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 infinitesimal “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 urgent problems practical problems mechanics, which the rapidly developing industry and technology posed to science, Newton and Leibniz created general methods finding the rate of change of functions (primarily in relation to the mechanical speed of movement of a body along a known trajectory), which led to the introduction of concepts such as 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 yrek”, “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 at 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 equal to the value increments 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 differential expressions 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 exact value increments.
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, such an 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.
Hence, 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 of 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.
Concept and geometric meaning of differential
Definition. The differential of a function at some point x is the main, linear part of the increment of the function.
The differential of the function y = f(x) is equal to the product of its derivative and the increment of the independent variable x (argument).
It is written like this:
Geometric meaning of differential. The differential of the function y = f(x) is equal to the increment in the ordinate of the tangent S drawn to the graph of this function at the point M(x; y), when x (the argument) changes by an amount (see figure).
Why can a differential be used in approximate calculations?
The differential 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. You can verify this by mentally moving the perpendicular lowered from point P (see figure) to the Ox axis, closer to the origin. Therefore, for small values (and for ) the increment of the function can be approximately replaced by its main part, i.e.
ABOUT different forms differential records
The differential of the function at the point x is denoted by
Hence,
, (2)
since 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 is the initial value of the argument, and is the incremented value, then the derivative in the differential expression is taken at the initial point x; in formula (1) this is not visible from the record.
The differential of a function can be written in another form:
(4)
Differential properties
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:
(C is a constant value) (5)
(6)
(7)
(9)
Formulas (5) – (9) are obtained from the corresponding formulas for the derivative by multiplying both sides of each equality by .
Application of differential in approximate calculations
The approximate equality established in the second paragraph
allows you to use a differential for approximate calculations of function values.
Let us write down the approximate equality in more detail. Because
Absolute and relative errors of approximate calculations
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 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:
If the exact number is unknown, then
Sometimes, before applying formula (11), it is necessary to first transform the original value. Typically, this is done for two purposes. Firstly, it is necessary to ensure that the value is sufficiently small compared to , since the smaller , the more accurate the result of the approximate calculation. Secondly, it is desirable that the value be calculated simply.
24. Application of function differential to approximate calculations
Applying differential to approximate calculations
The concept of differential suggests that if any process is close to linear in the nature of its change, then the increment of the function differs little from the differential. In addition, if a function has a finite derivative at some point x, then its increment and differential are also infinitesimal as , tending to zero:
Since the function being differentiated is continuous,
Because the product of a bounded function and an infinitesimal one as DX tends to zero is an infinitesimal function.
Moreover, these two infinitesimal functions are equivalent:
Equivalence makes it possible, with small increments of the argument, to approximately calculate
What can this formula give? Let the values of and be calculated relatively simply at some point. Then at another point, not far from , the following representation is possible:
Here the question remains open about the accuracy of the result obtained. This circumstance reduces the value of this approximate calculation formula, but it is generally useful and widely used in practice.
Let's look at an example. IN right triangle legs a=5 m and b=12 m. What will be the hypotenuse of this triangle if leg a is reduced by 0.2 m (Fig. 11.5, a)?
Let's find the original length of the hypotenuse:
.
After reducing leg a by 0.2 m, the hypotenuse will be equal to (Fig. 11.5, a)
Let us now apply formula (11.16) to approximately find c in connection with the decrease in leg a, considering a function of the form:
(B=Const);
In both cases, we received an approximate value of the desired value. But in the first case, the error arises as a result of approximate calculations, and in the second, comparatively simpler case, due to the use of an approximate formula (an error caused by approximate calculations can also be added to it). Note that when the leg a decreases by 0.2 m, the hypotenuse c decreases by approximately 0.08 m, and the approximate values we obtained differ by only 0.001 m.
Let's consider another situation: in the same triangle we reduce the hypotenuse c by 0.2 m, leaving leg b unchanged (Fig. 11.5, b). Let us determine how leg A will change in this case:
25. Application of the derivative to the study of functions and plotting of graphs
If on a certain interval the graph of a function is a continuous line, in other words, a line that can be drawn from a sheet of paper without a pencil, then such a function is called continuous on this interval. There are also functions that are not continuous. As an example, consider the graph of a function that, on the intervals and [s; b] is continuous, but at the point
x = c is discontinuous and therefore not continuous over the entire segment. All the functions we study in the school mathematics course are continuous functions on each interval on which they are defined.
Note that if a function has a derivative on a certain interval, then it is continuous on this interval.
The reverse statement is incorrect. A function that is continuous on an interval may not have a derivative at some points on that interval. For example, the function
y = |log 2 x| continuous on the interval x > 0, but at the point x = 1 it has no derivative, due to the fact that at this point the graph of the tangent function does not have a tangent.
Let's look at plotting graphs using the derivative.
Graph the function f(x) = x 3 – 2x 2 + x.
1) This function is defined for all x € R.
2) Let us find the intervals of monotonicity of the function under consideration and its extremum points using the derivative. The derivative is equal to f "(x) = 3x 2 – 4x + 1. Let's find stationary points:
3x 2 – 4x + 1 = 0, whence x 1 = 1/3, x 2 = 1.
To determine the sign of the derivative, we expand square trinomial 3x 2 – 4x + 1 for multipliers:
f "(x) = 3(x – 1/3)(x – 1). Therefore, on intervals x< 1/3 и х >1 derivative is positive; This means that the function increases over these intervals.
The derivative is negative at 1/3< х < 1; следовательно, функция убывает на этом интервале.
The point x 1 = 1/3 is a maximum point, since to the right of this point the function decreases, and to the left it increases. At this point the value of the function is f (1/3) = (1/3) 3 – 2(1/3) 2 + 1/3 = 4/27.
The minimum point is the point x2 = 1, since to the left of this point the function decreases and to the right it increases; its value at this minimum point is f (1) = 0.
3) When constructing a graph, the intersection points of the graph with the coordinate axes are usually found. Since f(0) = 0, the graph passes through the origin. Solving the equation f(0) = 0, we find the points of intersection of the graph with the abscissa axis:
x 3 – 2x 2 + x = 0, x(x 2 – 2x + 1) = 0, x(x – 1) 2 = 0, whence x = 0, x = 1.
4) For a more accurate plotting, let’s find the function values at two more points: f(-1/2) = -9/8, f(2) = 2.
5) Using the results of the study (points 1 – 4), we build a graph of the function y = x 3 – 2x 2 + x.
To construct a graph of a function, one usually first examines the properties of this function using its derivative according to a scheme similar to the scheme for solving Problem 1.
Thus, when examining the properties of a function, you need to find:
1) the scope of its definition;
2) derivative;
3) stationary points;
4) intervals of increase and decrease;
5) extremum points and function values at these points.
It is convenient to record the results of the study in the form of a table. Then, using the table, a graph of the function is plotted. To construct a graph more accurately, the points of its intersection with the coordinate axes and, if necessary, several more points of the graph are usually found.
If we are faced with an even or odd function, then to construct its graph it is enough to study the properties and construct its graph for x > 0, and then reflect it symmetrically relative to the ordinate axis (origin). For example, analyzing the function f(x) = x + 4/x, we come to the conclusion that this function is odd: f(-x) = -x + 4/(-x) = -(x + 4/x ) = -f(x). Having completed all the points of the plan, we build a graph of the function for x > 0, and a graph of this function for x< 0 получаем посредством симметричного отражения графика при х >0 relative to the origin.
For brevity, solving problems on constructing graphs of functions, most of the reasoning is carried out orally.
We also note that when solving some problems we may be faced with the need to study a function not over the entire domain of definition, but only over a certain interval, for example, if we need to plot, say, the function f(x) = 1 + 2x 2 – x 4 on segment [-1; 2].
26.An antiderivative of a function. Indefinite integral and its properties
Definition of antiderivative.
An antiderivative of a function f(x) on the interval (a; b) is a function F(x) such that the equality holds for any x from the given interval.
If we take into account the fact that the derivative of the constant C is equal to zero, then the equality is true 
. Thus, the function f(x) has a set of antiderivatives F(x)+C, for an arbitrary constant C, and these antiderivatives differ from each other by an arbitrary constant value.
Definition of an indefinite integral.
The entire set of antiderivatives of the function f(x) is called the indefinite integral of this function and is denoted 
.
The expression is called the integrand, and f(x) is called the integrand. The integrand represents the differential of the function f(x).
The action of finding an unknown function given its differential is called indefinite integration, because the result of integration is not one function F(x), but a set of its antiderivatives F(x)+C.
Based on the properties of the derivative, it is possible to formulate and prove the properties of the indefinite integral (properties of the antiderivative).
1. 
The derivative of the integration result is equal to the integrand.
2.
The indefinite integral of the differential of a function is equal to the sum of the function itself and an arbitrary constant.
3. 
, where k is an arbitrary constant.
The coefficient can be taken out as a sign of the indefinite integral.
4.
The indefinite integral of the sum/difference of functions is equal to the sum/difference of the indefinite integrals of functions.
Intermediate equalities of the first and second properties of the indefinite integral are given for clarification.
To prove the third and fourth properties, it is enough to find the derivatives of the right-hand sides of the equalities: 
These derivatives are equal to the integrands, which is a proof due to the first property. It is also used in the last transitions.
Thus, the integration problem is the inverse of the differentiation problem, and there is a very close connection between these problems:
· the first property allows one to check integration. To check the correctness of the integration performed, it is enough to calculate the derivative of the result obtained. If the function obtained as a result of differentiation turns out to be equal to the integrand, this will mean that the integration was carried out correctly;
· the second property of the indefinite integral allows one to find its antiderivative using a known differential of a function. The direct calculation of indefinite integrals is based on this property.
Let's look at an example.
Find the antiderivative of the function whose value is equal to one at x = 1.
We know from differential calculus that 
(just look at the table of derivatives of basic elementary functions). Thus, 
. By the second property 
. That is, we have many antiderivatives. For x = 1 we get the value . According to the condition, this value must be equal to one, therefore, C = 1. The desired antiderivative will take the form .
If the table of derivatives of basic elementary functions is rewritten in the form of differentials, then from it, using the second property of the indefinite integral, a table of antiderivatives can be compiled.
Related information.
As you can see, to find the differential you need to multiply the derivative by dx. This allows you to immediately write down the corresponding table for differentials from the table of formulas for derivatives.
Total differential for a function of two variables:
The total differential for a function of three variables is equal to the sum of partial differentials: d f(x,y,z)=d x f(x,y,z)dx+d y f(x,y,z)dy+d z f(x,y,z)dz
Definition . A function y=f(x) is called differentiable at a point x 0 if its increment at this point can be represented as ∆y=A∆x + α(∆x)∆x, where A is a constant and α(∆x) – infinitesimal as ∆x → 0.
The requirement that a function be differentiable at a point is equivalent to the existence of a derivative at this point, and A=f’(x 0).
Let f(x) be differentiable at the point x 0 and f "(x 0)≠0, then ∆y=f'(x 0)∆x + α∆x, where α= α(∆x) →0 at ∆x →0. The quantity ∆y and each term on the right side are infinitesimal quantities for ∆x→0. 
, that is, α(∆x)∆x is an infinitesimal of a higher order than f’(x 0)∆x.
, that is, ∆y~f’(x 0)∆x. Consequently, f’(x 0)∆x represents the main and at the same time linear relative to ∆x part of the increment ∆y (linear - meaning containing ∆x to the first power). This term is called the differential of the function y=f(x) at the point x 0 and is denoted dy(x 0) or df(x 0). So, for arbitrary values of x
dy=f′(x)∆x. (1)
Set dx=∆x, then
dy=f′(x)dx. (2)
Example. Find derivatives and differentials of these functions.
a) y=4 tan2 x
Solution:
differential: 
b) 
Solution:
differential: 
c) y=arcsin 2 (lnx)
Solution: 
differential: 
G) 
Solution:
= 
differential: 
Example. For the function y=x 3 find an expression for ∆y and dy for some values of x and ∆x.
Solution. ∆y = (x+∆x) 3 – x 3 = x 3 + 3x 2 ∆x +3x∆x 2 + ∆x 3 – x 3 = 3x 2 ∆x+3x∆x 2 +∆x 3 ; dy=3x 2 ∆x (we took the main linear part ∆y relative to ∆x). IN in this caseα(∆x)∆x = 3x∆x 2 + ∆x 3 .
Since I have not explained (at the moment) what the derivative of a function is, then it makes no sense to explain what the differential of a function is. In its most primitive formulation, a differential is “almost the same as a derivative.”
The derivative of a function is most often denoted by .
The differential of a function is standardly denoted by (that’s how it’s read – “de igrek”)
The differential of a function of one variable is written in the following form:
Another recording option: 
The simplest task: Find the differential of a function
1) First stage. Let's find the derivative:
2) Second stage. Let's write down the differential:
The differential of a function of one or more variables is most often used to approximate calculations.
In addition to other problems with differentials, from time to time there is also a “pure” task of finding the differential of a function. In addition, as for a derivative, for a differential there is the concept of a differential at a point. And we will also consider such examples.
Example 7
Find the differential of a function 
Before finding a derivative or differential, it is always advisable to see if it is possible to somehow simplify the function (or function notation) further to differentiation? Let's look at our example. First, you can transform the root:
(the fifth root refers specifically to the sine).
Secondly, we notice that under the sine we have a fraction, which obviously needs to be differentiated. The formula for differentiating a fraction is very cumbersome. Is it possible to get rid of the fraction? In this case, we can divide the numerator by the denominator term by term:
The function is complex. It contains two embeddings: the sine is embedded under the degree, and the expression is embedded under the sine. Let's find the derivative using the differentiation rule complex function 
twice:
Let’s write down the differential, and again present it in its original “beautiful” form:
When the derivative is a fraction, the icon is usually “stuck” at the very end of the numerator (it can also be on the right at the level of the fraction line).
Example 8
Find the differential of a function
This is an example for you to solve on your own.
The following two examples show how to find a differential at a point.
Example 9
Calculate the differential of a function 
at the point
Let's find the derivative: 
Again, the derivative seems to have been found. But we still have to substitute a number into this piece of shit, so let’s simplify the result as much as possible:
The work was not in vain, we write down the differential: 
Now let's calculate the differential at the point: 
There is no need to put a unit in the differential icon; it is from a slightly different story.
Definition of differential
Consider the function \(y = f\left(x \right),\) which is continuous in the interval \(\left[ (a,b) \right].\) Suppose that at some point \((x_0) \ in \left[ (a,b) \right]\) the independent variable receives an increment \(\Delta x.\) The increment of the function \(\Delta y,\) corresponding to such a change in the argument \(\Delta x,\) is expressed by the formula \[\Delta y = \Delta f\left(((x_0)) \right) = f\left(((x_0) + \Delta x) \right) - f\left(((x_0)) \right) .\] For any differentiable function, the increment \(\Delta y\) can be represented as the sum of two terms: \[\Delta y = A\Delta x + \omicron\left((\Delta x) \right),\] where the first term (so-called main part increment) linearly depends on the increment \(\Delta x,\) and the second term has a higher order of smallness relative to \(\Delta x.\) The expression \(A\Delta x\) is called differential function and is denoted by the symbol \(dy\) or \(df\left(((x_0)) \right).\)
Let's look at this idea of splitting the increment of the function \(\Delta y\) into two parts using a simple example. Let a square with side \((x_0) = 1 \,\text(m)\,\) be given (Figure \(1\)). Its area is obviously equal to \[(S_0) = x_0^2 = 1 \,\text(m)^2.\] If the side of the square is increased by \(\Delta x = 1\,\text(cm),\ ) then the exact value of the area of the enlarged square will be \ i.e. the area increment \(\Delta S\) is equal to \[ (\Delta S = S - (S_0) = 1.0201 - 1 = 0.0201\,\text(m)^2 ) = (201\,\text( cm)^2.) \] Now imagine this increment \(\Delta S\) in this form: \[\require(cancel) (\Delta S = S - (S_0) = (\left(((x_0) + \Delta x) \right)^2) - x_0^2 ) = (\cancel(x_0^2) + 2(x_0)\Delta x + (\left((\Delta x) \right)^2) - \ cancel(x_0^2) ) = (2(x_0)\Delta x + (\left((\Delta x) \right)^2) ) = (A\Delta x + \omicron\left((\Delta x) \right) ) = (dy + o\left((\Delta x) \right).) \] So, the increment of the function \(\Delta S\) consists of the main part (function differential), which is proportional to \(\Delta x\) and equal to \ and a term of higher order of smallness, in turn equal to \[\omicron\left((\Delta x) \right) = (\left((\Delta x) \right)^2) = (0.01^2) = 0.0001\,\text(m)^2 = 1\,\text(cm)^2.\] In sum, both of these terms amount to a total increase in the area of the square equal to \(200 + 1 = 201\,\text(cm)^2.\)
Note that in in this example the coefficient \(A\) is equal to the value of the derivative of the function \(S\) at the point \((x_0):\) \ It turns out that for any differentiable function the following is true theorem :
The coefficient \(A\) of the main part of the increment of the function at the point \((x_0)\) is equal to the value of the derivative \(f"\left(((x_0)) \right)\) at this point, i.e. the increment \( \Delta y\) is expressed by the formula \[ (\Delta y = A\Delta x + \omicron\left((\Delta x) \right) ) = (f"\left(((x_0)) \right)\Delta x + \omicron\left((\Delta x) \right).) \] Dividing both sides of this equality by \(\Delta x \ne 0,\) we have \[ (\frac((\Delta y))( (\Delta x)) = A + \frac((\omicron\left((\Delta x) \right)))((\Delta x)) ) = (f"\left(((x_0)) \right ) + \frac((\omicron\left((\Delta x) \right)))((\Delta x)).) \] In the limit at \(\Delta x \to 0\) we obtain the value of the derivative at the point \((x_0):\) \[ (y"\left(((x_0)) \right) = \lim\limits_(\Delta x \to 0) \frac((\Delta y))((\Delta x)) ) = (A = f"\left(((x_0)) \right).) \] Here we took into account that for a small value \(\omicron\left((\Delta x) \right)\) of a higher order of smallness than \(\Delta x,\) the limit is equal to \[\lim\limits_(\Delta x \to 0) \frac((\omicron\left((\Delta x) \right)))( (\Delta x)) = 0.\] If we assume that differential of the independent variable \(dx\) is equal to its increment \(\Delta x:\) \ then from the relation \ it follows that \ i.e. the derivative of a function can be represented as the ratio of two differentials.
Geometric meaning of the differential function
Figure \(2\) schematically shows the breakdown of the increment of the function \(\Delta y\) into main part\(A\Delta x\) (function differential) and a term of higher order of smallness \(\omicron\left((\Delta x) \right)\).
The tangent \(MN\) drawn to the curve of the function \(y = f\left(x \right)\) at the point \(M\), as is known, has an angle of inclination \(\alpha\), the tangent of which is equal to the derivative : \[\tan \alpha = f"\left(((x_0)) \right).\] When the argument changes to \(\Delta x\), the tangent receives an increment \(A\Delta x.\) This is a linear increment , formed by the tangent, is precisely the differential of the function. The rest of the total increment \(\Delta y\) (segment \(N(M_1)\)) corresponds to a “nonlinear” addition with a higher order of smallness relative to \(\Delta x\ ).
Differential properties
Let \(u\) and \(v\) be functions of the variable \(x\). The differential has the following properties:
- The constant coefficient can be taken out of the differential sign:
\(d\left((Cu) \right) = Cdu\), where \(C\) is a constant number.
- Differential of the sum (difference) of functions:
\(d\left((u \pm v) \right) = du \pm dv.\)
- Differential constant value equal to zero:
\(d\left(C \right) = 0.\)
- The differential of the independent variable \(x\) is equal to its increment:
\(dx = \Delta x.\)
- The differential of a linear function is equal to its increment:
\(d\left((ax + b) \right) = \Delta \left((ax + b) \right) = a\Delta x.\)
- Differential of the product of two functions:
\(d\left((uv) \right) = du \cdot v + u \cdot dv.\)
- Differential of the quotient of two functions:
\(d\left((\large\frac(u)(v)\normalsize) \right) = \large\frac((du \cdot v - u \cdot dv))(((v^2))) \normalsize.\)
- The differential of a function is equal to the product of the derivative and the differential of the argument:
\(dy = df\left(x \right) = f"\left(x \right)dx.\)
Invariance of differential shape
Let's consider the composition of two functions \(y = f\left(u \right)\) and \(u = g\left(x \right),\) i.e. complex function \(y = f\left((g\left(x \right)) \right).\) Its derivative is determined by the expression \[(y"_x) = (y"_u) \cdot (u"_x) ,\] where the subscript denotes the variable by which differentiation is performed.
The differential of the “external” function \(y = f\left(u \right)\) is written in the form \ The differential of the “internal” function \(u = g\left(x \right)\) can be represented in a similar way: \ If we substitute \(du\) into the previous formula, we get \ Since \((y"_x) = (y"_u) \cdot (u"_x),\) then \ It can be seen that in the case of a complex function we got the same form of an expression for the differential of a function, as in the case of a “simple” function. This property is called. invariance of the differential form .
