Each partial derivative (over x and by y) of a function of two variables is the ordinary derivative of a function of one variable with a fixed value of the other variable:

(where y= const),

(where x= const).

Therefore, partial derivatives are calculated from formulas and rules for calculating derivatives of functions of one variable, while considering the other variable as a constant (constant).

If you do not need an analysis of examples and the minimum theory necessary for this, but you only need a solution to your problem, then proceed to online partial derivative calculator .

If it’s hard to focus on keeping track of where the constant is in the function, then you can substitute any number in the draft solution of the example instead of a variable with a fixed value - then you can quickly calculate the partial derivative as the ordinary derivative of a function of one variable. It is only necessary not to forget to return the constant (a variable with a fixed value) to its place when finishing.

The property of partial derivatives described above follows from the definition of a partial derivative, which can be found in exam questions. Therefore, to get acquainted with the definition below, you can open the theoretical reference.

The concept of continuity of a function z= f(x, y) at a point is defined similarly to this concept for a function of one variable.

Function z = f(x, y) is called continuous at a point if

The difference (2) is called the total increment of the function z(it is obtained by incrementing both arguments).

Let the function z= f(x, y) and dot

If the function change z occurs when only one of the arguments changes, for example, x, with a fixed value of the other argument y, then the function will be incremented

called partial increment of the function f(x, y) on x.

Considering the function change z depending on the change of only one of the arguments, we actually pass to a function of one variable.

If there is a finite limit

then it is called the partial derivative of the function f(x, y) by argument x and is denoted by one of the symbols

(4)

The partial increment is defined similarly z on y:

and partial derivative f(x, y) on y:

(6)

Example 1

Solution. We find the partial derivative with respect to the variable "x":

(y fixed);

We find the partial derivative with respect to the variable "y":

(x fixed).

As you can see, it does not matter to what extent the variable that is fixed: in this case, it is just some number that is a factor (as in the case of the usual derivative) with the variable by which we find the partial derivative. If the fixed variable is not multiplied by the variable with respect to which we find the partial derivative, then this lonely constant, no matter to what extent, as in the case of an ordinary derivative, vanishes.

Example 2 Given a function

Find Partial Derivatives

(by x) and (by y) and calculate their values ​​at the point BUT (1; 2).

Solution. At a fixed y the derivative of the first term is found as the derivative of the power function ( table of derivative functions of one variable):

.

At a fixed x the derivative of the first term is found as the derivative of the exponential function, and the second - as the derivative of the constant:

Now we calculate the values ​​of these partial derivatives at the point BUT (1; 2):

You can check the solution of problems with partial derivatives on online partial derivative calculator .

Example 3 Find Partial Derivatives of Functions

Solution. In one step we find

(y x, as if the argument of sine were 5 x: in the same way, 5 appears before the sign of the function);

(x is fixed and is in this case a factor at y).

You can check the solution of problems with partial derivatives on online partial derivative calculator .

The partial derivatives of a function of three or more variables are defined similarly.

If each set of values ​​( x; y; ...; t) independent variables from the set D corresponds to one specific value u from many E, then u is called a function of variables x, y, ..., t and denote u= f(x, y, ..., t).

For functions of three or more variables, there is no geometric interpretation.

Partial derivatives of a function of several variables are also defined and calculated under the assumption that only one of the independent variables changes, while the others are fixed.

Example 4 Find Partial Derivatives of Functions

.

Solution. y and z fixed:

x and z fixed:

x and y fixed:

Find partial derivatives on your own and then see solutions

Example 5

Example 6 Find partial derivatives of a function.

The partial derivative of a function of several variables has the same mechanical meaning as the derivative of a function of one variable, is the rate at which the function changes relative to a change in one of the arguments.

Example 8 flow quantity P passengers railways can be expressed as a function

where P- the number of passengers, N- the number of residents of the corresponding points, R– distance between points.

Partial derivative of a function P on R equal to

shows that the decrease in the flow of passengers is inversely proportional to the square of the distance between the corresponding points for the same number of inhabitants in the points.

Partial derivative P on N equal to

shows that the increase in the flow of passengers is proportional to twice the number of inhabitants settlements with the same distance between points.

You can check the solution of problems with partial derivatives on online partial derivative calculator .

Full differential

The product of the partial derivative and the increment of the corresponding independent variable is called the partial differential. Partial differentials are denoted as follows:

The sum of partial differentials over all independent variables gives the total differential. For a function of two independent variables, the total differential is expressed by the equality

(7)

Example 9 Find the full differential of a function

Solution. The result of using formula (7):

A function that has a total differential at every point of some domain is called differentiable in that domain.

Find the total differential on your own and then see the solution

Just as in the case of a function of one variable, the differentiability of a function in a certain region implies its continuity in this region, but not vice versa.

Let us formulate without proof a sufficient condition for the differentiability of a function.

Theorem. If the function z= f(x, y) has continuous partial derivatives

in a given region, then it is differentiable in this region and its differential is expressed by formula (7).

It can be shown that, just as in the case of a function of one variable, the differential of the function is the main linear part of the increment of the function, so in the case of a function of several variables, the total differential is the main, linear with respect to increments of independent variables, part of the total increment of the function.

For a function of two variables, the total increment of the function has the form

(8)

where α and β are infinitesimal for and .

Partial derivatives of higher orders

Partial derivatives and functions f(x, y) are themselves some functions of the same variables and, in turn, may have derivatives with respect to different variables, which are called partial derivatives of higher orders.

Consider a function of two variables z=f(x, y) and its total increment at the point M 0 (x 0 , y 0)

Δ z \u003d f (x 0 + Δ x, y 0 + Δ y) - f (x 0, y 0).

Definition. If there are numbers P and Q such that the total increment can be represented as

Δz = PΔx + QΔy + ε Δρ,

where and ε→ 0 at Δρ→ 0 , then the expression PΔx + QΔy is called the total differential of the function z=f(x,y) at the point M0 (x0,y0).

In this case, the full increment of the function consists of two parts: the first part PΔx + QΔy is linear with respect to Δx and Δy, the second is an infinitesimal higher order in comparison with .

Total differential of a function z=f(x,y) denoted by dz, that is

dz = PΔx+QΔy.

A function that has a total differential at a given point is called differentiable at that point.

Theorem. If a u=f(M) differentiable at a point M0, then it is continuous in it.

Comment. The continuity of a function of two variables does not imply its differentiability.

Example. continuous in (0,0) , but has no partial derivative - does not exist. Similarly, there is no partial derivative with respect to y. Therefore, the function is not differentiable.

Theorem [necessary condition for differentiability]. If a z=f(x,y) differentiable at a point M0, then it has partial derivatives with respect to x and y, and

f′ x (x 0 ,y 0) = P, f′ y (x 0 , y 0) = Q.

Comment. Differentiability does not follow from the existence of partial derivatives. Example:

We have , but the function is not continuous, hence it is not differentiable.

Theorem [sufficient condition for differentiability]. If the first partial derivatives of the functions z=f(x,y) are defined in some neighborhood of the point M0 (x0,y0) and continuous at the point M0, then given function has a total differential at that point.

Comment. We have

Δ z \u003d f′ x (x 0, y 0)Δ x + f′ y (x 0, y 0)Δ y + ε Δρ,

where ε→ 0 at Δρ→ 0 . Consequently,

f(x 0 +Δ x,y 0 +Δ y) - f(x 0 ,y 0) ≈ f′ x (x 0 ,y 0)Δ x + f′ y (x 0 ,y 0)Δ y

f(x 0 +Δ x, y 0 +Δ y) ≈ f(x 0 ,y 0) + f′ x (x 0 , y 0)Δ x + f′ y (x 0 , y 0)Δ y.

This formula is used in approximate calculations.

At fixed Δx and Δy the total differential is a function of the variables x and y:

Let's put dx=Δx, dy=Δy and call these quantities differentials of independent variables.

Then we get the formula

that is, the total differential of the function is equal to the sum products of the first partial derivatives and the corresponding differentials of the arguments.

The total differential of a function of three variables is defined and expressed similarly. If a u=f(x, y, z) and there are numbers P, Q, R such that

Δu = PΔx+QΔy+RΔz+εΔρ, ε→ 0 at δρ→ 0 ,

then the total differential is the expression

du = PΔx+QΔy+RΔz.

If the first partial derivatives of this function are continuous, then

where dx=Δx, dz=Δz, dz=Δz.

Definition. The total second-order differential of some function is the total differential of its total differential.

If a z=f(x,y), dz=z′ x dx+z′ y dy, then

Tangent plane and surface normal

Consider the surface S, given by the equation

z=f(x, y).

Let f(x, y) has partial derivatives in some domain. Consider M 0 (x 0 , y 0).

- slope tangent at point M0 to the section of the surface by the plane y=y0, that is, to the line z=f(x,y 0). The tangent to this line is:

z-z 0 \u003d f′ x (x 0, y 0) (x-x 0), y=y 0.

Similarly, a section by a plane x=x0 gives the equation

z-z 0 =f′ y (x 0 , y 0)(y-y 0), x=x 0.

The plane containing both of these lines has the equation

z-z 0 \u003d f′ x (x 0, y 0)(x-x 0)+f′ y (x 0, y 0)(y-y 0)

and is called the tangent plane to the surface S at the point P 0 (x 0 , y 0 , z 0).

Note that the tangent plane equation can be rewritten as

z-z 0 =df.

In this way, geometric sense total differential: differential at a point M0 for increment (x-x 0 , y-y 0) is the increment of the applicate point of the tangent plane to the surface z=f(x,y) at the point (x0, y0) for the same increments.

The tangent plane has a normal vector at the point (x0, y0, z0) - \vec(n)=(f′ x (x 0 , y 0), f′ y (x 0 , y 0), -1). A line passing through a point P0 and having a direction vector \vec(n), is called the normal to the surface z=f(x,y) at this point. Her equations are:

Differentiation of complex functions

Let a differentiable function be given z=F(v, w), whose arguments are differentiable functions of variables x and y:

v=v(x, y), w=w(x, y).

If at the same time the function

z=F(v(x, y), w(x, y))=\Phi(x, y)

makes sense, then it is called a complex function of x and y.

Theorem. Partial derivatives z′ x, z'y complex function exist and are expressed by the formulas

If a v and w- differentiable functions of one variable t, that is

v=v(t), w=w(t),

and the function makes sense

z=F(v(t), w(t))=f(t),

then its derivative is expressed by the formula

This derivative is called the total derivative.

If a differentiable function is given

u=F(ξ, η, ζ),

whose arguments ξ=ξ(t), η=η(t), ζ=ζ(t)- differentiable functions of a variable t and function

u=F(ξ(t), η(t), ζ(t))

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
Solution:

differential:
b)
Solution:

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

differential:
G)
Solution:
=
differential:

Example. For the function y=x 3 find an expression for ∆y and dy for some values ​​of x and ∆x.
Solution. ∆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 .

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 complex function. We also need a derivative table 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 is 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. talking plain language, 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 .

Solution:

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) In the first term, both the numerator and the denominator contain 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 reached almost the end of the lesson, I will 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.

Collection output:

ON THE SECOND ORDER DIFFERENTIAL

Lovkov Ivan Yurievich

Moscow student state university information technologies, radio engineering and electronics, Russian Federation, 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:

In this way,

(5)

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

Comparing (1) and (5), we see that the formal formula for dz is preserved, 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 at 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:

, , , ,

, ,