Definition 1

If for each pair $ (x, y) $ of values ​​of two independent variables from a certain region a certain value of $ z $ is associated, then $ z $ is said to be a function of two variables $ (x, y) $. Notation: $ z = f (x, y) $.

With regard to the function $ z = f (x, y) $, consider the concepts of general (full) and partial increments of a function.

Let a function $ z = f (x, y) $ of two independent variables $ (x, y) $ be given.

Remark 1

Since the variables $ (x, y) $ are independent, one of them can change, while the other remains constant.

Let's give the variable $ x $ an increment of $ \ Delta x $, while keeping the value of the variable $ y $ unchanged.

Then the function $ z = f (x, y) $ will receive an increment, which will be called the partial increment of the function $ z = f (x, y) $ with respect to the variable $ x $. Designation:

Similarly, let's give the variable $ y $ an increment of $ \ Delta y $, while keeping the value of the variable $ x $ unchanged.

Then the function $ z = f (x, y) $ will receive an increment, which will be called the partial increment of the function $ z = f (x, y) $ with respect to the variable $ y $. Designation:

If the argument $ x $ is given the increment $ \ Delta x $, and the argument $ y $ - the increment $ \ Delta y $, then the full increment of the given function $ z = f (x, y) $ is obtained. Designation:

Thus, we have:

$ \ Delta _ (x) z = f (x + \ Delta x, y) -f (x, y) $ - partial increment of the function $ z = f (x, y) $ with respect to $ x $;

$ \ Delta _ (y) z = f (x, y + \ Delta y) -f (x, y) $ - partial increment of the function $ z = f (x, y) $ with respect to $ y $;

$ \ Delta z = f (x + \ Delta x, y + \ Delta y) -f (x, y) $ - full increment of the function $ z = f (x, y) $.

Example 1

Solution:

$ \ Delta _ (x) z = x + \ Delta x + y $ - partial increment of the function $ z = f (x, y) $ with respect to $ x $;

$ \ Delta _ (y) z = x + y + \ Delta y $ is the partial increment of the function $ z = f (x, y) $ with respect to $ y $.

$ \ Delta z = x + \ Delta x + y + \ Delta y $ - full increment of the function $ z = f (x, y) $.

Example 2

Calculate the quotient and total increment of the function $ z = xy $ at the point $ (1; 2) $ for $ \ Delta x = 0,1; \, \, \ Delta y = 0,1 $.

Solution:

By the definition of the private increment, we find:

$ \ Delta _ (x) z = (x + \ Delta x) \ cdot y $ - partial increment of the function $ z = f (x, y) $ with respect to $ x $

$ \ Delta _ (y) z = x \ cdot (y + \ Delta y) $ - partial increment of the function $ z = f (x, y) $ with respect to $ y $;

By the definition of the full increment, we find:

$ \ Delta z = (x + \ Delta x) \ cdot (y + \ Delta y) $ - full increment of the function $ z = f (x, y) $.

Hence,

\ [\ Delta _ (x) z = (1 + 0.1) \ cdot 2 = 2.2 \] \ [\ Delta _ (y) z = 1 \ cdot (2 + 0.1) = 2.1 \] \ [\ Delta z = (1 + 0.1) \ cdot (2 + 0.1) = 1.1 \ cdot 2.1 = 2.31. \]

Remark 2

The total increment of a given function $ z = f (x, y) $ is not equal to the sum of its partial increments $ \ Delta _ (x) z $ and $ \ Delta _ (y) z $. Mathematical notation: $ \ Delta z \ ne \ Delta _ (x) z + \ Delta _ (y) z $.

Example 3

Check assertion remark for function

Solution:

$ \ Delta _ (x) z = x + \ Delta x + y $; $ \ Delta _ (y) z = x + y + \ Delta y $; $ \ Delta z = x + \ Delta x + y + \ Delta y $ (obtained in example 1)

Find the sum of the partial increments of the given function $ z = f (x, y) $

\ [\ Delta _ (x) z + \ Delta _ (y) z = x + \ Delta x + y + (x + y + \ Delta y) = 2 \ cdot (x + y) + \ Delta x + \ Delta y. \]

\ [\ Delta _ (x) z + \ Delta _ (y) z \ ne \ Delta z. \]

Definition 2

If for each triple $ (x, y, z) $ of values ​​of three independent variables from a certain region a certain value of $ w $ is associated, then $ w $ is said to be a function of three variables $ (x, y, z) $ in this area.

Designation: $ w = f (x, y, z) $.

Definition 3

If for each collection $ (x, y, z, ..., t) $ of values ​​of independent variables from a certain region a certain value of $ w $ is associated, then $ w $ is said to be a function of the variables $ (x, y, z, ..., t) $ in this domain.

Notation: $ w = f (x, y, z, ..., t) $.

For a function of three or more variables, in the same way as for a function of two variables, partial increments are determined for each of the variables:

$ \ Delta _ (z) w = f (x, y, z + \ Delta z) -f (x, y, z) $ is the partial increment of the function $ w = f (x, y, z, ..., t ) $ by $ z $;

$ \ Delta _ (t) w = f (x, y, z, ..., t + \ Delta t) -f (x, y, z, ..., t) $ - partial increment of the function $ w = f (x, y, z, ..., t) $ by $ t $.

Example 4

Write the quotient and total increment of a function

Solution:

By the definition of the private increment, we find:

$ \ Delta _ (x) w = ((x + \ Delta x) + y) \ cdot z $ is the partial increment of the function $ w = f (x, y, z) $ with respect to $ x $

$ \ Delta _ (y) w = (x + (y + \ Delta y)) \ cdot z $ - partial increment of the function $ w = f (x, y, z) $ with respect to $ y $;

$ \ Delta _ (z) w = (x + y) \ cdot (z + \ Delta z) $ - partial increment of the function $ w = f (x, y, z) $ with respect to $ z $;

By the definition of the full increment, we find:

$ \ Delta w = ((x + \ Delta x) + (y + \ Delta y)) \ cdot (z + \ Delta z) $ - full increment of the function $ w = f (x, y, z) $.

Example 5

Calculate the quotient and total increment of the function $ w = xyz $ at the point $ (1; 2; 1) $ for $ \ Delta x = 0,1; \, \, \ Delta y = 0,1; \, \, \ Delta z = 0.1 $.

Solution:

By the definition of the private increment, we find:

$ \ Delta _ (x) w = (x + \ Delta x) \ cdot y \ cdot z $ - partial increment of the function $ w = f (x, y, z) $ with respect to $ x $

$ \ Delta _ (y) w = x \ cdot (y + \ Delta y) \ cdot z $ - partial increment of the function $ w = f (x, y, z) $ with respect to $ y $;

$ \ Delta _ (z) w = x \ cdot y \ cdot (z + \ Delta z) $ - partial increment of the function $ w = f (x, y, z) $ with respect to $ z $;

By the definition of the full increment, we find:

$ \ Delta w = (x + \ Delta x) \ cdot (y + \ Delta y) \ cdot (z + \ Delta z) $ - full increment of the function $ w = f (x, y, z) $.

Hence,

\ [\ Delta _ (x) w = (1 + 0.1) \ cdot 2 \ cdot 1 = 2.2 \] \ [\ Delta _ (y) w = 1 \ cdot (2 + 0.1) \ cdot 1 = 2.1 \] \ [\ Delta _ (y) w = 1 \ cdot 2 \ cdot (1 + 0.1) = 2.2 \] \ [\ Delta z = (1 + 0.1) \ cdot (2 + 0.1) \ cdot (1 + 0.1) = 1.1 \ cdot 2.1 \ cdot 1.1 = 2.541. \]

From a geometric point of view, the total increment of the function $ z = f (x, y) $ (by definition, $ \ Delta z = f (x + \ Delta x, y + \ Delta y) -f (x, y) $) is equal to the increment of the plot applicate function $ z = f (x, y) $ when passing from point $ M (x, y) $ to point $ M_ (1) (x + \ Delta x, y + \ Delta y) $ (Fig. 1).

Picture 1.

Let be NS- argument (independent variable); y = y (x)- function.

Let's take a fixed value of the argument x = x 0 and calculate the value of the function y 0 = y (x 0 ) ... Now we arbitrarily set increment (change) the argument and denote it  NS ( NS can be of any sign).

An incremental argument is a dot NS 0 +  NS... Suppose it also contains the value of the function y = y (x 0 +  NS)(see figure).

Thus, with an arbitrary change in the value of the argument, a change in the function is obtained, which is called incrementally function values:

and is not arbitrary, but depends on the form of the function and the value
.

Argument and function increments can be final, i.e. expressed as constant numbers, in this case they are sometimes called finite differences.

In economics, final increments are considered very often. For example, the table shows data on the length of the railway network of a certain state. Obviously, the net length increment is calculated by subtracting the previous value from the next one.

We will consider the length of the railway network as a function, the argument of which will be time (years).

By itself, the increment of the function (in this case, the length of the railway) of the network) poorly characterizes the change in the function. In our example, from the fact that 2,5>0,9 it cannot be concluded that the network grew faster in 2000-2003 years than in 2004 g, because the increment 2,5 refers to a three-year period, and 0,9 - by just one year. Therefore, it is quite natural that the increment of the function leads to the unit of change in the argument. Argument increments here are periods: 1996-1993=3; 2000-1996=4; 2003-2000=3; 2004-2003=1 .

We get what is called in the economic literature average annual growth.

It is possible to avoid the operation of converting the increment to the unit of change of the argument, if we take the values ​​of the function for the values ​​of the argument that differ by one, which is not always possible.

In mathematical analysis, in particular, in differential calculus, infinitesimal (BM) increments of an argument and a function are considered.

Differentiation of a function of one variable (derivative and differential) Derivative of a function

Argument and function increments at point NS 0 can be considered as comparable infinitesimal quantities (see topic 4, comparison of BM), i.e. BM of the same order.

Then their ratio will have a finite limit, which is defined as the derivative of the function at m NS 0 .

The limit of the ratio of the increment of the function to the BM increment of the argument at the point x = x 0 called derivative functions at a given point.

The symbolic designation of the derivative by a prime (or rather, by the Roman numeral I) was introduced by Newton. You can also use a subscript, which shows which variable is used to calculate the derivative, for example, ... Another notation is also widely used, proposed by the founder of the calculus of derivatives, the German mathematician Leibniz:
... You will learn more about the origin of this designation in the section Function differential and argument differential.

This number estimates speed changing the function passing through the point
.

Install geometric meaning derivative of the function at the point. For this purpose, we plot the function y = y (x) and mark the points on it that determine the change y (x) in the interim

The tangent to the graph of the function at the point M 0
we will consider the limiting position of the secant M 0 M on condition
(point M slides the function graph to a point M 0 ).

Consider
... Obviously,
.

If point M move along the graph of the function towards the point M 0 , then the value
will tend to a certain limit, which we denote
... Wherein.

Limiting angle  coincides with the angle of inclination of the tangent drawn to the graph of the function, incl. M 0 , so the derivative
numerically equal slope of the tangent at the specified point.

-

the geometric meaning of the derivative of a function at a point.

Thus, we can write down the equations of the tangent and normal ( normal Is a line perpendicular to the tangent) to the graph of the function at some point NS 0 :

Tangent -.

Normal -
.

Of interest are the cases when these straight lines are located horizontally or vertically (see topic 3, special cases of the position of a straight line on a plane). Then,

if
;

if
.

The definition of the derivative is called differentiation functions.

 If the function at the point NS 0 has a finite derivative, then it is called differentiable at this point. A function differentiable at all points of a certain interval is called differentiable on this interval.

 Theorem . If the function y = y (x) differentiable incl. NS 0 , then it is continuous at this point.

Thus, continuity- a necessary (but not sufficient) condition for the differentiability of a function.

Definition 1

If for each pair $ (x, y) $ of values ​​of two independent variables from a certain region a certain value of $ z $ is associated, then $ z $ is said to be a function of two variables $ (x, y) $. Notation: $ z = f (x, y) $.

In a relationship function$ z = f (x, y) $ consider the concepts of general (full) and partial increments of a function.

Let a function $ z = f (x, y) $ of two independent variables $ (x, y) $ be given.

Remark 1

Since the variables $ (x, y) $ are independent, one of them can change, while the other remains constant.

Let's give the variable $ x $ an increment of $ \ Delta x $, while keeping the value of the variable $ y $ unchanged.

Then the function $ z = f (x, y) $ will receive an increment, which will be called the partial increment of the function $ z = f (x, y) $ with respect to the variable $ x $. Designation:

Similarly, let's give the variable $ y $ an increment of $ \ Delta y $, while keeping the value of the variable $ x $ unchanged.

Then the function $ z = f (x, y) $ will receive an increment, which will be called the partial increment of the function $ z = f (x, y) $ with respect to the variable $ y $. Designation:

If the argument $ x $ is given the increment $ \ Delta x $, and the argument $ y $ - the increment $ \ Delta y $, then the full increment of the given function $ z = f (x, y) $ is obtained. Designation:

Thus, we have:

$ \ Delta _ (x) z = f (x + \ Delta x, y) -f (x, y) $ - partial increment of the function $ z = f (x, y) $ with respect to $ x $;

$ \ Delta _ (y) z = f (x, y + \ Delta y) -f (x, y) $ - partial increment of the function $ z = f (x, y) $ with respect to $ y $;

$ \ Delta z = f (x + \ Delta x, y + \ Delta y) -f (x, y) $ - full increment of the function $ z = f (x, y) $.

Example 1

Solution:

$ \ Delta _ (x) z = x + \ Delta x + y $ - partial increment of the function $ z = f (x, y) $ with respect to $ x $;

$ \ Delta _ (y) z = x + y + \ Delta y $ is the partial increment of the function $ z = f (x, y) $ with respect to $ y $.

$ \ Delta z = x + \ Delta x + y + \ Delta y $ - full increment of the function $ z = f (x, y) $.

Example 2

Calculate the quotient and total increment of the function $ z = xy $ at the point $ (1; 2) $ for $ \ Delta x = 0,1; \, \, \ Delta y = 0,1 $.

Solution:

By the definition of the private increment, we find:

$ \ Delta _ (x) z = (x + \ Delta x) \ cdot y $ - partial increment of the function $ z = f (x, y) $ with respect to $ x $

$ \ Delta _ (y) z = x \ cdot (y + \ Delta y) $ - partial increment of the function $ z = f (x, y) $ with respect to $ y $;

By the definition of the full increment, we find:

$ \ Delta z = (x + \ Delta x) \ cdot (y + \ Delta y) $ - full increment of the function $ z = f (x, y) $.

Hence,

\ [\ Delta _ (x) z = (1 + 0.1) \ cdot 2 = 2.2 \] \ [\ Delta _ (y) z = 1 \ cdot (2 + 0.1) = 2.1 \] \ [\ Delta z = (1 + 0.1) \ cdot (2 + 0.1) = 1.1 \ cdot 2.1 = 2.31. \]

Remark 2

The total increment of a given function $ z = f (x, y) $ is not equal to the sum of its partial increments $ \ Delta _ (x) z $ and $ \ Delta _ (y) z $. Mathematical notation: $ \ Delta z \ ne \ Delta _ (x) z + \ Delta _ (y) z $.

Example 3

Check assertion remark for function

Solution:

$ \ Delta _ (x) z = x + \ Delta x + y $; $ \ Delta _ (y) z = x + y + \ Delta y $; $ \ Delta z = x + \ Delta x + y + \ Delta y $ (obtained in example 1)

Find the sum of the partial increments of the given function $ z = f (x, y) $

\ [\ Delta _ (x) z + \ Delta _ (y) z = x + \ Delta x + y + (x + y + \ Delta y) = 2 \ cdot (x + y) + \ Delta x + \ Delta y. \]

\ [\ Delta _ (x) z + \ Delta _ (y) z \ ne \ Delta z. \]

Definition 2

If for each triple $ (x, y, z) $ of values ​​of three independent variables from a certain region a certain value of $ w $ is associated, then $ w $ is said to be a function of three variables $ (x, y, z) $ in this area.

Designation: $ w = f (x, y, z) $.

Definition 3

If for each collection $ (x, y, z, ..., t) $ of values ​​of independent variables from a certain region a certain value of $ w $ is associated, then $ w $ is said to be a function of the variables $ (x, y, z, ..., t) $ in this domain.

Notation: $ w = f (x, y, z, ..., t) $.

For a function of three or more variables, in the same way as for a function of two variables, partial increments are determined for each of the variables:

$ \ Delta _ (z) w = f (x, y, z + \ Delta z) -f (x, y, z) $ is the partial increment of the function $ w = f (x, y, z, ..., t ) $ by $ z $;

$ \ Delta _ (t) w = f (x, y, z, ..., t + \ Delta t) -f (x, y, z, ..., t) $ - partial increment of the function $ w = f (x, y, z, ..., t) $ by $ t $.

Example 4

Write the quotient and total increment of a function

Solution:

By the definition of the private increment, we find:

$ \ Delta _ (x) w = ((x + \ Delta x) + y) \ cdot z $ is the partial increment of the function $ w = f (x, y, z) $ with respect to $ x $

$ \ Delta _ (y) w = (x + (y + \ Delta y)) \ cdot z $ - partial increment of the function $ w = f (x, y, z) $ with respect to $ y $;

$ \ Delta _ (z) w = (x + y) \ cdot (z + \ Delta z) $ - partial increment of the function $ w = f (x, y, z) $ with respect to $ z $;

By the definition of the full increment, we find:

$ \ Delta w = ((x + \ Delta x) + (y + \ Delta y)) \ cdot (z + \ Delta z) $ - full increment of the function $ w = f (x, y, z) $.

Example 5

Calculate the quotient and total increment of the function $ w = xyz $ at the point $ (1; 2; 1) $ for $ \ Delta x = 0,1; \, \, \ Delta y = 0,1; \, \, \ Delta z = 0.1 $.

Solution:

By the definition of the private increment, we find:

$ \ Delta _ (x) w = (x + \ Delta x) \ cdot y \ cdot z $ - partial increment of the function $ w = f (x, y, z) $ with respect to $ x $

$ \ Delta _ (y) w = x \ cdot (y + \ Delta y) \ cdot z $ - partial increment of the function $ w = f (x, y, z) $ with respect to $ y $;

$ \ Delta _ (z) w = x \ cdot y \ cdot (z + \ Delta z) $ - partial increment of the function $ w = f (x, y, z) $ with respect to $ z $;

By the definition of the full increment, we find:

$ \ Delta w = (x + \ Delta x) \ cdot (y + \ Delta y) \ cdot (z + \ Delta z) $ - full increment of the function $ w = f (x, y, z) $.

Hence,

\ [\ Delta _ (x) w = (1 + 0.1) \ cdot 2 \ cdot 1 = 2.2 \] \ [\ Delta _ (y) w = 1 \ cdot (2 + 0.1) \ cdot 1 = 2.1 \] \ [\ Delta _ (y) w = 1 \ cdot 2 \ cdot (1 + 0.1) = 2.2 \] \ [\ Delta z = (1 + 0.1) \ cdot (2 + 0.1) \ cdot (1 + 0.1) = 1.1 \ cdot 2.1 \ cdot 1.1 = 2.541. \]

From a geometric point of view, the total increment of the function $ z = f (x, y) $ (by definition, $ \ Delta z = f (x + \ Delta x, y + \ Delta y) -f (x, y) $) is equal to the increment of the plot applicate function $ z = f (x, y) $ when passing from point $ M (x, y) $ to point $ M_ (1) (x + \ Delta x, y + \ Delta y) $ (Fig. 1).

Picture 1.

In life, we are not always interested in the exact values ​​of any quantities. Sometimes it is interesting to know the change in this value, for example, the average speed of the bus, the ratio of the amount of movement to the period of time, etc. To compare the value of a function at some point with the values ​​of the same function at other points, it is convenient to use concepts such as "function increment" and "argument increment".

The concepts of "function increment" and "argument increment"

Suppose x is some arbitrary point that lies in some neighborhood of the point x0. The increment of the argument at the point x0 is the difference x-x0. The increment is indicated as follows: ∆х.

• ∆x = x-x0.

Sometimes this value is also called the increment of the independent variable at the point x0. From the formula it follows: x = x0 + ∆x. In such cases, it is said that the initial value of the independent variable x0 received an increment ∆x.

If we change the argument, then the value of the function will also change.

• f (x) - f (x0) = f (x0 + ∆х) - f (x0).

The increment of the function f at the point x0, the difference f (x0 + ∆x) - f (x0) is called corresponding to the increment ∆x. The increment of a function is denoted as ∆f. Thus, we get, by definition:

• ∆f = f (x0 + ∆x) - f (x0).

Sometimes, ∆f is also called the increment of the dependent variable and ∆y is used to denote it if the function was, for example, y = f (x).

Geometric meaning of increment

Take a look at the following figure.

As you can see, the increment shows the change in the ordinate and abscissa of the point. And the ratio of the increment of the function to the increment of the argument determines the angle of inclination of the secant passing through the initial and final position of the point.

Consider examples of function and argument increments

Example 1. Find the increment of the argument ∆x and the increment of the function ∆f at the point x0, if f (x) = x 2, x0 = 2 a) x = 1.9 b) x = 2.1

Let's use the formulas given above:

a) ∆х = х-х0 = 1.9 - 2 = -0.1;

• ∆f = f (1.9) - f (2) = 1.9 2 - 2 2 = -0.39;

b) ∆x = x-x0 = 2.1-2 = 0.1;

• ∆f = f (2.1) - f (2) = 2.1 2 - 2 2 = 0.41.

Example 2. Calculate the increment ∆f for the function f (x) = 1 / x at the point x0, if the increment of the argument is equal to ∆x.

Again, we will use the formulas obtained above.

• ∆f = f (x0 + ∆x) - f (x0) = 1 / (x0-∆x) - 1 / x0 = (x0 - (x0 + ∆x)) / (x0 * (x0 + ∆x)) = - ∆x / ((x0 * (x0 + ∆x)).