Final work in the form of the Unified State Exam for 11th graders, it necessarily contains tasks on calculating limits, intervals of decreasing and increasing derivatives of a function, searching for extremum points and drawing graphs. Good knowledge of this topic allows you to correctly answer several exam questions and not experience difficulties in further professional training.
Fundamentals of differential calculus - one of the main topics of mathematics modern school. She studies the use of the derivative to study the dependencies of variables - it is through the derivative that one can analyze the increase and decrease of a function without resorting to a drawing.
Comprehensive preparation of graduates for passing the Unified State Exam on educational portal“Shkolkovo” will help you deeply understand the principles of differentiation - understand the theory in detail, study examples of solutions typical tasks and try your hand at independent work. We will help you close gaps in knowledge - clarify your understanding of the lexical concepts of the topic and the dependencies of quantities. Students will be able to review how to find intervals of monotonicity, which means the derivative of a function rises or decreases on a certain segment when boundary points are and are not included in the intervals found.
Before you begin directly solving thematic problems, we recommend that you first go to the “Theoretical Background” section and repeat the definitions of concepts, rules and tabular formulas. Here you can read how to find and write down each interval of increasing and decreasing function on the derivative graph.
All information offered is presented in the most accessible form for understanding, practically from scratch. The website provides materials for perception and assimilation in several various forms– reading, video viewing and direct training under the guidance of experienced teachers. Professional teachers will tell you in detail how to find the intervals of increasing and decreasing derivatives of a function using analytical and graphical methods. During the webinars, you will be able to ask any question you are interested in, both on theory and on solving specific problems.
Having remembered the main points of the topic, look at examples of increasing the derivative of a function, similar to the tasks in the exam options. To consolidate what you have learned, take a look at the “Catalog” - here you will find practical exercises for independent work. The tasks in the section have been selected different levels difficulties taking into account the development of skills. For example, each of them is accompanied by solution algorithms and correct answers.
By choosing the "Constructor" section, students will be able to practice studying the increase and decrease of the derivative of a function on real options Unified State Examination, constantly updated taking into account latest changes and innovations.
Create a ratio and calculate the limit.
Where did it come from? table of derivatives and differentiation rules? Thanks to the only limit. It seems like magic, but in reality it is sleight of hand and no fraud. In class What is a derivative? I started looking at specific examples, where, using the definition, I found the derivatives of linear and quadratic function. For the purpose of cognitive warm-up, we will continue to disturb table of derivatives, honing the algorithm and technique solutions:
Example 1
Essentially, we need to prove the special case of the derivative power function, which usually appears in the table: .
Solution technically formalized in two ways. Let's start with the first, already familiar approach: the ladder starts with a plank, and the derivative function starts with the derivative at a point.
Let's consider some(specific) point belonging to domain of definition function in which there is a derivative. Let us set the increment at this point (of course, within the scopeo/o
-I) and compose the corresponding increment of the function:
Let's calculate the limit:
The uncertainty 0:0 is eliminated by a standard technique, considered back in the first century BC. Multiply the numerator and denominator by the conjugate expression 
:
The technique for solving such a limit is discussed in detail at introductory lesson about the limits of functions.
Since you can choose ANY point of the interval as quality, then, having made the replacement, we get:
Answer
Once again let's rejoice at logarithms:
Example 2
Find the derivative of a function using the definition of derivative
Solution: Let's consider a different approach to promoting the same task. It is exactly the same, but more rational in terms of design. The idea is to get rid of the subscript at the beginning of the solution and use the letter instead of the letter.
Let's consider arbitrary point belonging to domain of definition function (interval) and set the increment in it. But here, by the way, as in most cases, you can do without any reservations, since the logarithmic function is differentiable at any point in the domain of definition.
Then the corresponding increment of the function is:
Let's find the derivative:
The simplicity of the design is balanced by the confusion that may arise for beginners (and not only). After all, we are accustomed to the fact that the letter “X” changes in the limit! But here everything is different: – antique statue, and – a live visitor, briskly walking along the museum corridor. That is, “x” is “like a constant.”
I will comment on the elimination of uncertainty step by step:
(1) We use the property of the logarithm 
.
(2) In parentheses, divide the numerator by the denominator term by term.
(3) In the denominator, we artificially multiply and divide by “x” to take advantage of remarkable limit 
, while as infinitesimal stands out.
Answer: by definition of derivative:
Or in short:
I propose to construct two more table formulas yourself:
Example 3
IN in this case it is convenient to immediately lead the composed increment to common denominator. Approximate sample completing the assignment at the end of the lesson (first method).
Example 3:Solution
: consider some point
, belonging to the domain of definition of the function
. Let us set the increment at this point
and compose the corresponding increment of the function:
Let's find the derivative at the point
:
Since as a
you can select any point
function domain
, That
And
Answer
:
by definition of derivative
Example 4
Find derivative by definition
And here everything needs to be reduced to wonderful limit. The solution is formalized in the second way.
A number of other tabular derivatives. Full list can be found in a school textbook, or, for example, the 1st volume of Fichtenholtz. I don’t see much point in copying proofs of differentiation rules from books - they are also generated by the formula.
Example 4:Solution
, belonging to
, and set the increment in it
Let's find the derivative:
Using a wonderful limit
Answer
:
by definition
Example 5
Find the derivative of a function 
, using the definition of derivative
Solution: we use the first design style. Let's consider some point belonging to , and specify the increment of the argument at it. Then the corresponding increment of the function is:
Perhaps some readers have not yet fully understood the principle by which increments need to be made. Take a point (number) and find the value of the function in it: 
, that is, into the function instead of"X" should be substituted. Now we also take a very specific number and also substitute it into the function instead of"iksa": . We write down the difference, and it is necessary put in brackets completely.
Compiled function increment It can be beneficial to immediately simplify. For what? Facilitate and shorten the solution to a further limit.
We use formulas, open the brackets and shorten everything that can be shortened: 
The turkey is gutted, no problem with the roast:
As a result: 
Since we can choose any real number as a value, we make the replacement and get 
.
Answer: by definition.
For verification purposes, let’s find the derivative using differentiation rules and tables:
It is always useful and pleasant to know the correct answer in advance, so it is better to differentiate the proposed function in a “quick” way, either mentally or in a draft, at the very beginning of the solution.
Example 6
Find the derivative of a function by definition of derivative
This is an example for you to solve on your own. The result is obvious:
Example 6:Solution
: consider some point
, belonging to
, and set the increment of the argument in it
. Then the corresponding increment of the function is:
Let's calculate the derivative:
Thus: 
Because as
you can choose any real number, then
And
Answer
:
by definition.
Let's go back to style #2:
Example 7
Let's find out immediately what should happen. By differentiation rule complex function
:
Solution: consider an arbitrary point belonging to , set the increment of the argument at it and compose the increment of the function:
Let's find the derivative: 
(1) Use trigonometric formula
.
(2) Under the sine we open the brackets, under the cosine we present similar terms.
(3) Under the sine we reduce the terms, under the cosine we divide the numerator by the denominator term by term.
(4) Due to the oddness of the sine, we take out the “minus”. Under the cosine we indicate that the term .
(5) We carry out artificial multiplication in the denominator in order to use first wonderful limit. Thus, the uncertainty is eliminated, let’s tidy up the result.
Answer: by definition
As you can see, the main difficulty of the problem under consideration rests on the complexity of the limit itself + a slight uniqueness of the packaging. In practice, both methods of design occur, so I describe both approaches in as much detail as possible. They are equivalent, but still, in my subjective impression, it is more advisable for dummies to stick to option 1 with “X-zero”.
Example 8
Using the definition, find the derivative of the function
Example 8:Solution
: consider an arbitrary point
, belonging to
, let us set the increment in it
and compose the increment of the function:
Let's find the derivative:
We use the trigonometric formula
and the first remarkable limit:
Answer
:
by definition
Let's look at a rarer version of the problem:
Example 9
Find the derivative of the function at the point using the definition of derivative.
Firstly, what should be the bottom line? Number
Let's calculate the answer in the standard way: 
Solution: from the point of view of clarity, this task is much simpler, since the formula instead considers a specific value.
Let's set the increment at the point and compose the corresponding increment of the function:
Let's calculate the derivative at a point:
We use a very rare tangent difference formula 
and once again we reduce the solution to the first wonderful limit:
Answer: by definition of derivative at a point.
The problem is not so difficult to solve and “in general view" - it is enough to replace with or simply depending on the design method. In this case, it is clear that the result will not be a number, but a derived function.
Example 10
Using the definition, find the derivative of the function 
at a point (one of which may turn out to be infinite), which I’m talking about general outline already told on theoretical lesson about derivative.
Some piecewise given functions are also differentiable at the “junction” points of the graph, for example, catdog 
has a common derivative and a common tangent (x-axis) at the point. Curve, but differentiable by ! Those interested can verify this for themselves using the example just solved.
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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 mathematical analysis. It arose 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) – law rectilinear movement
material point, That 
is the speed of this point at time t.
Geometric meaning of derivative.
If the function y=f(x) has a derivative at the point 
, That slope tangent to the graph of a function at a point 
equals 
.
Example.
Find the derivative of the function 
at the point 
=2:
1) Let's give it a point 
=2 increment 
. Note 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 derivatives of a 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 pointx, then at this point their sum is also differentiable, 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 by the divisor and the product of the dividend by 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 C=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)
Problem B9 gives a graph of a function or derivative from which you need to determine one of the following quantities:
- The value of the derivative at some point x 0,
- Maximum or minimum points (extremum points),
- Intervals of increasing and decreasing functions (intervals of monotonicity).
The functions and derivatives presented in this problem are always continuous, making the solution much easier. Despite the fact that the task belongs to the section of mathematical analysis, even the weakest students can do it, since no deep theoretical knowledge is required here.
To find the value of the derivative, extremum points and monotonicity intervals, there are simple and universal algorithms - all of them will be discussed below.
Read the conditions of problem B9 carefully to avoid making stupid mistakes: sometimes you come across quite lengthy texts, but important conditions, which influence the course of the decision, there are few.
Calculation of the derivative value. Two point method
If the problem is given a graph of a function f(x), tangent to this graph at some point x 0, and it is required to find the value of the derivative at this point, the following algorithm is applied:
- Find two “adequate” points on the tangent graph: their coordinates must be integer. Let's denote these points A (x 1 ; y 1) and B (x 2 ; y 2). Write down the coordinates correctly - this is a key point in the solution, and any mistake here will lead to an incorrect answer.
- Knowing the coordinates, it is easy to calculate the increment of the argument Δx = x 2 − x 1 and the increment of the function Δy = y 2 − y 1 .
- Finally, we find the value of the derivative D = Δy/Δx. In other words, you need to divide the increment of the function by the increment of the argument - and this will be the answer.
Let us note once again: points A and B must be looked for precisely on the tangent, and not on the graph of the function f(x), as often happens. The tangent line will necessarily contain at least two such points - otherwise the problem will not be composed correctly.
Consider points A (−3; 2) and B (−1; 6) and find the increments:
Δx = x 2 − x 1 = −1 − (−3) = 2; Δy = y 2 − y 1 = 6 − 2 = 4.
Let's find the value of the derivative: D = Δy/Δx = 4/2 = 2.
Task. The figure shows a graph of the function y = f(x) and a tangent to it at the point with the abscissa x 0. Find the value of the derivative of the function f(x) at the point x 0 .
Consider points A (0; 3) and B (3; 0), find the increments:
Δx = x 2 − x 1 = 3 − 0 = 3; Δy = y 2 − y 1 = 0 − 3 = −3.
Now we find the value of the derivative: D = Δy/Δx = −3/3 = −1.
Task. The figure shows a graph of the function y = f(x) and a tangent to it at the point with the abscissa x 0. Find the value of the derivative of the function f(x) at the point x 0 .
Consider points A (0; 2) and B (5; 2) and find the increments:
Δx = x 2 − x 1 = 5 − 0 = 5; Δy = y 2 − y 1 = 2 − 2 = 0.
It remains to find the value of the derivative: D = Δy/Δx = 0/5 = 0.
From the last example, we can formulate a rule: if the tangent is parallel to the OX axis, the derivative of the function at the point of tangency is zero. In this case, you don’t even need to count anything - just look at the graph.
Calculation of maximum and minimum points
Sometimes, instead of a graph of a function, Problem B9 gives a graph of the derivative and requires finding the maximum or minimum point of the function. In this situation, the two-point method is useless, but there is another, even simpler algorithm. First, let's define the terminology:
- The point x 0 is called the maximum point of the function f(x) if in some neighborhood of this point the following inequality holds: f(x 0) ≥ f(x).
- The point x 0 is called the minimum point of the function f(x) if in some neighborhood of this point the following inequality holds: f(x 0) ≤ f(x).
In order to find the maximum and minimum points from the derivative graph, just follow these steps:
- Redraw the derivative graph, removing all unnecessary information. As practice shows, unnecessary data only interferes with the decision. Therefore, we note on coordinate axis zeros of the derivative - that's all.
- Find out the signs of the derivative on the intervals between zeros. If for some point x 0 it is known that f'(x 0) ≠ 0, then only two options are possible: f'(x 0) ≥ 0 or f'(x 0) ≤ 0. The sign of the derivative is easy to determine from the original drawing: if the derivative graph lies above the OX axis, then f'(x) ≥ 0. And vice versa, if the derivative graph lies below the OX axis, then f'(x) ≤ 0.
- Again we check the zeros and signs of the derivative. Where the sign changes from minus to plus is the minimum point. Conversely, if the sign of the derivative changes from plus to minus, this is the maximum point. Counting is always done from left to right.
This scheme only works for continuous functions - there are no others in problem B9.
Task. The figure shows a graph of the derivative of the function f(x) defined on the interval [−5; 5]. Find the minimum point of the function f(x) on this segment.
Let's get rid of unnecessary information and leave only the boundaries [−5; 5] and zeros of the derivative x = −3 and x = 2.5. We also note the signs:
Obviously, at the point x = −3 the sign of the derivative changes from minus to plus. This is the minimum point.
Task. The figure shows a graph of the derivative of the function f(x) defined on the interval [−3; 7]. Find the maximum point of the function f(x) on this segment.
Let's redraw the graph, leaving only the boundaries [−3; 7] and zeros of the derivative x = −1.7 and x = 5. Let us note the signs of the derivative on the resulting graph. We have:
Obviously, at the point x = 5 the sign of the derivative changes from plus to minus - this is the maximum point.
Task. The figure shows a graph of the derivative of the function f(x), defined on the interval [−6; 4]. Find the number of maximum points of the function f(x) belonging to the segment [−4; 3].
From the conditions of the problem it follows that it is enough to consider only the part of the graph limited by the segment [−4; 3]. Therefore, we build a new graph on which we mark only the boundaries [−4; 3] and zeros of the derivative inside it. Namely, points x = −3.5 and x = 2. We get:
On this graph there is only one maximum point x = 2. It is at this point that the sign of the derivative changes from plus to minus.
A small note about points with non-integer coordinates. For example, in the last problem the point x = −3.5 was considered, but with the same success we can take x = −3.4. If the problem is compiled correctly, such changes should not affect the answer, since the points “without a fixed place of residence” do not directly participate in solving the problem. Of course, this trick won’t work with integer points.
Finding intervals of increasing and decreasing functions
In such a problem, like the maximum and minimum points, it is proposed to use the derivative graph to find areas in which the function itself increases or decreases. First, let's define what increasing and decreasing are:
- A function f(x) is said to be increasing on a segment if for any two points x 1 and x 2 from this segment the following statement is true: x 1 ≤ x 2 ⇒ f(x 1) ≤ f(x 2). In other words, the larger the argument value, the larger the function value.
- A function f(x) is called decreasing on a segment if for any two points x 1 and x 2 from this segment the following statement is true: x 1 ≤ x 2 ⇒ f(x 1) ≥ f(x 2). Those. higher value argument corresponds to the smaller value of the function.
Let us formulate sufficient conditions for increasing and decreasing:
- In order for a continuous function f(x) to increase on the segment , it is sufficient that its derivative inside the segment be positive, i.e. f’(x) ≥ 0.
- In order for a continuous function f(x) to decrease on the segment , it is sufficient that its derivative inside the segment be negative, i.e. f’(x) ≤ 0.
Let us accept these statements without evidence. Thus, we obtain a scheme for finding intervals of increasing and decreasing, which is in many ways similar to the algorithm for calculating extremum points:
- Remove all unnecessary information. In the original graph of the derivative, we are primarily interested in the zeros of the function, so we will leave only them.
- Mark the signs of the derivative at the intervals between zeros. Where f’(x) ≥ 0, the function increases, and where f’(x) ≤ 0, it decreases. If the problem sets restrictions on the variable x, we additionally mark them on a new graph.
- Now that we know the behavior of the function and the constraints, it remains to calculate the quantity required in the problem.
Task. The figure shows a graph of the derivative of the function f(x) defined on the interval [−3; 7.5]. Find the intervals of decrease of the function f(x). In your answer, indicate the sum of the integers included in these intervals.
As usual, let's redraw the graph and mark the boundaries [−3; 7.5], as well as zeros of the derivative x = −1.5 and x = 5.3. Then we note the signs of the derivative. We have:
Since the derivative is negative on the interval (− 1.5), this is the interval of decreasing function. It remains to sum all the integers that are inside this interval:
−1 + 0 + 1 + 2 + 3 + 4 + 5 = 14.
Task. The figure shows a graph of the derivative of the function f(x), defined on the interval [−10; 4]. Find the intervals of increase of the function f(x). In your answer, indicate the length of the largest of them.
Let's get rid of unnecessary information. Let us leave only the boundaries [−10; 4] and zeros of the derivative, of which there were four this time: x = −8, x = −6, x = −3 and x = 2. Let’s mark the signs of the derivative and get the following picture:
We are interested in the intervals of increasing function, i.e. such where f’(x) ≥ 0. There are two such intervals on the graph: (−8; −6) and (−3; 2). Let's calculate their lengths:
l 1 = − 6 − (−8) = 2;
l 2 = 2 − (−3) = 5.
Since we need to find the length of the largest of the intervals, we write down the value l 2 = 5 as an answer.
(\large\bf Derivative of a function)
Consider the function y=f(x), specified on the interval (a, b). Let x- any fixed point of the interval (a, b), A Δx- an arbitrary number such that the value x+Δx also belongs to the interval (a, b). This number Δx called argument increment.
Definition. Function increment y=f(x) at the point x, corresponding to the argument increment Δx, let's call the number
Δy = f(x+Δx) - f(x).
We believe that Δx ≠ 0. Consider at a given fixed point x the ratio of the function increment at this point to the corresponding argument increment Δx
We will call this relation the difference relation. Since the value x we consider fixed, the difference ratio is a function of the argument Δx. This function is defined for all argument values Δx, belonging to some sufficiently small neighborhood of the point Δx=0, except for the point itself Δx=0. Thus, we have the right to consider the question of the existence of a limit of the specified function at Δx → 0.
Definition. Derivative of a function y=f(x) at a given fixed point x called the limit at Δx → 0 difference ratio, that is
Provided that this limit exists.
Designation. y′(x) or f′(x).
Geometric meaning of derivative: Derivative of a function f(x) at this point x equal to the tangent of the angle between the axis Ox and a tangent to the graph of this function at the corresponding point:
f′(x 0) = \tgα.
Mechanical meaning of derivative: The derivative of the path with respect to time is equal to the speed of rectilinear motion of the point:
Equation of a tangent to a line y=f(x) at the point M 0 (x 0 ,y 0) takes the form
y-y 0 = f′(x 0) (x-x 0).
The normal to a curve at some point is the perpendicular to the tangent at the same point. If f′(x 0)≠ 0, then the equation of the normal to the line y=f(x) at the point M 0 (x 0 ,y 0) is written like this:
The concept of differentiability of a function
Let the function y=f(x) defined over a certain interval (a, b), x- some fixed argument value from this interval, Δx- any increment of the argument such that the value of the argument x+Δx ∈ (a, b).
Definition. Function y=f(x) called differentiable at a given point x, if increment Δy this function at the point x, corresponding to the argument increment Δx, can be represented in the form
Δy = A Δx +αΔx,
Where A- some number independent of Δx, A α - argument function Δx, which is infinitesimal at Δx→ 0.
Since the product of two infinitesimal functions αΔx is an infinitesimal of a higher order than Δx(property of 3 infinitesimal functions), then we can write:
Δy = A Δx +o(Δx).
Theorem. In order for the function y=f(x) was differentiable at a given point x, it is necessary and sufficient that it has a finite derivative at this point. At the same time A=f′(x), that is
Δy = f′(x) Δx +o(Δx).
The operation of finding the derivative is usually called differentiation.
Theorem. If the function y=f(x) x, then it is continuous at this point.
Comment. From the continuity of the function y=f(x) at this point x, generally speaking, the differentiability of the function does not follow f(x) at this point. For example, the function y=|x|- continuous at a point x=0, but has no derivative.
Concept of differential function
Definition. Function differential y=f(x) the product of the derivative of this function and the increment of the independent variable is called x:
dy = y′ Δx, df(x) = f′(x) Δx.
For function y=x we get dy=dx=x′Δx = 1· Δx= Δx, that is dx=Δx- the differential of an independent variable is equal to the increment of this variable.
Thus, we can write
dy = y′ dx, df(x) = f′(x) dx
Differential dy and increment Δy functions y=f(x) at this point x, both corresponding to the same argument increment Δx, generally speaking, are not equal to each other.
Geometric meaning of differential: The differential of a function is equal to the increment of the ordinate of the tangent to the graph of this function when the argument is incremented Δx.
Rules of differentiation
Theorem. If each of the functions u(x) And v(x) differentiable at a given point x, then the sum, difference, product and quotient of these functions (quotient provided that v(x)≠ 0) are also differentiable at this point, and the formulas hold:
Consider the complex function y=f(φ(x))≡ F(x), Where y=f(u), u=φ(x). In this case u called intermediate argument, x - independent variable.
Theorem. If y=f(u) And u=φ(x) are differentiable functions of their arguments, then the derivative of a complex function y=f(φ(x)) exists and is equal to the product of this function with respect to the intermediate argument and the derivative of the intermediate argument with respect to the independent variable, i.e.
Comment. For a complex function that is a superposition of three functions y=F(f(φ(x))), the differentiation rule has the form
y′ x = y′ u u′ v v′ x,
where are the functions v=φ(x), u=f(v) And y=F(u)- differentiable functions of their arguments.
Theorem. Let the function y=f(x) increases (or decreases) and is continuous in some neighborhood of the point x 0. Let, in addition, this function be differentiable at the indicated point x 0 and its derivative at this point f′(x 0) ≠ 0. Then in some neighborhood of the corresponding point y 0 =f(x 0) the inverse is defined for y=f(x) function x=f -1 (y), and the indicated inverse function differentiable at the corresponding point y 0 =f(x 0) and for its derivative at this point y the formula is valid
Derivatives table
Invariance of the form of the first differential
Let's consider the differential of a complex function. If y=f(x), x=φ(t)- functions of their arguments are differentiable, then the derivative of the function y=f(φ(t)) expressed by the formula
y′ t = y′ x x′ t.
By definition dy=y′ t dt, then we get
dy = y′ t dt = y′ x · x′ t dt = y′ x (x′ t dt) = y′ x dx,
dy = y′ x dx.
So, we have proven
Property of invariance of the form of the first differential of a function: as in the case when the argument x is an independent variable, and in the case when the argument x itself is a differentiable function of the new variable, the differential dy functions y=f(x) is equal to the derivative of this function multiplied by the differential of the argument dx.
Application of differential in approximate calculations
We have shown that the differential dy functions y=f(x), generally speaking, is not equal to the increment Δy this function. However, up to an infinitesimal function of a higher order of smallness than Δx, the approximate equality is valid
Δy ≈ dy.
The ratio is called the relative error of the equality of this equality. Because Δy-dy=o(Δx), then the relative error of this equality becomes as small as desired with decreasing |Δх|.
Considering that Δy=f(x+δ x)-f(x), dy=f′(x)Δx, we get f(x+δ x)-f(x) ≈ f′(x)Δx or
f(x+δ x) ≈ f(x) + f′(x)Δx.
This approximate equality allows with error o(Δx) replace function f(x) in a small neighborhood of the point x(i.e. for small values Δx) linear function of the argument Δx, standing on the right side.
Higher order derivatives
Definition. Second derivative (or second order derivative) of a function y=f(x) is called the derivative of its first derivative.
Notation for the second derivative of a function y=f(x):
Mechanical meaning of the second derivative. If the function y=f(x) describes the law of motion of a material point in a straight line, then the second derivative f″(x) equal to the acceleration of a moving point at the moment of time x.
The third and fourth derivatives are determined similarly.
Definition. n th derivative (or derivative n-th order) functions y=f(x) is called the derivative of it n-1 th derivative:
y (n) =(y (n-1))′, f (n) (x)=(f (n-1) (x))′.
Designations: y″′, y IV, y V etc.