How to apply the concept of differential to approximate calculations. Application of differential in approximate calculations. Absolute and relative error of calculations
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 a function, the difference
is an infinitesimal value at . Expressing the increment of the function from equality (1), 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 more high order smaller than the first, since their ratio tends to zero as
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 the proportion of the increment this part makes up. Therefore, for small values (and for ) the increment of the function can be approximately replaced by its main part, i.e.
This main part of the increment of the function is called the differential of this function at the point x and denote
Hence,
(5)
So, the differential of the function 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– initial value of the argument,
The incremented value, then the derivative in the differential expression is taken at the starting point x; in formula (5) this is evident from the record, in formula (4) it is not.
The differential of a function can be written in another form:
Geometric meaning 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 the amount.
Differential properties. Invariance of differential shape
In this and the next paragraphs, we will consider each of the functions to be differentiable for all considered values of its arguments.
The differential has properties similar to those of the derivative:
(WITH - constant) (8)
(9)
(10)
(12)
Formulas (8) – (12) are obtained from the corresponding formulas for the derivative by multiplying both sides of each equality by .
Consider the differential complex function. Let be a complex function:
Differential
this function, using the formula for the derivative of a complex function, can be written in the form
But there is a differential function, 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, expressing 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(invariance) of the differential shape.
We emphasize that in formula (13) cannot be replaced by , since
for any function except linear.
Example 2. Write the differential of the function
in two ways, expressing it: through the differential of the intermediate variable and through the differential of the variable x. Check the matching of the resulting expressions.
Solution. Let's put
and the differential will be written in the form
Substituting into this equality
We get
Application of differential in approximate calculations
The approximate equality established in the first paragraph
allows you to use a differential for approximate calculations of function values.
Let us write down 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= log 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) will take the form
we get
(tabular value
).
Using the approximate value of a 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 the approximate number is absolute value 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 the table value of the root
behind exact number, we estimate using formulas (16) and (17) the absolute and relative errors of the approximate value:
23. The concept of differential function. Properties. Application of differential in approx.y calculations.
Concept of differential function
Let the function y=ƒ(x) have a nonzero derivative at the point x.
Then, according to the theorem about the connection between a function, its limit and an infinitesimal function, we can write у/х=ƒ"(x)+α, where α→0 at ∆х→0, or ∆у=ƒ"(x) ∆х+α ∆х.
Thus, the increment of the function ∆у is the sum of two terms ƒ"(x) ∆x and a ∆x, which are infinitesimal for ∆x→0. In this case, the first term is infinitely small function of the same order as ∆х, since 
and the second term is an infinitesimal function of a higher order than ∆x:
Therefore, the first term ƒ"(x) ∆x is called the main part of the increment functions ∆у.
Function differential y=ƒ(x) at point x is called main part its increment, equal to the product of the derivative of the function and the increment of the argument, and is denoted dу (or dƒ(x)):
dy=ƒ"(x) ∆x. (1)
The dу differential is also called first order differential. Let's find the differential of the independent variable x, i.e. the differential of the function y=x.
Since y"=x"=1, then, according to formula (1), we have dy=dx=∆x, i.e. the differential of the independent variable is equal to the increment of this variable: dx=∆x.
Therefore, formula (1) can be written as follows:
dy=ƒ"(х)dх, (2)
in other words, the differential of a function is equal to the product of the derivative of this function and the differential of the independent variable.
From formula (2) follows the equality dy/dx=ƒ"(x). Now the notation
the derivative dy/dx can be considered as the ratio of the differentials dy and dx.
Differentialhas the following main properties.
1. d(With)=0.
2. d(u+w-v)= du+dw-dv.
3. d(uv)=du·v+u·dv.
d(Withu)=Withd(u).
4. 
.
5. y= f(z), , ,
The form of the differential is invariant (unchanging): it is always equal to the product of the derivative of the function and the differential of the argument, regardless of whether the argument is simple or complex.
Applying differential to approximate calculations
As is already known, the increment ∆у of the function y=ƒ(x) at point x can be represented as ∆у=ƒ"(x) ∆х+α ∆х, where α→0 at ∆х→0, or ∆у= dy+α ∆х. Discarding the infinitesimal α ∆х of a higher order than ∆х, we obtain an approximate equality
∆у≈dy, (3)
Moreover, this equality is more accurate, the smaller ∆х.
This equality allows us to approximately calculate the increment of any differentiable function with great accuracy.
The differential is usually much simpler to find than the increment of a function, so formula (3) is widely used in computing practice.
24. Antiderivative function and indefiniteth integral.
THE CONCEPT OF A PRIMITIVE FUNCTION AND AN INDEMNITE INTEGRAL
Function F (X) is called antiderivative function for this function f (X) (or, in short, antiderivative this function f (X)) on a given interval, if on this interval . Example. The function is an antiderivative of the function on the entire number axis, since for any X. Note that, together with a function, an antiderivative for is any function of the form , where WITH- an arbitrary constant number (this follows from the fact that the derivative of a constant is equal to zero). This property also holds in the general case.
Theorem 1. If and are two antiderivatives for the function f
(X) in a certain interval, then the difference between them in this interval is equal to a constant number. From this theorem it follows that if any antiderivative is known F (X) of this function f
(X), then the entire set of antiderivatives for f
(X) is exhausted by functions F (X)
+ WITH. Expression F (X)
+ WITH, Where F (X) - antiderivative of the function f
(X) And WITH- an arbitrary constant, called indefinite integral
from function f
(X) and is denoted by the symbol, and f
(X) is called integrand function
;
- integrand
,
X - integration variable
;
∫
-
sign of the indefinite integral
. Thus, by definition 
If . The question arises: for everyone
functions f
(X) there is an antiderivative, and therefore an indefinite integral? Theorem 2. If the function f
(X) continuous
on [ a ; b], then on this segment for the function f
(X) there is an antiderivative
. Below we will talk about antiderivatives only for continuous functions. Therefore, the integrals we consider later in this section exist.
25. Properties of the indefiniteAndintegral. Integrals from basic elementary functions.
Properties of the indefinite integral
In the formulas below f And g- variable functions x, F- antiderivative of function f, a, k, C- constant values.
Integrals of elementary functions
List of integrals of rational functions
(the antiderivative of zero is a constant; within any limits of integration, the integral of zero is equal to zero)
List of integrals of logarithmic functions
List of integrals of exponential functions
List of integrals of irrational functions 
("long logarithm")
list of integrals of trigonometric functions , list of integrals of inverse trigonometric functions
26. Substitution methods variable, method of integration by parts in the indefinite integral.
Variable replacement method (substitution method)
The method of integration by substitution involves introducing a new integration variable (that is, substitution). In this case, the given integral is reduced to a new integral, which is tabular or reducible to it. Common methods there is no selection of substitutions. The ability to correctly determine substitution is acquired through practice.
Suppose we need to calculate the integral. Let us make the substitution where is a function that has a continuous derivative.
Then 
and based on the invariance property of the integration formula for the indefinite integral, we obtain integration formula by substitution:
Integration by parts
Integration by parts - applying the following formula for integration:
In particular, with the help n-multiple application of this formula we find the integral
where is a polynomial of degree.
30. Properties of a definite integral. Newton–Leibniz formula.
Basic properties of the definite integral
Properties of a definite integral
Newton–Leibniz formula.
Let the function f (x) is continuous on the closed interval [ a, b]. If F (x) - antiderivative functions f (x) on the[ a, b], That
Consider the widespread problem on approximate calculation of the value of a function using a differential.
Here and further we will talk about first-order differentials; for brevity, we will often simply say “differential”. The problem of approximate calculations using differentials has a strict solution algorithm, and, therefore, no special difficulties should arise. The only thing is that there are small pitfalls that will also be cleaned up. So feel free to dive in head first.
In addition, the section contains formulas for finding the absolute and relative errors of calculations. The material is very useful, since errors have to be calculated in other problems.
To successfully master the examples, you need to be able to find derivatives of functions at least at an intermediate level, so if you are completely at a loss with differentiation, please start with finding the derivative at a point and with finding the differential at the point. From technical means you will need a micro calculator with various mathematical functions. You can use the capabilities of MS Excel, but in this case it is less convenient.
The lesson consists of two parts:
– Approximate calculations using the differential value of a function of one variable at a point.
– Approximate calculations using full differential values of a function of two variables at a point.
The task under consideration is closely related to the concept of differential, but since we do not yet have a lesson on the meaning of derivatives and differentials, we will limit ourselves to a formal consideration of examples, which is quite enough to learn how to solve them.
Approximate calculations using the differential of a function of one variable
In the first paragraph, the function of one variable rules. As everyone knows, it is denoted by y or through f(x). For this task it is much more convenient to use the second notation. Let's move straight to a popular example that is often encountered in practice:
Example 1
Solution: Please copy down in your notebook the working formula for an approximate calculation using a differential:
Let's start to figure it out, everything is simple here!
The first step is to create a function. According to the condition, it is proposed to calculate the cube root of the number: , so the corresponding function has the form: .
We need to use the formula to find the approximate value.
Let's look at left side formulas, and the thought comes to mind that the number 67 must be represented in the form. What's the easiest way to do this? I recommend the following algorithm: calculate this value on a calculator:
– it turned out to be 4 with a tail, this is an important guideline for the solution.
As x 0 select a “good” value, so that the root is removed completely. Naturally this meaning x 0 should be as close as possible to 67.
In this case x 0 = 64. Indeed, .
Note: When with selectionx 0 there is still a difficulty, just look at the calculated value (in this case 
), take the nearest integer part (in this case 4) and raise it to the required power (in this case ). As a result, the desired selection will be made x 0 = 64.
If x 0 = 64, then the increment of the argument: .
So, the number 67 is represented as a sum 
First we calculate the value of the function at the point x 0 = 64. Actually, this has already been done earlier:
The differential at a point is found by the formula:
– You can also copy this formula into your notebook.
From the formula it follows that you need to take the first derivative:
And find its value at the point x 0:
.
Thus:
All is ready! According to the formula:
The approximate value found is quite close to the value 4.06154810045 calculated using a microcalculator.
Answer:
Example 2
Calculate approximately by replacing the increments of the function with its differential.
This is an example for independent decision. An approximate sample of the final design and an answer at the end of the lesson. For beginners, I recommend first calculating the exact value on a microcalculator to find out what number to take as x 0, and which one – for Δ x. It should be noted that Δ x V in this example will be negative.
Some may have wondered why this task is needed if everything can be calmly and more accurately calculated on a calculator? I agree, the task is stupid and naive. But I’ll try to justify it a little. Firstly, the task illustrates the meaning of the differential function. Secondly, in ancient times, a calculator was something like a personal helicopter in modern times. I myself saw how a computer the size of a room was thrown out of one of the institutes somewhere in 1985-86 (radio amateurs came running from all over the city with screwdrivers, and after a couple of hours only the case remained from the unit). We also had antiques in our physics department, although they were smaller in size – about the size of a desk. This is how our ancestors struggled with methods of approximate calculations. A horse-drawn carriage is also transport.
One way or another, the problem remains in the standard course of higher mathematics, and it will have to be solved. This is the main answer to your question =).
Example 3
Calculate approximately the value of a function using a differential 
at the point x= 1.97. Calculate a more accurate function value at a point x= 1.97 using a microcalculator, estimate the absolute and relative error of calculations.
In fact, this task can easily be reformulated as follows: “Calculate the approximate value 
using a differential"
Solution: We use the familiar formula:
In this case, a ready-made function is already given: 
. Once again, I would like to draw your attention to the fact that to denote a function, instead of “game” it is more convenient to use f(x).
Meaning x= 1.97 must be represented in the form x 0 = Δ x. Well, it’s easier here, we see that the number 1.97 is very close to “two”, so it suggests itself x 0 = 2. And, therefore: .
Let's calculate the value of the function at the point x 0 = 2:
Using formula , let's calculate the differential at the same point.
We find the first derivative:
And its meaning at the point x 0 = 2:
Thus, the differential at the point:
As a result, according to the formula:
The second part of the task is to find the absolute and relative error of the calculations.
Differential functions at a point 
called the main one, linear with respect to the increment of the argument 
part of the function increment 
, equal to the product of the derivative of the function at the point 
for the increment of the independent variable:
.
Hence the increment of the function 
different from its differential 
to an infinitesimal value and for sufficiently small values we can consider 
or
The given formula is used in approximate calculations, and the smaller 
, the more accurate the formula.
Example 3.1. Calculate approximately 
Solution. Consider the function 
. This power function and its derivative 
As 
you need to take a number that satisfies the following conditions:
Meaning 
known or fairly easily calculated;
Number 
should be as close to the number 33.2 as possible.
In our case, these requirements are satisfied by the number 
= 32, for which 
=
2,
= 33,2 -32 = 1,2.
Using the formula, we find the required number:
+
.
Example 3.2. Find the time it takes to double a bank deposit if the bank interest rate for the year is 5% per annum.
Solution. Over the course of a year, the contribution increases by 
once and for 
years, the contribution will increase by 
once. Now we need to solve the equation: 
=2. Taking logarithms, we get where 
. We obtain an approximate formula for calculating 
. Believing 
, we'll find 
and in accordance with the approximate formula. In our case 
And 
. From here. Because 
, find the time to double the contribution 
years.
Self-test questions
1. Give the definition of the differential of a function at a point.
2. Why is the formula used for calculations approximate?
3. What conditions must the number satisfy? 
included in the above formula?
Tasks for independent work
Calculate approximate value 
, replacing at the point 
function increment 
its differential.
Table 3.1
|
Option number |
|
|
|
4 .Studying functions and constructing their graphs
If a function of one variable is given as a formula 
, then the domain of its definition is such a set of values of the argument 
, on which the function values are defined.
Example 4.1. Function value 
are defined only for non-negative values of the radical expression: 
. Hence the domain of definition of the function is the half-interval, since the value of the trigonometric function 
satisfy the inequality: -1 
1.
Function 
called even, if for any values 
from its domain of definition the equality
,
And odd, if another relation is true:
.
In other cases the function is called function general view.
Example 4.4. Let
.
Let's check: . Thus, this function is even.
For function 
right. Hence this function is odd.
Sum of previous functions 
is a function of general form, since the function is not equal 
And 
.
Asymptote function graphics 
is a straight line that has the property that the distance from a point ( 
;
) of the plane up to this straight line tends to zero as the graph point moves indefinitely from the origin. There are vertical (Fig. 4.1), horizontal (Fig. 4.2) and oblique (Fig. 4.3) asymptotes.
Rice. 4.1. Schedule 
Rice. 4.2. Schedule 
Rice. 4.3. Schedule 
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 a point is infinite or does not exist), or at the ends of its domain of definition 
, If 
– finite numbers.
If the function 
is defined on the entire number line and there is a finite limit 
, or 
, then the straight line given by the equation 
, is a right-handed horizontal asymptote, and the straight line 
- left-sided horizontal asymptote.
If there are finite limits
And 
,
then it's straight 
is the slant asymptote of the graph of the function. Oblique asymptote can also be right-sided ( 
) or left-handed ( 
).
Function 
is called increasing on the set 
, if for any 
, such that 
>
, the inequality holds: 
>
(decreasing if: 
<
). 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 valid: if the derivative of a differentiable function inside the set 
is positive (negative), then the function increases (decreases) 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 
inequality holds 
(
)
. Function value at a point 
called maximum (minimum). The maximum and minimum functions are united 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 equals zero ( 
) or did not exist.
The points at which the derivative of a function is equal to zero are called stationary function points. There does not have to be an extremum of the function at a stationary point. To find extrema, it is necessary to additionally examine the stationary points of the function, for example, by using sufficient conditions for the extremum.
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, то
is 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 
inequality holds:
.
Fig.4.4. Graph of a convex function
If the second derivative of a twice differentiable function 
positive (negative) within the set 
, then the function is concave (convex) on the set 
.
The inflection point of the graph of a continuous function 
called the point separating the intervals in which the function is convex and concave.
Second derivative 
twice differentiable function at an inflection point 
is equal to zero, that is 
= 0.
If the second derivative when passing through a certain 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:
By analogy with the linearization of a function of one variable, when approximately calculating the values of a function of several variables that is differentiable at a certain point, one can replace its increment with a differential. Thus, you can 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. We'll 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 from u And v. Let's 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 functions under consideration have partial derivatives with respect to all their arguments.
Let's 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) u Let us divide both sides of equality (2.7) by Δ v, and equalities (2.8) – on Δ u→ and move to the limit, respectively, at Δ v→ 0 and Δ 0. Let us take into account that due to the continuity of functions And X at
. Hence,
Let
x
=
x(Let's consider some special cases.),
y
=
y(Let's consider some special cases.).
t f
(x,
y)
Then the function Let's consider some special cases. is actually a function of one variable X And X, and it is possible, using formulas (2.9) and replacing the partial derivatives in them u
By v And Let's consider some special cases. to ordinary derivatives with respect to x(Let's consider some special cases.)
And
y(Let's consider some special cases.)
(of course, provided that the functions are differentiable 
:
(2.10)
) , get the expression for Let's consider some special cases. Let us now assume that as X acts as a variable X And X, that is related by the relation y = y (x). f In this case, as in the previous case, the function is a function of one variable X. Let's consider some special cases.
=
x
Using formula (2.10) with 
and given that
.
(2.11)
, we get that f Let us pay attention to the fact that this formula contains two derivatives of the function X by argument : 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: X By X(In the final result we substitute expressions for u And v).
as functions z = Let's find the complete derivative of the function x + y sin ( y = ²), where x.
cos
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 By v:
(2.12)
Therefore, the differential form is preserved for arguments u And v same as for functions of these arguments X And X, that is, is invariant(unchangeable).
Implicit functions, conditions for their existence. Differentiation of implicit functions. Partial derivatives and differentials of higher orders, their properties.
Definition 3.1. Function X 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 X as a unique (and, moreover, continuous) function of X. For example, the equation of the ellipse
sets X as a two-valued function of X:
For 
The conditions for the existence of a unique 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 X as a single-valued function of X: y = f(x) ;
b) at x = x 0 this function takes the value X 0 : f (x 0 ) = y 0 ;
c) function f (x) continuous.
Let us find, if the specified conditions are met, the derivative of the function y = f (x) , and it is possible, using formulas (2.9) and replacing the partial derivatives in them X.
Theorem 3.2.
Let the function X 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 area D, containing a point (x,y), whose coordinates satisfy equation (3.1), and at this point 
. Then the function X from X has a derivative
(3.2)
Example. We'll find 
, If 
. We'll 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 variables X And X. 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 X and get eight partial derivatives of the 3rd order, etc. Let us define derivatives of higher orders as follows:
Definition 3.2.Partial derivativen -th order a function 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 vicinity, then at this point
(3.3)
Consequence. This property is true for derivatives of any order and for functions of any number of variables.