First order differential in approximate calculations. Application of the differential in approximate calculations

Rice. 4.2. Schedule

Rice. 4.3. Schedule

The vertical asymptotes of a function should be sought either at discontinuity points of the second kind (at least one of the one-sided limits of the function at the point is infinite or does not exist), or at the ends of its domain of definition
, If
are final numbers.

If the function
is defined on the whole number line and there is a finite limit
, or
, then the straight line given by the equation
, is the right-hand horizontal asymptote, and the straight line
is the left-hand horizontal asymptote.

If there are limits

And
,

then straight
is the oblique asymptote of the graph of the function. The oblique asymptote can also be right handed (
) or left-handed (
).

Function
is called increasing on the set
, if for any
, such that >, the following inequality holds:
>
(decreasing if at the same time:
<
). A bunch of
in this case is called the monotonicity interval of the function.

The following sufficient condition for the monotonicity of a function is true: if the derivative of a differentiable function inside the set
is positive (negative), then the function is increasing (decreasing) on ​​this set.

Example 4.5. Given a function
. Find its intervals of increase and decrease.

Solution. Let's find its derivative
. It's obvious that >0 at >3 and <0 при<3. Отсюда функция убывает на интервале (
;3) and increases by (3;
).

Dot called a point local maximum (minimum) functions
, if in some neighborhood of the point the inequality
(
) . Function value at point called maximum (minimum). The maximum and minimum of a function are combined by a common name extremum functions.

In order for the function
had an extremum at the point it is necessary that its derivative at this point be equal to zero (
) or did not exist.

The points where the derivative of a function is zero are called stationary function points. At a stationary point, there should not necessarily be an extremum of the function. To find the extrema, it is required to additionally investigate the stationary points of the function, for example, by using sufficient extremum conditions.

The first of them is that if, when passing through a stationary point from left to right, the derivative of the differentiable function changes sign from plus to minus, then a local maximum is reached at the point. If the sign changes from minus to plus, then this is the minimum point of the function.

If the sign of the derivative does not change when passing through the point under study, then there is no extremum at this point.

The second sufficient condition for the extremum of a function at a stationary point uses the second derivative of the function: if
<0, тоis the maximum point, and if
>0, then - minimum point. At
=0 the question about the type of extremum remains open.

Function
called convex (concave)) on the set
, if for any two values
the following inequality holds:

.

Fig.4.4. Graph of a convex function

If the second derivative of a twice differentiable function
positive (negative) inside the set
, then the function is concave (convex) on the set
.

The inflection point of a graph of a continuous function
is called the point separating the intervals in which the function is convex and concave.

Second derivative
doubly differentiable function at an inflection point equals zero, that is
= 0.

If the second derivative when passing through some point changes its sign, then is the inflection point of its graph.

When studying a function and plotting its graph, it is recommended to use the following scheme:

Concept of differential

Let the function y = f(x) is differentiable for some value of the variable x. Therefore, at the point x there is a finite derivative

Then, by definition of the limit of the function, the difference

is an infinitesimal quantity at . Expressing from equality (1) the increment of the function, we obtain

(2)

(the value does not depend on , i.e., remains constant at ).

If , then on the right side of equality (2) the first term is linear with respect to . Therefore, when

it is infinitesimal of the same order of smallness as . The second term is an infinitesimal of a higher order of smallness than the first, since their ratio tends to zero at

Therefore, they say that the first term of formula (2) is the main, relatively linear part of the increment of the function; the smaller , the larger share of the increment is this part. Therefore, for small values ​​(and for ), the increment of the function can be approximately replaced by its principal part, i.e.

This main part of the increment of the function is called the differential of the given function at the point x and denote

Hence,

(5)

So the function differential y=f(x) is equal to the product of its derivative and the increment of the independent variable.

Comment. It must be remembered that if x is the initial value of the argument,

Accumulated value, then the derivative in the expression of the differential is taken at the starting point x; in formula (5) this can be seen from the record, in formula (4) it is not.

The differential of a function can be written in another form:

The geometric meaning of the differential. Function differential y=f(x) is equal to the increment of the ordinate of the tangent drawn to the graph of this function at the point ( x; y), when it changes x by size.

differential properties. Differential shape invariance

In this and the next sections, each of the functions will be considered differentiable for all considered values ​​of its arguments.

The differential has properties similar to those of the derivative:

(C is a constant value) (8)

(9)

(12)

Formulas (8) - (12) are obtained from the corresponding formulas for the derivative by multiplying both parts of each equality by .

Consider the differential of a complex function. Let be a complex function:

Differential

of this function, using the formula for the derivative of a complex function, can be written as

But there is a function differential, so

(13)

Here the differential is written in the same form as in formula (7), although the argument is not an independent variable, but a function. Therefore, the expression of the differential of a function as the product of the derivative of this function and the differential of its argument is valid regardless of whether the argument is an independent variable or a function of another variable. This property is called invariance(constancy) of the form of the differential.

We emphasize that in formula (13) cannot be replaced by , since

for any function except linear.

Example 2 Write function differential

in two ways, expressing it: through the differential of the intermediate variable and through the differential of the variable x. Check if the received expressions match.

Solution. Let's put

and the differential can be written as

Substituting into this equality

We get

Application of the differential in approximate calculations

The approximate equality established in the first section

allows you to use the differential for approximate calculations of function values.

Let us write the approximate equality in more detail. Because

Example 3 Using the concept of differential, calculate approximately ln 1.01.

Solution. The number ln 1.01 is one of the values ​​of the function y=ln x. Formula (15) in this case takes the form

Hence,

which is a very good approximation: table value ln 1.01 = 0.0100.

Example 4 Using the concept of differential, calculate approximately

Solution. Number
is one of the function values

Since the derivative of this function

then formula (15) takes the form

we get

(table value

).

Using the approximate value of the number, you need to be able to judge the degree of its accuracy. For this purpose, its absolute and relative errors are calculated.

The absolute error of an approximate number is equal to the absolute value of the difference between the exact number and its approximate value:

The relative error of an approximate number is the ratio of the absolute error of this number to the absolute value of the corresponding exact number:

Multiplying by 4/3, we find

Taking a table root value

for the exact number, we estimate by formulas (16) and (17) the absolute and relative errors of the approximate value:

By analogy with the linearization of a function of one variable, in the approximate calculation of the values ​​of a function of several variables, differentiable at some point, its increment can be replaced by a differential. Thus, it is possible to find the approximate value of a function of several (for example, two) variables using the formula:

Example.

Calculate approximate value
.

Consider the function
and choose X 0 = 1, at 0 = 2. Then Δ x = 1.02 - 1 = 0.02; Δ y= 1.97 - 2 = -0.03. Let's find
,

Therefore, given that f ( 1, 2) = 3, we get:

Differentiation of complex functions.

Let the function arguments z = f (x, y) u And v: x = x (u, v), y = y (u, v). Then the function f there is also a function u And v. Find out how to find its partial derivatives with respect to the arguments u And v, without making a direct substitution

z = f (x(u, v), y(u, v)). In this case, we will assume that all the considered functions have partial derivatives with respect to all their arguments.

Set the argument u increment Δ u, without changing the argument v. Then

If you set the increment only to the argument v, we get: . (2.8)

We divide both sides of equality (2.7) by Δ u, and equalities (2.8) on Δ v and pass to the limit, respectively, for Δ u→ 0 and ∆ v→ 0. In this case, we take into account that, due to the continuity of the functions X And at. Hence,

Let's consider some particular cases.

Let x = x(t), y = y(t). Then the function f (x, y) is actually a function of one variable t, and it is possible, using formulas (2.9) and replacing the partial derivatives in them X And at By u And v to the usual derivatives with respect to t(of course, under the condition of differentiability of the functions x(t) And y(t) ) , get an expression for :

(2.10)

Let us now assume that as t favored variable X, that is X And at related by the ratio y = y(x). In this case, as in the previous case, the function f is a function of one variable X. Using formula (2.10) for t = x and given that
, we get that

. (2.11)

Note that this formula contains two derivatives of the function f by argument X: on the left is the so-called total derivative, in contrast to the private one on the right.

Examples.

Then from formula (2.9) we obtain:

(In the final result we substitute the expressions for X And at how to functions u And v).

Let's find the total derivative of the function z = sin( x + y²), where y = cos x.

Invariance of the differential form.

Using formulas (2.5) and (2.9), we express the total differential of the function z = f (x, y) , Where x = x(u, v), y = y(u, v), through differentials of variables u And v:

(2.12)

Therefore, the form of the differential is preserved for the arguments u And v the same as for the functions of these arguments X And at, that is, is invariant(unchanged).

Implicit functions, conditions for their existence. Differentiation of implicit functions. Partial derivatives and differentials of higher orders, their properties.

Definition 3.1. Function at from X, defined by the equation

F(x,y)= 0 , (3.1)

called implicit function.

Of course, not every equation of the form (3.1) determines at as a single-valued (and, moreover, continuous) function of X. For example, the ellipse equation

sets at as a two-valued function of X:
For

The conditions for the existence of a single-valued and continuous implicit function are determined by the following theorem:

Theorem 3.1 (no proof). Let be:

a) in some neighborhood of the point ( X 0 , y 0 ) equation (3.1) defines at as a single-valued function of X: y = f(x) ;

b) when x = x 0 this function takes the value at 0 : f (x 0 ) = y 0 ;

c) function f (x) continuous.

Let us find, under the specified conditions, the derivative of the function y = f (x) By X.

Theorem 3.2. Let the function at from X is given implicitly by equation (3.1), where the function F (x, y) satisfies the conditions of Theorem 3.1. Let, in addition,
- continuous functions in some domain D containing point (x, y), whose coordinates satisfy equation (3.1), and at this point
. Then the function at from X has a derivative

(3.2)

Example. Let's find , If
. Let's find
,
.

Then from formula (3.2) we obtain:
.

Derivatives and differentials of higher orders.

Partial derivative functions z = f (x, y) are, in turn, functions of the variables X And at. Therefore, one can find their partial derivatives with respect to these variables. Let's designate them like this:

Thus, four partial derivatives of the 2nd order are obtained. Each of them can be differentiated again according to X and by at and get eight partial derivatives of the 3rd order, etc. We define higher-order derivatives as follows:

Definition 3.2.private derivativen -th order functions of several variables is called the first derivative of the derivative ( n– 1)th order.

Partial derivatives have an important property: the result of differentiation does not depend on the order of differentiation (for example,
). Let's prove this statement.

Theorem 3.3. If the function z = f (x, y) and its partial derivatives
defined and continuous at a point M (x, y) and in some of its neighborhood, then at this point

(3.3)

Consequence. This property is valid for derivatives of any order and for functions of any number of variables.

Approximate value of the function increment

For sufficiently small increments of the function is approximately equal to its differential, i.e. Dy » dy and, therefore,

Example 2 Find the approximate value of the increment of the function y= when the argument x changes from the value x 0 =3 to x 1 =3.01.

Solution. We use formula (2.3). To do this, we calculate

X 1 - x 0 \u003d 3.01 - 3 \u003d 0.01, then

Do » .

Approximate value of a function at a point

In accordance with the definition of the increment of the function y = f(x) at the point x 0, when the argument Dx (Dx®0) is incremented, Dy = f(x 0 + Dx) - f(x 0) and formula (3.3) can be written

f(x 0 + Dx) » f(x 0) + . (3.4)

Particular cases of formula (3.4) are the expressions:

(1 + Dx) n » 1 + nDx (3.4a)

ln(1 + Dx) » Dx (3.4b)

sinDx » Dx (3.4v)

tgDx » Dx (3.4g)

Here, as before, it is assumed that Dx®0.

Example 3 Find the approximate value of the function f (x) \u003d (3x -5) 5 at the point x 1 \u003d 2.02.

Solution. For calculations, we use formula (3.4). Let's represent x 1 as x 1 = x 0 + Dx. Then x 0 = 2, Dx = 0.02.

f(2.02)=f(2 + 0.02) » f(2) +

f(2) = (3 × 2 - 5) 5 = 1

15 × (3 × 2 - 5) 4 = 15

f(2.02) = (3 × 2.02 - 5) 5 » 1 + 15 × 0.02 = 1.3

Example 4 Calculate (1.01) 5 , , ln(1.02), ln .

Solution

1. Let us use formula (3.4a). To do this, we represent (1.01) 5 as (1+0.01) 5 .

Then, assuming Dx = 0.01, n = 5, we get

(1.01) 5 = (1 + 0.01) 5 » 1 + 5 × 0.01 = 1.05.

2. Representing in the form (1 - 0.006) 1/6, according to (3.4a), we obtain

(1 - 0.006) 1/6 "1 + .

3. Considering that ln(1.02) = ln(1 + 0.02) and assuming Dx=0.02, by formula (3.4b) we obtain

ln(1.02) = ln(1 + 0.02) » 0.02.

4. Similarly

ln = ln(1 - 0.05) 1/5 = .

Find approximate increments of functions

155. y = 2x 3 + 5 when the argument x changes from x 0 = 2 to x 1 = 2.001

156. y \u003d 3x 2 + 5x + 1 for x 0 \u003d 3 and Dx \u003d 0.001

157. y \u003d x 3 + x - 1 with x 0 \u003d 2 and Dx \u003d 0.01

158. y \u003d ln x at x 0 \u003d 10 and Dx \u003d 0.01

159. y \u003d x 2 - 2x with x 0 \u003d 3 and Dx \u003d 0.01

Find approximate values ​​of functions

160. y \u003d 2x 2 - x + 1 at x 1 \u003d 2.01

161. y \u003d x 2 + 3x + 1 at x 1 \u003d 3.02

162.y= at point x 1 = 1.1

163. y \u003d at the point x 1 \u003d 3.032

164. y \u003d at the point x 1 \u003d 3.97

165. y \u003d sin 2x at x 1 \u003d 0.015

Calculate approximately

166. (1,025) 10 167. (9,06) 2 168.(1,012) 3

169. (9,95) 3 170. (1,005) 10 171. (0,975) 4

172. 173. 174.

175. 176. 177.

178 ln(1.003×e) 179 ln(1.05) 5 180 ln

181.ln0.98 182.ln 183.ln(e 2 ×0.97)

Exploring functions and plotting

Signs of monotonicity of a function

Theorem 1 (necessary condition for increasing (decreasing) functions) . If a differentiable function y = f(x), xн(a; b) increases (decreases) on the interval (a; b), then for any x 0 н(a; b).

Theorem 2 (sufficient condition for increasing (decreasing) function) . If the function y = f(x), xн(a; b) has a positive (negative) derivative at each point of the interval (a; b), then this function increases (decreases) on this interval.

Function extremes

Definition 1. The point x 0 is called the maximum (minimum) point of the function y \u003d f (x) if for all x from some d-neighborhood of the point x 0 the inequality f (x)< f(x 0) (f(x) >f(x 0)) for x ¹ x 0 .

Theorem 3 (Farm) (necessary condition for the existence of an extremum) . If the point x 0 is the extremum point of the function y = f(x) and there is a derivative at this point, then

Theorem 4 (the first sufficient condition for the existence of an extremum) . Let the function y = f(x) be differentiable in some d-neighborhood of the point x 0 . Then:

1) if the derivative, when passing through the point x 0, changes sign from (+) to (-), then x 0 is the maximum point;

2) if the derivative, when passing through the point x 0, changes sign from (-) to (+), then x 0 is the minimum point;

3) if the derivative does not change sign when passing through the point x 0, then at the point x 0 the function does not have an extremum.

Definition 2. The points at which the derivative of a function vanishes or does not exist are called critical points of the first kind.

using the first derivative

1. Find the domain of definition D(f) of the function y = f(x).

2. Calculate the first derivative

3. Find critical points of the first kind.

4. Place the critical points in the domain D(f) of the function y = f(x) and determine the sign of the derivative in the intervals into which the critical points divide the domain of the function.

5. Select the maximum and minimum points of the function and calculate the values ​​of the function at these points.

Example 1 Investigate the function y \u003d x 3 - 3x 2 for an extremum.

Solution. In accordance with the algorithm for finding the extremum of a function using the first derivative, we have:

1. D(f): xн(-¥; ¥).

2. .

3. 3x 2 - 6x = 0 z x = 0, x = 2 are critical points of the first kind.

Derivative when passing through the point x = 0

changes sign from (+) to (-), hence it is a point

Maximum. When passing through the point x \u003d 2, it changes sign from (-) to (+), therefore this is the minimum point.

5. ymax = f(0) = 0 3 × 3 × 0 2 = 0.

Maximum coordinates (0; 0).

y min \u003d f (2) \u003d 2 3 - 3 × 2 2 \u003d -4.

Minimum coordinates (2; -4).

Theorem 5 (the second sufficient condition for the existence of an extremum) . If the function y = f(x) is defined and twice differentiable in some neighborhood of the point x 0 , and , then at the point x 0 the function f(x) has a maximum if and a minimum if .

Algorithm for finding the extremum of a function

using the second derivative

1. Find the domain of definition D(f) of the function y = f(x).

2. Calculate the first derivative

208. f(x) = 209. f(x) =

210. f(x) = x (log x - 2) 211. f(x) = x log 2 x + x + 4

ButΔ y = Δ f(X 0) is the function increment, and f (X 0) Δ x = df(X 0) is the function differential.

Therefore, we finally get

Theorem 1. Let the function y = f(X) at point x 0 has a finite derivative f (X 0)≠0. Then for sufficiently small values Δ x, approximate equality (1) takes place, which becomes arbitrarily exact for Δ x→ 0.

Thus, the differential of a function at a point X 0 is approximately equal to the increment of the function at that point.

Because then from equality (1) we obtain

at Δ x→ 0 (2)

at x→ X 0 (2  )

Since the equation of the tangent to the graph of the function y= f(x) at the point X 0 has the form

That approximate equalities (1)-(2) geometrically mean that near the point x=x 0 graph of the function y \u003d f(X) is approximately replaced by the tangent to the curve y = f(X).

For sufficiently small values, the total increment of the function and the differential differ insignificantly, i.e. . This circumstance is used for approximate calculations.

Example 1 Calculate approximately .

Solution. Consider a function and set X 0 = 4, X= 3.98. Then Δ x =x –x 0 = – 0,02, f(x 0)= 2. Since , then f (X 0)=1/4=0.25. Therefore, according to formula (2), we finally obtain: .

Example 2 Using the differential of the function, determine how approximately the value of the function will change y=f(X)=(3x 3 +5)∙tg4 x when decreasing the value of its argument X 0 = 0 by 0.01.

Solution. By virtue of (1), the change in the function y = f(X) at the point X 0 is approximately equal to the differential of the function at this point for sufficiently small values ​​of D x:

Calculate the differential of the function df(0). We have D x= -0.01. Because f (X)= 9x 2 tg4 x + ((3x 3 +5)/ cos 2 4 x)∙4, then f (0)=5∙4=20 and df(0)=f (0)∙Δ x= 20 (–0.01) = –0.2.

Therefore, Δ f(0) ≈ –0.2, i.e. when decreasing the value X 0 = 0 function argument by 0.01 function value itself y=f(X) will decrease by approximately 0.2.

Example 3 Let the demand function for a product be . It is required to find the quantity demanded for a product at a price p 0 \u003d 3 den. and establish how approximately the demand will increase with a decrease in the price of goods by 0.2 monetary units.

Solution. At a price p 0 \u003d 3 den. volume of demand Q 0 =D(p 0)=270/9=30 units goods. Price change Δ p= -0.2 den. units Due to (1) Δ Q (p 0) ≈ dQ (p 0). Let us calculate the differential of the volume of demand for the product.

Since then D (3) = –20 and

demand volume differential dQ(3) = D  (3)∙Δ p= –20 (–0.2) = 4. Therefore, Δ Q(3) ≈ 4, i.e. when the price of goods decreases p 0 \u003d 3 by 0.2 monetary units. the volume of demand for the product will increase by approximately 4 units of goods and will become equal to approximately 30 + 4 = 34 units of goods.

Questions for self-examination

1. What is called the differential of a function?

2. What is the geometric meaning of the differential of a function?

3. List the main properties of the function differential.

3. Write formulas that allow you to find the approximate value of a function using its differential.