’ = 0,
http://pandia.ru/text/78/481/images/image168.gif" width="64" height="41 src=">
Hence,
http://pandia.ru/text/78/481/images/image171.gif" width="20" height="24"> is true for both one and the other function.
5.7. First order total differential. Invariance of the form of a first order differential
Let's substitute the expressions for http://pandia.ru/text/78/481/images/image173.gif" width="23" height="41 src="> defined by equalities (*) (see case 1 in clause 5.6 "Rules for differentiation of complex and implicit functions. Total derivative") into the total differential formula.
Gif" width="33" height="19 src=">.gif" width="33" height="19 src=">.gif" width="140" height="44 src=">
Then the formula for the first order total differential of a function of two variables has the form
http://pandia.ru/text/78/481/images/image180.gif" width="139" height="41 src=">
Comparing the last equality with the formula for the first differential of a function of two independent variables, we can say that the expression for the complete first-order differential of a function of several variables has the same form as it would have if u and v were independent variables.
In other words, the form of the first differential is invariant, that is, it does not depend on whether the variables u and v are independent variables or depend on other variables.
EXAMPLE
Find the first order total differential of a complex function
z=u2v3, u=x2 sin y, v=x3·ey.
Solution. Using the formula for the first order total differential, we have
dz = 2uv3 du+3u2v2 dv =
2uv3 (2x sin y·dx+x2·cos y·dy)+3u2v2·(3x2·ey·dx+x3·ey·dy).
This expression can be rewritten like this
dz=(2uv3 2x siny+3u2v2 3x2 ey) dx+(2uv3x2 cozy+3u2v2x3 ey) dy=
The invariance property of a differential allows us to extend the rule for finding the differential of a sum, product, and quotient to the case of a function of several variables:
http://pandia.ru/text/78/481/images/image183.jpg" width="409" height="46 src=">
http://pandia.ru/text/78/481/images/image185.gif" width="60" height="41 src=">. This
the function will be homogeneous of the third degree for all real x, y and t. The same function will be any homogeneous polynomial in x and y of the third degree, i.e. such a polynomial in each term of which the sum of the exponents xn is equal to three:
http://pandia.ru/text/78/481/images/image187.jpg" width="229" height="47 src=">
are homogeneous functions of degrees 1, 0 and (- 1) respectively..jpg" width="36" height="15">. Indeed,
http://pandia.ru/text/78/481/images/image191.jpg" width="363" height="29 src=">
Assuming t=1, we find
http://pandia.ru/text/78/481/images/image193.jpg" width="95" height="22 src=">
Partial derivatives http://pandia.ru/text/78/481/images/image195.jpg" width="77" height="30 src=">), in general
In other words, they are functions of the variables x and y. Therefore, partial derivatives can again be found from them. Consequently, there are four second-order partial derivatives of a function of two variables, since each of the functions and can be differentiated with respect to both x and y.
The second partial derivatives are denoted as follows:
is the derivative of the nth order; here the function z was first differentiated p times with respect to x, and then n - p times with respect to y.
For a function of any number of variables, partial derivatives of higher orders are determined similarly.
P R And m e r 1. Calculate second order partial derivatives of a function
http://pandia.ru/text/78/481/images/image209.jpg" width="600" height="87 src=">
Example 2. Calculate and http://pandia.ru/text/78/481/images/image212.jpg" width="520" height="97 src=">
Example 3. Calculate if
http://pandia.ru/text/78/481/images/image215.jpg" width="129" height="36 src=">
x, f"y, f"xy and f"yx are defined and continuous at the point M(x, y) and in some of its neighborhood, then at this point
http://pandia.ru/text/78/481/images/image218.jpg" width="50 height=28" height="28">.jpg" width="523" height="128 src=">
Hence,
http://pandia.ru/text/78/481/images/image222.jpg" width="130" height="30 src=">
Solution.
Mixed derivatives are equal.
5.10. Higher order differentials of a functionnvariables.
Total differential d u functions of several variables are in turn a function of the same variables, and we can determine the total differential of this last function. Thus, we will obtain a second-order differential d2u of the original function and, which will also be a function of the same variables, and its complete differential will lead us to a third-order differential d3u of the original function, etc.
Let us consider in more detail the case of the function u=f(x, y) of two variables x and y and assume that the variables x and y are independent variables. A-priory
http://pandia.ru/text/78/481/images/image230.jpg" width="463" height="186 src=">
Calculating d3u in exactly the same way, we get
http://pandia.ru/text/78/481/images/image232.jpg" width="347" height="61 src="> (*)-
Moreover, this formula should be understood as follows: the amount worth parentheses, must be raised to the power n, using Newton's Binomial Formula, after which the exponents of y and http://pandia.ru/text/78/481/images/image235.jpg" width="22" height="21 src=" >.gif" width="22" height="27"> with direction cosines cos α, cos β (α + β = 90°). On the vector, consider the point M1(x + Δx; y + Δy). When moving from point M to point M1, the function z = f(x; y) will receive a full increment
http://pandia.ru/text/78/481/images/image239.jpg" width="133 height=27" height="27"> tending to zero (see figure).
http://pandia.ru/text/78/481/images/image241.jpg" width="324" height="54 src=">
where http://pandia.ru/text/78/481/images/image243.gif" width="76" height="41 src=">, and therefore we get:
http://pandia.ru/text/78/481/images/image245.gif" width="24" height="41 src="> for Δs->0 is called the product
water function z = f(x; y) at the point (x; y) in the direction of the vector and is denoted
http://pandia.ru/text/78/481/images/image247.jpg" width="227" height="51 src="> (*)
Thus, knowing the partial derivatives of the function
z = f(x; y) you can find the derivative of this function in any direction, and each partial derivative is a special case of the directional derivative.
EXAMPLE Find the derivative of a function
http://pandia.ru/text/78/481/images/image249.jpg" width="287" height="56 src=">
http://pandia.ru/text/78/481/images/image251.jpg" width="227" height="59 src=">
http://pandia.ru/text/78/481/images/image253.gif" width="253 height=62" height="62">
Consequently, the function z = f(x;y) increases in a given direction.
5. 12 . Gradient
The gradient of a function z = f(x; y) is a vector whose coordinates are the corresponding partial derivatives of this function
http://pandia.ru/text/78/481/images/image256.jpg" width="205" height="56 src=">
i.e..jpg" width="89" height="33 src=">
at point M(3;4).
Solution.
http://pandia.ru/text/78/481/images/image259.jpg" width="213" height="56 src=">
