If the function
differentiable at the point 
,
then its increment can be represented as the sum of two terms
. These terms are infinitesimal functions at 
.The first term is linear with respect to 
,the second is an infinitesimal of a higher order than 
.Really,
.
Thus, the second term at 
tends to zero faster when finding the increment of the function 
the first term plays the main role 
or (since 
)
.
Definition
.
Main part of function increment
at the point 
, linear with respect to
,called differential
functions
at this point and is designateddyordf(x)
. (2)
Thus, we can conclude: the differential of the independent variable coincides with its increment, that is 
.
Relationship (2) now takes the form
(3)
Comment . Formula (3) for brevity is often written in the form
(4)
Geometric meaning of differential
Consider the graph of the differentiable function 
. Points 
and belong to the graph of the function. At the point M tangent drawn TO to the graph of a function whose angle is with the positive direction of the axis 
denote by 
. Let's draw straight lines MN
parallel to the axis Ox
And 
parallel to the axis Oy. The increment of the function is equal to the length of the segment 
. From a right triangle 
, in which 
, we get
The above considerations allow us to conclude:
Function differential
at the point 
is represented by the increment of the ordinate of the tangent to the graph of this function at its corresponding point 
.
Relationship between differential and derivative
Consider formula (4)
.
Let us divide both sides of this equality by dx, Then
.
Thus, the derivative of a function is equal to the ratio of its differential to the differential of the independent variable.
Often this attitude 
treated simply as a symbol denoting the derivative of a function at by argument X.
Convenient notations for the derivative are also:
,
and so on.
The entries are also used
,
,
especially convenient when taking the derivative of a complex expression.
2. Differential of sum, product and quotient.
Since the differential is obtained from the derivative by multiplying it by the differential of the independent variable, then, knowing the derivatives of the basic elementary functions, as well as the rules for finding derivatives, one can come to similar rules for finding differentials.
1 0 . The differential of the constant is zero
.
2 0 . The differential of an algebraic sum of a finite number of differentiable functions is equal to the algebraic sum of the differentials of these functions
3 0 . The differential of the product of two differentiable functions is equal to the sum of the products of the first function by the differential of the second and the second function by the differential of the first
.
Consequence. The constant multiplier can be taken out of the differential sign
.
Example. Find the differential of the function.
Solution: Let's write this function in the form
,
then we get
.
4. Functions defined parametrically, their differentiation.
Definition
.
Function 
is said to be given parametrically if both variables X And
at
each are defined separately as single-valued functions of the same auxiliary variable - parametert:
Wheretvaries within 
.
Comment
. Parametric specification of functions is widely used in theoretical mechanics, where the parameter t
denotes time, and the equations 
represent the laws of change in projections of a moving point 
on the axis 
And 
.
Comment . Let us present the parametric equations of a circle and an ellipse.
a) Circle with center at the origin and radius r has parametric equations:
Where 
.
b) Let us write the parametric equations for the ellipse:
Where 
.
By excluding the parameter t From the parametric equations of the lines under consideration, one can arrive at their canonical equations.
Theorem
. If the function y from argument
x is given parametrically by the equations 
, Where 
And 
differentiable with respect totfunctions and 
, That
.
Example. Find the derivative of a function at from X, given by parametric equations.
Solution. 
.
24.1. 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 D у/D x=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. Moreover, the first term is an infinitesimal function of the same order as ∆x, 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 the point x is called the main part of 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. (24.1)
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 (24.1), 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 (24.1) can be written as follows:
dy=ƒ"(х)dх, (24.2)
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 (24.2) 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.
<< Пример 24.1
Find the differential of the function ƒ(x)=3x 2 -sin(l+2x).
Solution: Using the formula dy=ƒ"(x) dx we find
dy=(3x 2 -sin(l+2x))"dx=(6x-2cos(l+2x))dx.
<< Пример 24.2
Find the differential of a function
Calculate dy for x=0, dx=0.1.
Solution:
Substituting x=0 and dx=0.1, we get
24.2. Geometric meaning of the differential function
Let's find out the geometric meaning of the differential.
To do this, let’s draw a tangent MT to the graph of the function y=ƒ(x) at the point M(x; y) and consider the ordinate of this tangent for the point x+∆x (see Fig. 138). In the figure ½ AM½ =∆х, |AM 1 |=∆у. From the right triangle MAV we have:
But, according to the geometric meaning of the derivative, tga=ƒ"(x). Therefore, AB=ƒ"(x) ∆x.
Comparing the obtained result with formula (24.1), we obtain dy=AB, i.e. the differential of the function y=ƒ(x) at point x is equal to the increment in the ordinate of the tangent to the graph of the function at this point, when x receives an increment ∆x.
This is the geometric meaning of the differential.
24.3 Basic theorems about differentials
The basic theorems about differentials can be easily obtained using the connection between the differential and the derivative of a function (dy=f"(x)dx) and the corresponding theorems about derivatives.
For example, since the derivative of the function y=c is equal to zero, then the differential of a constant value is equal to zero: dy=с"dx=0 dx=0.
Theorem 24.1. The differential of the sum, product and quotient of two differentiable functions is determined by the following formulas:
Let us prove, for example, the second formula. By definition of differential we have:
d(uv)=(uv)" dx=(uv" +vu" )dx=vu" dx+uv" dx=udv+vdu
Theorem 24.2. The differential of a complex function is equal to the product of the derivative of this function with respect to the intermediate argument and the differential of this intermediate argument.
Let y=ƒ(u) and u=φ(x) be two differentiable functions that form a complex function y=ƒ(φ(x)). Using the theorem on the derivative of a complex function, we can write
y" x =y" u u" x.
Multiplying both sides of this equality by dx, we learn y" x dx=y" u u" x dx. But y" x dx=dy and u" x dx=du. Consequently, the last equality can be rewritten as follows:
dy=y" u du.
Comparing the formulas dy=y" x dx and dy=y" u du, we see that the first differential of the function y=ƒ(x) is determined by the same formula regardless of whether its argument is an independent variable or is a function of another argument.
This property of a differential is called invariance (immutability) of the form of the first differential.
The formula dy=y" x dx in appearance coincides with the formula dy=y" u du, but there is a fundamental difference between them: in the first formula x is an independent variable, therefore dx=∆x, in the second formula there is a function of x , therefore, generally speaking, du≠∆u.
Using the definition of a differential and the basic theorems about differentials, it is easy to convert a table of derivatives into a table of differentials.
For example: d(cosu)=(cosu)" u du=-sinudu
24.4. Differential table
24.5. 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 the approximate equality
∆у≈dy, (24.3)
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 (24.3) is widely used in computing practice.
<< Пример 24.3
Find the approximate value of the increment of the function y=x 3 -2x+1 at x=2 and ∆x=0.001.
Solution: We apply formula (24.3): ∆у≈dy=(x 3 -2x+1)" ∆x=(3x 2 -2) ∆x.
So, ∆у» 0.01.
Let's see what error was made by calculating the differential of a function instead of its increment. To do this, we find ∆у:
∆у=((x+∆x) 3 -2(x+∆x)+1)-(x 3 -2x+1)=x 3 +3x 2 ∆x+3x (∆x) 2 +(∆x ) 3 -2x-2 ∆x+1-x 3 +2x-1=∆x(3x 2 +3x ∆x+(∆x) 2 -2);
The absolute error of the approximation is
|∆у-dy|=|0.010006-0.011=0.000006.
Substituting the values of ∆у and dy into equality (24.3), we obtain
ƒ(x+∆x)-ƒ(x)≈ƒ"(x)∆x
ƒ(х+∆х)≈ƒ(х)+ƒ"(х) ∆х. (24.4)
Formula (24.4) is used to calculate approximate values of functions.
<< Пример 24.4
Calculate approximately arctan(1.05).
Solution: Consider the function ƒ(x)=arctgx. According to formula (24.4) we have:
arctg(x+∆х)≈arctgx+(arctgx)" ∆х,
i.e. 
Since x+∆x=1.05, then for x=1 and ∆x=0.05 we get:
It can be shown that the absolute error of formula (24.4) does not exceed the value M (∆x) 2, where M is the largest value of |ƒ"(x)| on the segment [x;x+∆x].
<< Пример 24.5
What distance will a body travel during free fall on the Moon in 10.04 s from the start of the fall? Equation of free fall of a body
H=g l t 2 /2, g l =1.6 m/s 2.
Solution: We need to find H(10,04). Let's use the approximate formula (ΔH≈dH)
H(t+∆t)≈H(t)+H"(t) ∆t. At t=10 s and ∆t=dt=0.04 s, H"(t)=g l t, we find
Problem (for independent solution). A body with mass m=20 kg moves at speed ν=10.02 m/s. Calculate approximately the kinetic energy of the body
24.6. Higher order differentials
Let y=ƒ(x) be a differentiable function, and let its argument x be independent variable. Then its first differential dy=ƒ"(x)dx is also a function of x; the differential of this function can be found.
The differential of the differential of the function y=ƒ(x) is called her second differential(or second-order differential) and is denoted by d 2 y or d 2 ƒ(x).
So, by definition d 2 y=d(dy). Let's find the expression for the second differential of the function y=ƒ(x).
Since dx=∆х does not depend on x, then when differentiating we consider dx constant:
d 2 y=d(dy)=d(f"(x)dx)=(ƒ"(x)dx)" dx=f"(x)dx dx=f"(x)(dx) 2 i.e. .
d 2 y=ƒ"(х)dх 2. (24.5)
Here dx 2 stands for (dx) 2.
The third order differential is defined and found similarly
d 3 y=d(d 2 y)=d(ƒ"(x)dx 2)≈f"(x)(dx) 3.
And, in general, a differential of the nth order is a differential from a differential of the (n-1)th order: d n y=d(d n-l y)=f (n) (x)(dx) n .
From here we find that, In particular, for n=1,2,3
accordingly we get:
that is, the derivative of a function can be considered as the ratio of its differential of the appropriate order to the corresponding degree of the differential of the independent variable.
Note that all the above formulas are valid only if x is an independent variable. If the function y=ƒ(x), where x is function of some other independent variable, then differentials of the second and higher orders do not have the property of form invariance and are calculated using other formulas. Let us show this using the example of a second order differential.
Using the product differential formula (d(uv)=vdu+udv), we obtain:
d 2 y=d(f"(x)dx)=d(ƒ"(x))dx+ƒ"(x) d(dx)=ƒ"(x)dx dx+ƒ"(x) d 2 x , i.e.
d 2 y=ƒ"(x)dx 2 +ƒ"(x) d 2 x. (24.6)
Comparing formulas (24.5) and (24.6), we are convinced that in the case of a complex function, the second-order differential formula changes: the second term ƒ"(x) d 2 x appears.
It is clear that if x is an independent variable, then
d 2 x=d(dx)=d(l dx)=dx d(l)=dx 0=0
and formula (24.6) goes into formula (24.5).
<< Пример 24.6
Find d 2 y if y = e 3x and x is an independent variable.
Solution: Since y"=3e 3x, y"=9e 3x, then according to formula (24.5) we have d 2 y=9e 3x dx 2.
<< Пример 24.7
Find d 2 y if y=x 2 and x=t 3 +1 and t is an independent variable.
Solution: We use formula (24.6): since
y"=2x, y"=2, dx=3t 2 dt, d 2 x=6tdt 2 ,
That d 2 y=2dx 2 +2x 6tdt 2 =2(3t 2 dt) 2 +2(t 3 +1)6tdt 2 =18t 4 dt 2 +12t 4 dt 2 +12tdt 2 =(30t 4 +12t)dt 2
Another solution: y=x 2, x=t 3 +1. Therefore, y=(t 3 +1) 2. Then according to formula (24.5)
d 2 y=y ¢¢ dt 2,
d 2 y=(30t 4 +12t)dt 2 .
Differential function y=ƒ(x) at point x is called the main part of 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.
Main differentials:
The differential of a function has properties similar to those of the derivative.
- Differential constant equal to zero:
dc = 0, c = const. - Differential of the sum of differentiable functions equal to the sum of the differentials of the terms:
Consequence. If two differentiable functions differ by a constant term, then their differentials are equal
d(u+c) = du (c= const).
- Product differential of two differentiable functions is equal to the product of the first function and the differential of the second plus the product of the second by the differential of the first:
d(uv) = udv + vdu.
Consequence. The constant multiplier can be taken out of the differential sign
d(cu) = cdu (c = const).
- Differential of quotient u/v of two differentiable functions u = u(x) and v = v(x) is determined by the formula
- The property of independence of the form of a differential from the choice of an independent variable (invariance of the form of a differential): the differential of a function is equal to the product of the derivative and the differential of an argument independent of whether this argument is an independent variable or a function of another independent variable.
Derivatives and differentials of higher orders.
Let the derivative of some function f differentiable. Then the derivative of the derivative of this function is called second derivative functions f and is designated f". Thus,
f"(x) = (f"(x))" .
If differentiable ( n- 1)th derivative of the function f, then her n th derivative is called the derivative of ( n- 1)th derivative of the function f and is designated f(n). So,
f(n)(x) = (f(n-1)(x))" , n ϵ N, f(0)(x) = f(x).
Number n called order of derivative.
Differential n-th order functions f called differential from differential ( n- 1)th order of the same function. Thus,
d n f(x) = d(d n -1 f(x)), d 0 f(x) = f(x), n ϵ N.
If x is the independent variable, then
dx= const and d 2 x = d 3 x = ... = d n x = 0.
In this case the formula is valid
d n f(x) = f (n) (x)(dx)n.
Derivatives n-th order from the basic elementary functions
The formulas are valid
Application of derivatives to the study of functions.
Basic theorems for differentiation of functions:
Rolle's theorem
Let the function f: [a, b] → R is continuous on the segment [ a, b], and has a finite or infinite derivative within that segment. Let, in addition, f(a) = f(b). Then inside the segment [ a, b] there is a point ξ such that f"(ξ ) = 0.
Lagrange's theorem
If the function f: [a, b] → R is continuous on the segment [ a, b] and has a finite or infinite derivative at interior points of this segment, then such that f(b) - f(a) = f"(ξ )(b - a).
Cauchy's theorem
If each of the functions f And g is continuous on [ a, b] and has a finite or infinite derivative on ] a, b[and if, in addition, the derivative g"(x) ≠ 0 on ] a, b[, then such that the formula is valid
If you additionally require that g(a) ≠ g(b), then the condition g"(x) ≠ 0 can be replaced with a less stringent one:
Being inextricably linked, both of them have been actively used for several centuries in solving almost all problems that arose in the process of human scientific and technical activity.
The emergence of the concept of differential
The famous German mathematician Gottfried Wilhelm Leibniz, one of the creators (along with Isaac Newton) of differential calculus, was the first to explain what a differential is. Before this, mathematicians of the 17th century. a very fuzzy and vague idea was used of some infinitely small “indivisible” part of any known function, which represented a very small constant value, but not equal to zero, less than which the values of the function simply cannot be. From here it was only one step to the introduction of the concept of infinitesimal increments of arguments of functions and the corresponding increments of the functions themselves, expressed through the derivatives of the latter. And this step was taken almost simultaneously by the two above-mentioned great scientists.
Based on the need to solve pressing practical problems of mechanics, which were posed to science by rapidly developing industry and technology, Newton and Leibniz created general methods for finding the rate of change of functions (primarily in relation to the mechanical speed of a body along a known trajectory), which led to the introduction of such concepts as as the derivative and differential of a function, and also found an algorithm for solving the inverse problem of how to find the distance traveled using a known (variable) speed, which led to the emergence of the concept of integral.
In the works of Leibniz and Newton, the idea first appeared that differentials are the main parts of increments of functions Δy proportional to the increments of arguments Δx, which can be successfully used to calculate the values of the latter. In other words, they discovered that the increment of a function can be at any point (within the domain of its definition) expressed through its derivative as Δу = y"(x) Δх + αΔх, where α Δх is the remainder term tending to zero as Δх→ 0, much faster than Δx itself.
According to the founders of mathematical analysis, differentials are precisely the first terms in the expressions for increments of any functions. Not yet having a clearly formulated concept of the limit of sequences, they intuitively understood that the value of the differential tends to the derivative of the function as Δх→0 - Δу/Δх→ y"(x).
Unlike Newton, who was primarily a physicist and considered the mathematical apparatus as an auxiliary tool for the study of physical problems, Leibniz paid more attention to this toolkit itself, including a system of visual and understandable notations for mathematical quantities. It was he who proposed the generally accepted notation for the differentials of the function dy = y"(x)dx, the argument dx and the derivative of the function in the form of their ratio y"(x) = dy/dx.
Modern definition
What is a differential from the point of view of modern mathematics? It is closely related to the concept of increment of a variable. If the variable y first takes the value y = y 1 and then y = y 2, then the difference y 2 ─ y 1 is called the increment of y.
The increment can be positive. negative and equal to zero. The word “increment” is denoted by Δ, the notation Δу (read “delta y”) denotes the increment of the value y. so Δу = y 2 ─ y 1 .
If the value Δу of an arbitrary function y = f (x) can be represented in the form Δу = A Δх + α, where A has no dependence on Δх, i.e. A = const for a given x, and the term α for Δх→0 tends to it is even faster than Δx itself, then the first (“main”) term, proportional to Δx, is for y = f (x) a differential, denoted dy or df(x) (read “de igrek”, “de ef from x "). Therefore, differentials are the “main” components of function increments that are linear with respect to Δx.
Mechanical interpretation
Let s = f (t) be the distance of the rectilinearly moving vehicle from the initial position (t is the travel time). The increment Δs is the path of the point during the time interval Δt, and the differential ds = f" (t) Δt is the path that the point would have covered in the same time Δt if it had maintained the speed f"(t) achieved by time t . For an infinitesimal Δt, the imaginary path ds differs from the true Δs by an infinitesimal amount, which has a higher order relative to Δt. If the speed at moment t is not zero, then ds gives an approximate value of the small displacement of the point.
Geometric interpretation
Let line L be the graph of y = f(x). Then Δ x = MQ, Δу = QM" (see figure below). The tangent MN splits the segment Δy into two parts, QN and NM." The first is proportional to Δх and is equal to QN = MQ∙tg (angle QMN) = Δх f "(x), i.e. QN is the differential dy.
The second part NM" gives the difference Δу ─ dy, with Δх→0 the length NM" decreases even faster than the increment of the argument, i.e. its order of smallness is higher than that of Δх. In the case under consideration, for f "(x) ≠ 0 (the tangent is not parallel to OX), the segments QM" and QN are equivalent; in other words, NM" decreases faster (its order of smallness is higher) than the total increment Δу = QM". This can be seen in the figure (as M "approaches M, the segment NM" constitutes an ever smaller percentage of the segment QM").
So, graphically, the differential of an arbitrary function is equal to the increment of the ordinate of its tangent.
Derivative and differential
Coefficient A in the first term of the expression for the increment of a function is equal to the value of its derivative f "(x). Thus, the following relation holds - dy = f "(x)Δx, or df (x) = f "(x)Δx.
It is known that the increment of an independent argument is equal to its differential Δх = dx. Accordingly, we can write: f "(x) dx = dy.
Finding (sometimes called “solving”) differentials follows the same rules as for derivatives. A list of them is given below.
What is more universal: the increment of an argument or its differential
Some clarifications need to be made here. Representing a differential by the value f "(x)Δx is possible when considering x as an argument. But the function can be complex, in which x can be a function of some argument t. Then representing the differential by the expression f "(x)Δx is, as a rule, impossible; except for the case of linear dependence x = at + b.
As for the formula f "(x)dx = dy, then both in the case of an independent argument x (then dx = Δx) and in the case of a parametric dependence of x on t, it represents a differential.
For example, the expression 2 x Δx represents for y = x 2 its differential when x is the argument. Let us now put x = t 2 and consider t as an argument. Then y = x 2 = t 4.
This expression is not proportional to Δt and therefore now 2xΔx is not a differential. It can be found from the equation y = x 2 = t 4. It turns out to be equal to dy=4t 3 Δt.
If we take the expression 2xdx, then it represents the differential y = x 2 for any argument t. Indeed, for x = t 2 we obtain dx = 2tΔt.
This means 2xdx = 2t 2 2tΔt = 4t 3 Δt, i.e., the expressions of the differentials written in terms of two different variables coincided.
Replacing increments with differentials
If f "(x) ≠ 0, then Δу and dy are equivalent (for Δх→0); if f "(x) = 0 (which means dy = 0), they are not equivalent.
For example, if y = x 2, then Δу = (x + Δх) 2 ─ x 2 = 2xΔх + Δх 2, and dy = 2xΔх. If x=3, then we have Δу = 6Δх + Δх 2 and dy = 6Δх, which are equivalent due to Δх 2 →0, at x=0 the values Δу = Δх 2 and dy=0 are not equivalent.
This fact, together with the simple structure of the differential (i.e., linearity with respect to Δx), is often used in approximate calculations, under the assumption that Δy ≈ dy for small Δx. Finding the differential of a function is usually easier than calculating the exact value of the increment.
For example, we have a metal cube with an edge x = 10.00 cm. When heated, the edge lengthened by Δx = 0.001 cm. How much did the volume V of the cube increase? We have V = x 2, so dV = 3x 2 Δx = 3∙10 2 ∙0/01 = 3 (cm 3). The increase in volume ΔV is equivalent to the differential dV, so ΔV = 3 cm 3 . A full calculation would give ΔV = 10.01 3 ─ 10 3 = 3.003001. But in this result all the figures except the first are unreliable; this means that it doesn’t matter, you need to round it to 3 cm 3.
Obviously, this approach is useful only if it is possible to estimate the magnitude of the error introduced by it.
Function differential: examples
Let's try to find the differential of the function y = x 3 without finding the derivative. Let's give the argument an increment and define Δу.
Δу = (Δх + x) 3 ─ x 3 = 3x 2 Δх + (3xΔх 2 + Δх 3).
Here the coefficient A = 3x 2 does not depend on Δx, so the first term is proportional to Δx, while the other term 3xΔx 2 + Δx 3 at Δx→0 decreases faster than the increment of the argument. Therefore, the term 3x 2 Δx is the differential y = x 3:
dy=3x 2 Δх=3x 2 dx or d(x 3) = 3x 2 dx.
In this case, d(x 3) / dx = 3x 2.
Let us now find dy of the function y = 1/x through its derivative. Then d(1/x) / dx = ─1/x 2. Therefore dy = ─ Δx/x 2.
Differentials of basic algebraic functions are given below.
Approximate calculations using differential
It is often not difficult to calculate the function f (x), as well as its derivative f "(x) at x=a, but doing the same in the vicinity of the point x=a is not easy. Then the approximate expression comes to the rescue
f(a + Δх) ≈ f "(a)Δх + f(a).
It gives an approximate value of the function for small increments Δх through its differential f "(a)Δх.
Consequently, this formula gives an approximate expression for the function at the end point of a certain section of length Δx in the form of the sum of its value at the starting point of this section (x=a) and the differential at the same starting point. The error in this method of determining the value of a function is illustrated in the figure below.
However, the exact expression for the value of the function for x=a+Δх is also known, given by the finite increment formula (or, in other words, the Lagrange formula)
f(a+ Δх) ≈ f "(ξ) Δх + f(a),
where the point x = a+ ξ is located on the segment from x = a to x = a + Δx, although its exact position is unknown. The exact formula allows you to estimate the error of the approximate formula. If we put ξ = Δx /2 in the Lagrange formula, then although it ceases to be accurate, it usually gives a much better approximation than the original expression through the differential.
Estimating the error of formulas using a differential
In principle, they are inaccurate and introduce corresponding errors into the measurement data. They are characterized by a marginal or, in short, maximum error - a positive number that is obviously greater than this error in absolute value (or, in extreme cases, equal to it). The limit is the quotient of it divided by the absolute value of the measured quantity.
Let the exact formula y= f (x) be used to calculate the function y, but the value of x is the result of a measurement and therefore introduces an error into y. Then, to find the maximum absolute error │Δу│function y, use the formula
│Δу│≈│dy│=│ f "(x)││Δх│,
where │Δх│is the maximum error of the argument. The value │Δу│ should be rounded upward, because The very replacement of the calculation of the increment with the calculation of the differential is inaccurate.
