When solving various problems of geometry, mechanics, physics and other branches of knowledge, the need arose using the same analytical process from this function y=f(x) get a new function called derivative function(or simply derivative) of a given function f(x) and is designated by the symbol
The process by which from a given function f(x) get a new feature f" (x), called differentiation and it consists of the following three steps: 1) give the argument x increment
x and determine the corresponding increment of the function
y = f(x+
x) -f(x); 2) make up a relation 
3) counting x constant and
x0, we find 
, which we denote by f" (x), as if emphasizing that the resulting function depends only on the value x, at which we go to the limit. Definition:
Derivative y " =f " (x)
given function y=f(x)
for a given x is called the limit of the ratio of the increment of a function to the increment of the argument, provided that the increment of the argument tends to zero, if, of course, this limit exists, i.e. finite. Thus, 
, or 
Note that if for some value x, for example when x=a, attitude 
at
x0 does not tend to the finite limit, then in this case they say that the function f(x) at x=a(or at the point x=a) has no derivative or is not differentiable at the point x=a.
2. Geometric meaning of the derivative.
Consider the graph of the function y = f (x), differentiable in the vicinity of the point x 0
f(x)
Let's consider an arbitrary straight line passing through a point on the graph of a function - point A(x 0, f (x 0)) and intersecting the graph at some point B(x;f(x)). Such a line (AB) is called a secant. From ∆ABC: AC = ∆x; ВС =∆у; tgβ=∆y/∆x.
Since AC || Ox, then ALO = BAC = β (as corresponding for parallel). But ALO is the angle of inclination of the secant AB to the positive direction of the Ox axis. This means tanβ = k - slope straight AB.
Now we will reduce ∆х, i.e. ∆х→ 0. In this case, point B will approach point A according to the graph, and secant AB will rotate. The limiting position of the secant AB at ∆x→ 0 will be a straight line (a), called the tangent to the graph of the function y = f (x) at point A.
If we go to the limit as ∆x → 0 in the equality tgβ =∆y/∆x, we get 
ortg =f "(x 0), since 
-angle of inclination of the tangent to the positive direction of the Ox axis 
, by definition of a derivative. But tg = k is the angular coefficient of the tangent, which means k = tg = f "(x 0).
So, the geometric meaning of the derivative is as follows:
Derivative of a function at point x 0 equal to the slope of the tangent to the graph of the function drawn at the point with the abscissa x 0 .
3. Physical meaning of the derivative.
Consider the movement of a point along a straight line. Let the coordinate of a point at any time x(t) be given. It is known (from a physics course) that the average speed over a period of time is equal to the ratio of the distance traveled during this period of time to the time, i.e.
Vav = ∆x/∆t. Let's go to the limit in the last equality as ∆t → 0.
lim Vav (t) = (t 0) - instantaneous speed at time t 0, ∆t → 0.
and lim = ∆x/∆t = x"(t 0) (by definition of derivative).
So, (t) =x"(t).
The physical meaning of the derivative is as follows: derivative of the functiony = f(x) at pointx 0 is the rate of change of the functionf(x) at pointx 0
The derivative is used in physics to find velocity from a known function of coordinates versus time, acceleration from a known function of velocity versus time.
(t) = x"(t) - speed,
a(f) = "(t) - acceleration, or
If the law of motion of a material point in a circle is known, then one can find the angular velocity and angular acceleration during rotational motion:
φ = φ(t) - change in angle over time,
ω = φ"(t) - angular velocity,
ε = φ"(t) - angular acceleration, or ε = φ"(t).
If the law of mass distribution of an inhomogeneous rod is known, then the linear density of the inhomogeneous rod can be found:
m = m(x) - mass,
x , l - length of the rod,
p = m"(x) - linear density.
Using the derivative, problems from the theory of elasticity and harmonic vibrations are solved. So, according to Hooke's law
F = -kx, x – variable coordinate, k – spring elasticity coefficient. Putting ω 2 =k/m, we obtain the differential equation of the spring pendulum x"(t) + ω 2 x(t) = 0,
where ω = √k/√m oscillation frequency (l/c), k - spring stiffness (H/m).
An equation of the form y" + ω 2 y = 0 is called the equation of harmonic oscillations (mechanical, electrical, electromagnetic). The solution to such equations is the function
y = Asin(ωt + φ 0) or y = Acos(ωt + φ 0), where
A - amplitude of oscillations, ω - cyclic frequency,
φ 0 - initial phase.
The operation of finding the derivative is called differentiation.
As a result of solving problems of finding derivatives of the simplest (and not very simple) functions by defining the derivative as the limit of the ratio of the increment to the increment of the argument, a table of derivatives appeared and exactly certain rules differentiation. The first to work in the field of finding derivatives were Isaac Newton (1643-1727) and Gottfried Wilhelm Leibniz (1646-1716).
Therefore, in our time, to find the derivative of any function, you do not need to calculate the above-mentioned limit of the ratio of the increment of the function to the increment of the argument, but you only need to use the table of derivatives and the rules of differentiation. The following algorithm is suitable for finding the derivative.
To find the derivative, you need an expression under the prime sign break down simple functions into components and determine what actions (product, sum, quotient) these functions are related. Further derivatives elementary functions we find in the table of derivatives, and the formulas for the derivatives of the product, sum and quotient are in the rules of differentiation. The derivative table and differentiation rules are given after the first two examples.
Example 1. Find the derivative of a function
Solution. From the rules of differentiation we find out that the derivative of a sum of functions is the sum of derivatives of functions, i.e.
From the table of derivatives we find out that the derivative of "x" is equal to one, and the derivative of sine is equal to cosine. We substitute these values into the sum of derivatives and find the derivative required by the condition of the problem:
Example 2. Find the derivative of a function
Solution. We differentiate as a derivative of a sum in which the second term has a constant factor; it can be taken out of the sign of the derivative:
If questions still arise about where something comes from, they are usually cleared up after familiarizing yourself with the table of derivatives and the simplest rules of differentiation. We are moving on to them right now.
Table of derivatives of simple functions
| 1. Derivative of a constant (number). Any number (1, 2, 5, 200...) that is in the function expression. Always equal to zero. This is very important to remember, as it is required very often | |
| 2. Derivative of the independent variable. Most often "X". Always equal to one. This is also important to remember for a long time | |
| 3. Derivative of degree. When solving problems, you need to convert non-square roots into powers. | |
| 4. Derivative of a variable to the power -1 | |
| 5. Derivative square root | |
| 6. Derivative of sine | |
| 7. Derivative of cosine |  |
| 8. Derivative of tangent |  |
| 9. Derivative of cotangent |  |
| 10. Derivative of arcsine |  |
| 11. Derivative of arccosine |  |
| 12. Derivative of arctangent |  |
| 13. Derivative of arc cotangent |  |
| 14. Derivative of the natural logarithm | |
| 15. Derivative of a logarithmic function |  |
| 16. Derivative of the exponent | |
| 17. Derivative of an exponential function |
Rules of differentiation
| 1. Derivative of a sum or difference |  |
| 2. Derivative of the product |  |
| 2a. Derivative of an expression multiplied by a constant factor | |
| 3. Derivative of the quotient |  |
| 4. Derivative of a complex function |  |
Rule 1.If the functions
are differentiable at some point, then the functions are differentiable at the same point
and
those. the derivative of an algebraic sum of functions is equal to the algebraic sum of the derivatives of these functions.
Consequence. If two differentiable functions differ by a constant term, then their derivatives are equal, i.e.
Rule 2.If the functions
are differentiable at some point, then their product is differentiable at the same point
and
those. The derivative of the product of two functions is equal to the sum of the products of each of these functions and the derivative of the other.
Corollary 1. The constant factor can be taken out of the sign of the derivative:
Corollary 2. The derivative of the product of several differentiable functions is equal to the sum of the products of the derivative of each factor and all the others.
For example, for three multipliers:
Rule 3.If the functions
differentiable at some point And , then at this point their quotient is also differentiableu/v , and
those. the derivative of the quotient of two functions is equal to a fraction, the numerator of which is the difference between the products of the denominator and the derivative of the numerator and the numerator and the derivative of the denominator, and the denominator is the square of the former numerator.
Where to look for things on other pages
When finding the derivative of a product and a quotient in real problems, it is always necessary to apply several differentiation rules at once, therefore more examples for these derivatives - in the article"Derivative of the product and quotient of functions".
Comment. You should not confuse a constant (that is, a number) as a term in a sum and as a constant factor! In the case of a term, its derivative is equal to zero, and in the case of a constant factor, it is taken out of the sign of the derivatives. This typical mistake, which occurs on initial stage studying derivatives, but as they solve several one- and two-part examples, the average student no longer makes this mistake.
And if, when differentiating a product or quotient, you have a term u"v, in which u- a number, for example, 2 or 5, that is, a constant, then the derivative of this number will be equal to zero and, therefore, the entire term will be equal to zero (this case is discussed in example 10).
Other common mistake - mechanical solution derivative of a complex function as a derivative of a simple function. That's why derivative of a complex function a separate article is devoted. But first we will learn to find derivatives of simple functions.
Along the way, you can’t do without transforming expressions. To do this, you may need to open the manual in new windows. Actions with powers and roots And Operations with fractions .
If you are looking for solutions to derivatives of fractions with powers and roots, that is, when the function looks like 
, then follow the lesson “Derivative of sums of fractions with powers and roots.”
If you have a task like 
, then you will take the lesson “Derivatives of simple trigonometric functions”.
Step-by-step examples - how to find the derivative
Example 3. Find the derivative of a function
Solution. We define the parts of the function expression: the entire expression represents a product, and its factors are sums, in the second of which one of the terms contains a constant factor. We apply the product differentiation rule: the derivative of the product of two functions is equal to the sum of the products of each of these functions by the derivative of the other:
Next, we apply the rule of differentiation of the sum: the derivative of the algebraic sum of functions is equal to the algebraic sum of the derivatives of these functions. In our case, in each sum the second term has a minus sign. In each sum we see both an independent variable, the derivative of which is equal to one, and a constant (number), the derivative of which is equal to zero. So, “X” turns into one, and minus 5 turns into zero. In the second expression, "x" is multiplied by 2, so we multiply two by the same unit as the derivative of "x". We obtain the following derivative values:
We substitute the found derivatives into the sum of products and obtain the derivative of the entire function required by the condition of the problem:
Example 4. Find the derivative of a function
Solution. We are required to find the derivative of the quotient. We apply the formula for differentiating the quotient: the derivative of the quotient of two functions is equal to a fraction, the numerator of which is the difference between the products of the denominator and the derivative of the numerator and the numerator and the derivative of the denominator, and the denominator is the square of the former numerator. We get:
We have already found the derivative of the factors in the numerator in example 2. Let us also not forget that the product, which is the second factor in the numerator in the current example, is taken with a minus sign:
If you are looking for solutions to problems in which you need to find the derivative of a function, where there is a continuous pile of roots and powers, such as, for example, 
, then welcome to class "Derivative of sums of fractions with powers and roots" .
If you need to learn more about the derivatives of sines, cosines, tangents and others trigonometric functions, that is, when the function looks like , then a lesson for you "Derivatives of simple trigonometric functions" .
Example 5. Find the derivative of a function
Solution. In this function we see a product, one of the factors of which is the square root of the independent variable, the derivative of which we familiarized ourselves with in the table of derivatives. Using the rule for differentiating the product and the tabular value of the derivative of the square root, we obtain:
Example 6. Find the derivative of a function
Solution. In this function we see a quotient whose dividend is the square root of the independent variable. Using the rule of differentiation of quotients, which we repeated and applied in example 4, and the tabulated value of the derivative of the square root, we obtain:
To get rid of a fraction in the numerator, multiply the numerator and denominator by .
Derivative of a function of one variable.
Introduction.
Real methodological developments intended for students of the Faculty of Industrial and Civil Engineering. They were compiled in relation to the mathematics course program in the section “Differential calculus of functions of one variable.”
The developments represent a single methodological guide, including: brief theoretical information; “standard” problems and exercises with detailed solutions and explanations for these solutions; test options.
There are additional exercises at the end of each paragraph. This structure of developments makes them suitable for independent mastery of the section with minimal assistance from the teacher.
§1. Definition of derivative.
Mechanical and geometric meaning
derivative.
The concept of derivative is one of the most important concepts of mathematical analysis. It arose back in the 17th century. The formation of the concept of derivative is historically associated with two problems: the problem of the speed of alternating motion and the problem of the tangent to a curve.
These problems, despite their different contents, lead to the same mathematical operation that must be performed on a function. This operation has received a special name in mathematics. It is called the operation of differentiation of a function. The result of the differentiation operation is called the derivative.
So, the derivative of the function y=f(x) at the point x0 is the limit (if it exists) of the ratio of the increment of the function to the increment of the argument 
at 
.
The derivative is usually denoted as follows: 
.
Thus, by definition
The symbols are also used to denote derivatives 
.
Mechanical meaning of derivative.
If s=s(t) is the law of rectilinear motion of a material point, then 
is the speed of this point at time t.
Geometric meaning derivative.
If the function y=f(x) has a derivative at the point 
, then the angular coefficient of the tangent to the graph of the function at the point 
equals 
.
Example.
Find the derivative of the function 
at the point 
=2:
1) Let's give it a point 
=2 increment 
. Notice, that.
2) Find the increment of the function at the point 
=2:
3) Let’s create the ratio of the increment of the function to the increment of the argument:
Let us find the limit of the ratio at 
:
.
Thus, 
.
§ 2. Derivatives of some
simplest functions.
The student needs to learn how to calculate derivatives of specific functions: y=x,y= 
and in generaly= 
.
Let's find the derivative of the function y=x.
those. (x)′=1.
Let's find the derivative of the function
Derivative 
Let 
Then
It is easy to notice a pattern in the expressions for the derivatives of the power function 
with n=1,2,3.
Hence,
. (1)
This formula is valid for any real n.
In particular, using formula (1), we have:
;
.
Example.
Find the derivative of the function
.
.
This function is a special case of a function of the form
at 
.
Using formula (1), we have
.
Derivatives of the functions y=sin x and y=cos x.
Let y=sinx.
Divide by ∆x, we get
Passing to the limit at ∆x→0, we have
Let y=cosx.
Passing to the limit at ∆x→0, we obtain
;
.
(2)
§3. Basic rules of differentiation.
Let's consider the rules of differentiation.
Theorem1 . If the functions u=u(x) and v=v(x) are differentiable at a given point x, then their sum is differentiable at this point, and the derivative of the sum is equal to the sum of the derivatives of the terms: (u+v)"=u"+v".(3 )
Proof: consider the function y=f(x)=u(x)+v(x).
The increment ∆x of the argument x corresponds to the increments ∆u=u(x+∆x)-u(x), ∆v=v(x+∆x)-v(x) of the functions u and v. Then the function y will increase
∆y=f(x+∆x)-f(x)=
=--=∆u+∆v.
Hence,
So, (u+v)"=u"+v".
Theorem2. If the functions u=u(x) and v=v(x) are differentiable at a given pointx, then their product is differentiable at the same point. In this case, the derivative of the product is found by the following formula: (uv)"=u"v+uv". ( 4)
Proof: Let y=uv, where u and v are some differentiable functions of x. Let's give x an increment of ∆x; then u will receive an increment of ∆u, v will receive an increment of ∆v, and y will receive an increment of ∆y.
We have y+∆y=(u+∆u)(v+∆v), or
y+∆y=uv+u∆v+v∆u+∆u∆v.
Therefore, ∆y=u∆v+v∆u+∆u∆v.
From here 
Passing to the limit at ∆x→0 and taking into account that u and v do not depend on ∆x, we will have
Theorem 3. The derivative of the quotient of two functions is equal to a fraction, the denominator of which is equal to the square of the divisor, and the numerator is the difference between the product of the derivative of the dividend and the divisor and the product of the dividend and the derivative of the divisor, i.e.
If 
That 
(5)
Theorem 4. The derivative of a constant is zero, i.e. if y=C, where C=const, then y"=0.
Theorem 5. The constant factor can be taken out of the sign of the derivative, i.e. if y=Cu(x), where С=const, then y"=Cu"(x).
Example 1.
Find the derivative of the function
.
This function has the form 
, whereu=x,v=cosx. Applying the differentiation rule (4), we find
.
Example 2.
Find the derivative of the function
.
Let's apply formula (5).
Here 
;
.
Tasks.
Find the derivatives of the following functions:
;
11)
2)
;
12)
;
3)
13)
4)
14)
5)
15)
6)
16)
7
)
17)
8)
18)
9)
19)
10)
20)
What is a derivative?
Definition and meaning of a derivative function
Many will be surprised by the unexpected placement of this article in my author’s course on the derivative of a function of one variable and its applications. After all, as it has been since school: the standard textbook first of all gives the definition of a derivative, its geometric, mechanical meaning. Next, students find derivatives of functions by definition, and, in fact, only then they perfect the technique of differentiation using derivative tables.
But from my point of view, the following approach is more pragmatic: first of all, it is advisable to UNDERSTAND WELL limit of a function, and, in particular, infinitesimal quantities. The fact is that the definition of derivative is based on the concept of limit, which is poorly considered in the school course. That is why a significant part of young consumers of the granite of knowledge do not understand the very essence of the derivative. Thus, if you have little knowledge of differential calculus or a wise brain for long years successfully got rid of this baggage, please start with function limits. At the same time, master/remember their solution.
The same practical sense dictates that it is advantageous first learn to find derivatives, including derivatives of complex functions. Theory is theory, but, as they say, you always want to differentiate. In this regard, it is better to work through the listed basic lessons, and maybe master of differentiation without even realizing the essence of their actions.
I recommend starting with the materials on this page after reading the article. The simplest problems with derivatives, where, in particular, the problem of the tangent to the graph of a function is considered. But you can wait. The fact is that many applications of the derivative do not require understanding it, and it is not surprising that the theoretical lesson appeared quite late - when I needed to explain finding increasing/decreasing intervals and extrema functions. Moreover, he was on the topic for quite a long time. Functions and graphs”, until I finally decided to put it earlier.
Therefore, dear teapots, do not rush to absorb the essence of the derivative like hungry animals, because the saturation will be tasteless and incomplete.
The concept of increasing, decreasing, maximum, minimum of a function
Many teaching aids lead to the concept of derivative using some practical problems, and I also came up with interesting example. Imagine that we are about to travel to a city that can be reached in different ways. Let’s immediately discard the curved winding paths and consider only straight highways. However, straight-line directions are also different: you can get to the city along a smooth highway. Or along a hilly highway - up and down, up and down. Another road goes only uphill, and another one goes downhill all the time. Extreme enthusiasts will choose a route through a gorge with a steep cliff and a steep climb.
But whatever your preferences, it is advisable to know the area or at least locate it topographic map. What if such information is missing? After all, you can choose, for example, a smooth path, but as a result stumble upon a ski slope with cheerful Finns. It is not a fact that a navigator or even a satellite image will provide reliable data. Therefore, it would be nice to formalize the relief of the path using mathematics.
Let's look at some road (side view): 
Just in case, I remind you of an elementary fact: travel happens from left to right. For simplicity, we assume that the function continuous in the area under consideration.
What are the features of of this schedule?
At intervals 
function increases, that is, each next value of it more previous one. Roughly speaking, the schedule is on down up(we climb the hill). And on the interval the function decreases– each next value less previous, and our schedule is on top down(we go down the slope).
Let's also pay attention to special points. At the point we reach maximum, that is exists such a section of the path where the value will be the largest (highest). At the same point it is achieved minimum, And exists its neighborhood in which the value is the smallest (lowest).
We will look at more strict terminology and definitions in class. about the extrema of the function, but for now let's study one more important feature: at intervals 
the function increases, but it increases at different speeds. And the first thing that catches your eye is that the graph soars up during the interval much more cool, than on the interval . Is it possible to measure the steepness of a road using mathematical tools?
Rate of change of function
The idea is this: let's take some value (read "delta x"), which we'll call argument increment, and let’s start “trying it on” to various points on our path: 
1) Let's look at the leftmost point: passing the distance, we climb the slope to a height (green line). The quantity is called function increment, and in in this case this increment is positive (the difference in values along the axis is Above zero). Let's create a ratio that will be a measure of the steepness of our road. Obviously, this is a very specific number, and since both increments are positive, then .
Attention! Designations are ONE symbol, that is, you cannot “tear off” the “delta” from the “X” and consider these letters separately. Of course, the comment also concerns the function increment symbol.
Let's explore the nature of the resulting fraction more meaningfully. Let us initially be at a height of 20 meters (at the left black point). Having covered the distance of meters (left red line), we will find ourselves at an altitude of 60 meters. Then the increment of the function will be 
meters (green line) and: . Thus, on every meter this section of the road height increases average by 4 meters...forgot your climbing equipment? =) In other words, the constructed relationship characterizes the AVERAGE RATE OF CHANGE (in this case, growth) of the function.
Note : The numerical values of the example in question correspond only approximately to the proportions of the drawing.
2) Now let's go the same distance from the rightmost black point. Here the rise is more gradual, so the increment (crimson line) is relatively small, and the ratio compared to the previous case will be very modest. Relatively speaking, 
meters and function growth rate is . That is, here for every meter of the path there are average half a meter of rise.
3) A little adventure on the mountainside. Let's look at the top black dot, located on the ordinate axis. Let's assume that this is the 50 meter mark. We overcome the distance again, as a result of which we find ourselves lower - at the level of 30 meters. Since the movement is carried out top down(in the “counter” direction of the axis), then the final the increment of the function (height) will be negative: 
meters (brown segment in the drawing). And in this case we are already talking about rate of decrease Features: 
, that is, for every meter of path of this section, the height decreases average by 2 meters. Take care of your clothes at the fifth point.
Now let's ask ourselves the question: what value of the “measuring standard” is best to use? It’s completely understandable, 10 meters is very rough. A good dozen hummocks can easily fit on them. Why are there bumps, there might be some down there? deep gorge, and after a few meters - its other side with a further steep rise. Thus, with a ten-meter we will not get an intelligible description of such sections of the path through the ratio .
From the above discussion the following conclusion follows: how less value , the more accurately we describe the road topography. Moreover, the following facts are true:
– For anyone lifting points 
you can select a value (even if very small) that fits within the boundaries of a particular rise. This means that the corresponding height increment will be guaranteed to be positive, and the inequality will correctly indicate the growth of the function at each point of these intervals.
- Likewise, for any slope point there is a value that will fit completely on this slope. Consequently, the corresponding increase in height is clearly negative, and the inequality will correctly show the decrease in the function at each point of the given interval.
– A particularly interesting case is when the rate of change of the function is zero: . Firstly, zero height increment () is a sign of a smooth path. And secondly, there are other interesting situations, examples of which you see in the figure. Imagine that fate has brought us to the very top of a hill with soaring eagles or the bottom of a ravine with croaking frogs. If you take a small step in any direction, the change in height will be negligible, and we can say that the rate of change of the function is actually zero. This is exactly the picture observed at the points.
Thus, we have come to an amazing opportunity to perfectly accurately characterize the rate of change of a function. After all mathematical analysis allows you to direct the increment of the argument to zero: , that is, make it infinitesimal.
As a result, another logical question arises: is it possible to find for the road and its schedule another function, which would let us know about all the flat sections, ascents, descents, peaks, valleys, as well as the rate of growth/decrease at each point along the way?
What is a derivative? Definition of derivative.
Geometric meaning of derivative and differential
Please read carefully and not too quickly - the material is simple and accessible to everyone! It’s okay if in some places something doesn’t seem very clear, you can always return to the article later. I will say more, it is useful to study the theory several times in order to thoroughly understand all the points (the advice is especially relevant for “techie” students who have higher mathematics plays a significant role in the educational process).
Naturally, in the very definition of the derivative at a point we replace it with:
What have we come to? And we came to the conclusion that for the function according to the law 
is put in accordance other function, which is called derivative function(or simply derivative).
The derivative characterizes rate of change functions How? The idea runs like a red thread from the very beginning of the article. Let's consider some point domain of definition functions Let the function be differentiable at a given point. Then:
1) If , then the function increases at the point . And obviously there is interval(even a very small one), containing a point at which the function grows, and its graph goes “from bottom to top”.
2) If , then the function decreases at the point . And there is an interval containing a point at which the function decreases (the graph goes “top to bottom”).
3) If , then infinitely close near a point the function maintains its speed constant. This happens, as noted, with a constant function and at critical points of the function, in particular at minimum and maximum points.
A bit of semantics. What does the verb “differentiate” mean in a broad sense? To differentiate means to highlight a feature. By differentiating a function, we “isolate” the rate of its change in the form of a derivative of the function. What, by the way, is meant by the word “derivative”? Function happened from function.
The terms are very successfully interpreted by the mechanical meaning of the derivative
:
Let us consider the law of change in the coordinates of a body, depending on time, and the function of the speed of movement of a given body. The function characterizes the rate of change of body coordinates, therefore it is the first derivative of the function with respect to time: . If the concept of “body movement” did not exist in nature, then there would be no derivative concept of "body speed".
The acceleration of a body is the rate of change of speed, therefore: 
. If the initial concepts of “body motion” and “body speed” did not exist in nature, then there would not exist derivative concept of “body acceleration”.