Partial derivatives of functions of two variables.
Concept and examples of solutions

In this lesson, we will continue our acquaintance with the function of two variables and consider, perhaps, the most common thematic task - finding partial derivatives of the first and second order, as well as the total differential of the function. Part-time students, as a rule, face partial derivatives in the 1st year in the 2nd semester. Moreover, according to my observations, the task of finding partial derivatives is almost always found in the exam.

In order to effectively study the following material, you necessary be able to more or less confidently find the "usual" derivatives of a function of one variable. You can learn how to handle derivatives correctly in the lessons How to find the derivative? and Derivative of a compound function. We also need a table of derivatives of elementary functions and differentiation rules, it is most convenient if it is at hand in printed form. You can find reference material on the page Mathematical formulas and tables.

Let's quickly repeat the concept of a function of two variables, I will try to limit myself to the bare minimum. A function of two variables is usually written as , with the variables being called independent variables or arguments.

Example: - a function of two variables.

Sometimes the notation is used. There are also tasks where the letter is used instead of a letter.

From a geometric point of view, a function of two variables is most often a surface of three-dimensional space (a plane, a cylinder, a ball, a paraboloid, a hyperboloid, etc.). But, in fact, this is already more analytical geometry, and we have mathematical analysis on the agenda, which my university teacher never let me write off is my “horse”.

We turn to the question of finding partial derivatives of the first and second orders. I have some good news for those of you who have had a few cups of coffee and are in the mood for unimaginably difficult material: partial derivatives are almost the same as the "ordinary" derivatives of a function of one variable.

For partial derivatives, all the rules of differentiation and the table of derivatives of elementary functions are valid. There are only a couple of small differences that we will get to know right now:

... yes, by the way, for this topic I did create small pdf book, which will allow you to "fill your hand" in just a couple of hours. But, using the site, you, of course, will also get the result - just maybe a little slower:

Example 1

Find partial derivatives of the first and second order of a function

First, we find the partial derivatives of the first order. There are two of them.

Notation:
or - partial derivative with respect to "x"
or - partial derivative with respect to "y"

Let's start with . When we find the partial derivative with respect to "x", then the variable is considered a constant (constant number).

Comments on the actions taken:

(1) The first thing we do when finding the partial derivative is to conclude all function in parentheses under the dash with subscript.

Attention important! Subscripts DO NOT LOSE in the course of the solution. In this case, if you draw a “stroke” somewhere without, then the teacher, at least, can put it next to the task (immediately bite off part of the score for inattention).

(2) Use the rules of differentiation , . For a simple example like this one, both rules can be applied in the same step. Pay attention to the first term: since is considered a constant, and any constant can be taken out of the sign of the derivative, then we take it out of brackets. That is, in this situation, it is no better than a regular number. Now let's look at the third term: here, on the contrary, there is nothing to take out. Since it is a constant, it is also a constant, and in this sense it is no better than the last term - the “seven”.

(3) We use tabular derivatives and .

(4) We simplify, or, as I like to say, "combine" the answer.

Now . When we find the partial derivative with respect to "y", then the variableconsidered a constant (constant number).

(1) We use the same differentiation rules , . In the first term we take out the constant beyond the sign of the derivative, in the second term nothing can be taken out because it is already a constant.

(2) We use the table of derivatives of elementary functions. Mentally change in the table all "X" to "Y". That is, this table is equally valid for (and indeed for almost any letter). In particular, the formulas we use look like this: and .

What is the meaning of partial derivatives?

At their core, 1st order partial derivatives resemble "ordinary" derivative:

- This functions, which characterize rate of change functions in the direction of the axes and respectively. So, for example, the function characterizes the steepness of "climbs" and "slopes" surfaces in the direction of the abscissa axis, and the function tells us about the "relief" of the same surface in the direction of the ordinate axis.

! Note : here refers to directions that are parallel coordinate axes.

For the sake of better understanding, let's consider a specific point of the plane and calculate the value of the function (“height”) in it:
- and now imagine that you are here (ON THE VERY surface).

We calculate the partial derivative with respect to "x" at a given point:

The negative sign of the "X" derivative tells us about descending functions at a point in the direction of the x-axis. In other words, if we make a small-small (infinitesimal) step towards the tip of the axis (parallel to this axis), then go down the slope of the surface.

Now we find out the nature of the "terrain" in the direction of the y-axis:

The derivative with respect to "y" is positive, therefore, at a point along the axis, the function increases. If it’s quite simple, then here we are waiting for an uphill climb.

In addition, the partial derivative at a point characterizes rate of change functions in the relevant direction. The greater the resulting value modulo- the steeper the surface, and vice versa, the closer it is to zero, the flatter the surface. So, in our example, the "slope" in the direction of the abscissa axis is steeper than the "mountain" in the direction of the ordinate axis.

But those were two private paths. It is quite clear that from the point at which we are, (and in general from any point of the given surface) we can move in some other direction. Thus, there is an interest in compiling a general "navigation chart" that would tell us about the "landscape" of the surface. if possible at every point scope of this function in all available ways. I will talk about this and other interesting things in one of the next lessons, but for now, let's get back to the technical side of the issue.

We systematize the elementary applied rules:

1) When we differentiate by , then the variable is considered a constant.

2) When differentiation is carried out according to, then is considered a constant.

3) The rules and the table of derivatives of elementary functions are valid and applicable for any variable (or any other) with respect to which differentiation is carried out.

Step two. We find partial derivatives of the second order. There are four of them.

Notation:
or - the second derivative with respect to "x"
or - the second derivative with respect to "y"
or - mixed derivative "x by y"
or - mixed derivative "Y with X"

There are no problems with the second derivative. In simple terms, the second derivative is the derivative of the first derivative.

For convenience, I will rewrite the first-order partial derivatives already found:

First we find the mixed derivatives:

As you can see, everything is simple: we take the partial derivative and differentiate it again, but in this case, already by “y”.

Similarly:

In practical examples, you can focus on the following equality:

Thus, through mixed derivatives of the second order, it is very convenient to check whether we have found the partial derivatives of the first order correctly.

We find the second derivative with respect to "x".
No inventions, we take and differentiate it by "X" again:

Similarly:

It should be noted that when finding , you need to show increased attention, since there are no miraculous equalities to test them.

The second derivatives also find wide practical application, in particular, they are used in the problem of finding extrema of a function of two variables. But everything has its time:

Example 2

Calculate the first order partial derivatives of the function at the point . Find derivatives of the second order.

This is an example for self-solving (answers at the end of the lesson). If you have difficulty differentiating roots, go back to the lesson How to find the derivative? In general, pretty soon you will learn how to find similar derivatives on the fly.

We fill our hand with more complex examples:

Example 3

Check that . Write the total differential of the first order.

Solution: We find partial derivatives of the first order:

Pay attention to the subscript: next to the "x" it is not forbidden to write in brackets that it is a constant. This mark can be very useful for beginners to make it easier to navigate the solution.

Further comments:

(1) We take out all the constants outside the sign of the derivative. In this case, and , and, hence, their product is considered a constant number.

(2) Do not forget how to properly differentiate the roots.

(1) We take all the constants out of the sign of the derivative, in this case the constant is .

(2) Under the prime, we have the product of two functions, therefore, we need to use the product differentiation rule .

(3) Do not forget that is a complex function (although the simplest of the complex ones). We use the corresponding rule: .

Now we find mixed derivatives of the second order:

This means that all calculations are correct.

Let's write the total differential. In the context of the task under consideration, it makes no sense to tell what the total differential of a function of two variables is. It is important that this very differential very often needs to be written down in practical problems.

Total First Order Differential functions of two variables has the form:

In this case:

That is, in the formula you just need to stupidly just substitute the already found partial derivatives of the first order. Differential icons and in this and similar situations, if possible, it is better to write in numerators:

And at the repeated request of readers, full differential of the second order.

It looks like this:

CAREFULLY find the "single-letter" derivatives of the 2nd order:

and write down the "monster", carefully "attaching" the squares, the product and not forgetting to double the mixed derivative:

It's okay if something seemed difficult, you can always return to derivatives later, after you pick up the differentiation technique:

Example 4

Find first order partial derivatives of a function . Check that . Write the total differential of the first order.

Consider a series of examples with complex functions:

Example 5

Find partial derivatives of the first order of the function .

Decision:

Example 6

Find first order partial derivatives of a function .
Write down the total differential.

This is an example for self-solving (answer at the end of the lesson). I won't post the complete solution because it's quite simple.

Quite often, all of the above rules are applied in combination.

Example 7

Find first order partial derivatives of a function .

(1) We use the rule of differentiating the sum

(2) The first term in this case is considered a constant, since there is nothing in the expression that depends on "x" - only "y". You know, it's always nice when a fraction can be turned into zero). For the second term, we apply the product differentiation rule. By the way, in this sense, nothing would change if a function were given instead - it is important that here the product of two functions, EACH of which depends on "X", and therefore, you need to use the rule of differentiation of the product. For the third term, we apply the rule of differentiation of a complex function.

(1) The first term in both the numerator and the denominator contains a “y”, therefore, you need to use the rule for differentiating the quotient: . The second term depends ONLY on "x", which means it is considered a constant and turns into zero. For the third term, we use the rule of differentiation of a complex function.

For those readers who courageously made it almost to the end of the lesson, I’ll tell you an old Mekhmatov anecdote for detente:

Once an evil derivative appeared in the space of functions and how it went to differentiate everyone. All functions scatter in all directions, no one wants to turn! And only one function does not escape anywhere. The derivative approaches it and asks:

"Why aren't you running away from me?"

- Ha. But I don't care, because I'm "e to the power of x", and you can't do anything to me!

To which the evil derivative with an insidious smile replies:

- This is where you are wrong, I will differentiate you by “y”, so be zero for you.

Who understood the joke, he mastered the derivatives, at least for the "troika").

Example 8

Find first order partial derivatives of a function .

This is a do-it-yourself example. A complete solution and a sample design of the problem are at the end of the lesson.

Well, that's almost all. Finally, I cannot help but please mathematicians with one more example. It's not even about amateurs, everyone has a different level of mathematical training - there are people (and not so rare) who like to compete with more difficult tasks. Although, the last example in this lesson is not so much complicated as cumbersome in terms of calculations.

Definition: The total differential of a function several variables is called the sum of all its partial differentials:

Example 1: .

Decision:

Since the partial derivatives of this function are equal:

Then we can immediately write the partial differentials of these functions:

, ,

Then the total differential of the function will look like:

.

Example 2 Find the full differential of a function

Decision:

This function is complex, i.e. can be imagined as

We find partial derivatives:

Full Differential:

The analytical meaning of the total differential is that the total differential of a function of several variables is the main part of the total increment of this function, that is, there is an approximate equality: ∆z≈dz.

However, it must be remembered that these approximate equalities are valid only for small differentials dx and dy of the arguments of the function z=f(x,y).

The use of the total differential in approximate calculations is based on the use of the formula ∆z≈dz.

Indeed, if in this formula the increment ∆z of the function is represented as , and the total differential as , then we get:

≈ ,

The resulting formula can be used to approximately find the "new" value of a function of two variables, which it takes with sufficiently small increments of both of its arguments.

Example. Find the approximate value of a function , with the following values ​​of its arguments: 1.01, .

Decision.

Substituting the partial derivatives of the functions found earlier in the formula, we get:

When substituting the values ​​x=1, ∆x=0.01, y=2, ∆y=0.02, we get:

scalar field.

If at each point of some region of space D the function U(p)=U(x,y,z) is given, then it is said that a scalar field is given in the region D.

If, for example, U(x, y, z) denotes the temperature at the point M(x, y, z), then we say that a scalar temperature field is given. If the region D is filled with liquid or gas and U(x,y,z) denotes pressure, then there is a scalar pressure field. If the arrangement of charges or massive bodies is given in space, then one speaks of a potential field.

The scalar field is called stationary, if the function U(x,y,z) does not change with time: U(x,y,z) ≠ f(t).

Any stationary field is characterized by:

1) the level surface of the scalar field

2) the rate of change of the field in a given direction.

Level surface scalar field is the locus of points at which the function U(x,y,z) takes a constant value, that is, U(x,y,z) = const. The collection of these points forms a certain surface. If we take another constant, we get another surface.

Example: Let a scalar field be given. An example of such a field is the electric potential field of a point electric charge (+q). Here, the level surfaces are the equipotential surfaces , that is, spheres in the center of which there is a charge that creates a field.

The direction of greatest increase of a scalar function is given by a vector called gradient and is denoted by the symbol (or ).

The gradient of the function is found in terms of the partial derivatives of this function and is always perpendicular to the level surface of the scalar field at a given point:

, where

Unit vectors respectively along the axes OX, OY, OZ

The derivative of the function U(x,y,z) in any other direction (λ) is determined by the formula:

, where

α, β, γ are the angles between the coordinate axes OX, OY, OZ and direction respectively.

As you can see, to find the differential, you need to multiply the derivative by dx. This allows you to immediately write the corresponding table for differentials from the table of formulas for derivatives.

Total differential for a function of two variables:

The total differential for a function of three variables is equal to the sum of partial differentials: d f(x,y,z)=d x f(x,y,z)dx+d y f(x,y,z)dy+d z f(x,y,z)dz

Definition . A function y=f(x) is called differentiable at a point x 0 if its increment at this point can be represented as ∆y=A∆x + α(∆x)∆x, where A is a constant and α(∆x) is infinitely small as ∆x → 0.
The requirement that a function be differentiable at a point is equivalent to the existence of a derivative at this point, with A=f'(x 0).

Let f(x) be differentiable at a point x 0 and f "(x 0)≠0 , then ∆y=f'(x 0)∆x + α∆x, where α= α(∆x) →0 as ∆x → 0. The quantity ∆y and each term on the right-hand side are infinitesimal values ​​as ∆x→0. Let's compare them: , that is, α(∆x)∆x is an infinitesimal higher order than f’(x 0)∆x.
, that is, ∆y~f’(x 0)∆x. Therefore, f’(x 0)∆x is the main and at the same time linear with respect to ∆x part of the increment ∆y (linear means containing ∆x to the first degree). This term is called the differential of the function y \u003d f (x) at the point x 0 and denoted dy (x 0) or df (x 0). So, for arbitrary x
dy=f′(x)∆x. (one)
Let dx=∆x, then
dy=f′(x)dx. (2)

Example. Find derivatives and differentials of these functions.
a) y=4tg2x
Decision:

differential:
b)
Decision:

differential:
c) y=arcsin 2 (lnx)
Decision:

differential:
G)
Decision:
=
differential:

Example. For the function y=x 3 find an expression for ∆y and dy for some values ​​of x and ∆x.
Decision. ∆y = (x+∆x) 3 – x 3 = x 3 + 3x 2 ∆x +3x∆x 2 + ∆x 3 – x 3 = 3x 2 ∆x+3x∆x 2 +∆x 3 dy=3x 2 ∆x (we took the main linear part of ∆y with respect to ∆x). In this case, α(∆x)∆x = 3x∆x 2 + ∆x 3 .

Collection output:

ON THE SECOND ORDER DIFFERENTIAL

Lovkov Ivan Yurievich

student of the Moscow State University of Information Technologies, Radio Engineering and Electronics, RF, Serpukhov

E- mail: alkasardancer@ rambler. en

Taperechkina Vera Alekseevna

cand. Phys.-Math. Sciences, Associate Professor, Moscow State University of Information Technologies, Radio Engineering and Electronics, Russian Federation, Serpukhov

ABOUT SECOND-ORDER DIFFERENTIAL

Lovkov Ivan

student of Moscow State University of Information Technologies, Radio Engineering and Electronics, Russia, Serpukhov

Vera Taperechkina

candidate of Physical and Mathematical Sciences, associate professor of Moscow State University of Information Technologies, Radio Engineering and Electronics, Russia, Serpukhov

ANNOTATION

The paper considers methods for finding derivatives and differentials of the first and second orders for complex functions of two variables.

ABSTRACT

Calculation methods of derivative and first and second differentials for composite functions of two variables.

Keywords: partial derivatives; differential.

keywords: partial derivatives; differential.

1. Introduction.

Let us formulate some facts from the theory of functions of several variables, which we will need below.

Definition: A function z=f(u, v) is called differentiable at a point (u, v) if its increment Δz can be represented as:

The linear part of the increment is called the total differential and is denoted dz.

Theorem (sufficient condition for differentiability) cf.

If in some neighborhood of m.(u, v) there exist continuous partial derivatives and , then the function f(u, v) is differentiable at this point and

(du=Δu, dv=Δv). (one)

Definition: The second differential of the function z=f(u, v) at a given point (u, v) is the first differential of the first differential of the function f(u, v), i.e.

From the definition of the second differential z=f(u, v), where u and v are independent variables, it follows

Thus, the formula is valid:

When deriving the formula, the Schwartz theorem on the equality of mixed derivatives was used. This equality is valid provided that are defined in a neighborhood of m.(u, v) and continuous in m.(u, v). see

The formula for finding the 2nd differential can be written symbolically in the following form: – formal squaring of the bracket with subsequent formal multiplication on the right by f(x y) gives the previously obtained formula . Similarly, the formula for the 3rd differential is valid:

And generally speaking:

Where the formal raising to the nth power is performed according to Newton's binomial formula:

;

Note that the first differential of a function of two variables has the form invariance property. That is, if u and v are independent variables, then for the function z=f(u, v), according to (1)

Let now u=u(x y), v=v(x y), then z=f(u(x y), v(x y)), x and y are independent variables, then

Using well-known formulas for the derivative of a complex function:

Then from (3) and (4) we get:

Thus,

(5)

where - the first differential of the function u, - the first differential of the function v.

Comparing (1) and (5), we see that the formula for dz remains formally written, but if in (1) du=Δu, dv=Δv are increments of independent variables, then in (5) du and dv are differentials of functions u and v.

2. The second differential of a compound function of two variables.

First of all, we show that the second differential does not have the form invariance property.

Let z=z(u, v) in the case of independent variables u and v, the second differential is found by the formula (2)

Let now u=u(x y), v=v(x y), z=z(u(x y), v(x y)), where x and y are independent variables. Then

.

So, we finally got:

Formulas (2) and (6) do not coincide in form, therefore, the second differential does not have the invariance property.

Previously, partial derivative formulas of the 1st order were derived for a complex function z=f(u, v), where u=u(x y), v=v(x y), where x and y are independent variables, see .

We derive formulas for calculating partial derivatives and a second-order differential for the function z=f(u, v), u=u(x y), v=v(x y), where x and y are independent variables.

For functions u(x y), v(x y) of independent variables x, y, we have the formulas:

Let us substitute formulas (8) into (6).

Thus, we have obtained a formula for the second-order differential of a complex function of two variables.

Comparing the coefficients for second-order partial derivatives of a complex function of two variables in (2) and (9), we obtain the formulas:

Example 1 cm

Let z=f(u, v), u=xy, v=. Find the second differential.

Solution: calculate partial derivatives:

, , , ,

, ,