About company
List of guide Izofatova Nina Mitrofanovna - Director

The history of the Kaliningrad College of Trade and Economics is a page in the history of the region, which has been written since 1946. Since then, more than 25,000 specialists have graduated from the college.

Since 2004, the college has become an experimental platform for the Moscow Institute for the Development of Secondary vocational education on the topic "Dissemination of European experience in the creation and organization of Adult Education Centers and Centers open education in the region". For ten years he has been a member of the Russian Marketing Association, has the status of a college of social orientation. The latter was assigned to the college by the regional administration for the constant support of socially unprotected students, teachers, pensioners, military personnel and their families, working teachers and employees.

Training of students at the Kaliningrad College of Trade and Economics is conducted at five faculties: technology and service, marketing management, law, economics and accounting, non-traditional forms of education. The educational field of the college includes sixteen specialties. These include cooking technology, food commerce, trade commerce, management, marketing, legal accountancy, banking, hospitality management, finance, tourism, and more.

Center in the college career guidance and preparation of applicants. At the faculty of non-traditional forms of education, you can not only improve your skills, but also acquire a new specialty on the job. The current Open Education Center is focused on providing assistance in vocational training in more than twenty specialties. Here you can improve your skills, undergo retraining. The methods are very diverse: business games, trainings, seminars, exercises, open meetings, conferences, project work All this allows listeners to assimilate the proposed material as much as possible.

Cooperation with Kaliningrad state university, Kaliningrad State technical university, Baltic state academy allows the college to train specialists whose knowledge becomes capital and the main resource economic development region. During the years of this interaction higher education on special faculty with a reduced period of study received more than two hundred graduates. All of them are in demand by the economic complex of the region, many have entered the elite of the business corps of the region.

The Kaliningrad College of Trade and Economics has established communication and is actively cooperating with Denmark, Sweden, Germany, Poland, and Finland. The team participates in international educational projects. Their subject matter is diverse, it includes such important topics as "Helping the Kaliningrad authorities in the development of small and medium-sized businesses", "Helping officers and unemployed members of their families in obtaining civilian specialties for subsequent employment", "Training teachers in andragogy and developing entrepreneurship training programs". activities in Kaliningrad” and the like.

In 1999, within the framework of an international project, thanks to the efforts of Lidia Ivanovna Motolyanets, Deputy Director for academic work- an imitation firm was created - an enterprise model that reflects the activities of a real trade organization, an effective specialized form of advanced training for personnel at all levels working in the field of small business.

The mission of the collective - to guarantee an education that will meet the needs of society, and contribute to the formation of a whole person - is being fully implemented. Kaliningrad College of Trade and Economics means professionalism, responsibility and prestige.

KTEK
PCC of Economics and Accounting

15 copies, 2006

Introduction. 4

The concept of a derivative. 5

Private derivatives. eleven

Inflection points. 16

Solution exercises. 17

Test. 20

Answers to exercises.. 21

Literature. 23

Introduction

f(x x, then called marginal product; If g(x) g(x) g′(x) called marginal cost.

For example, Let the function u=u(t) u while working t. ∆t=t 1 - t 0:

z cf. =

z cf. at ∆t→ 0: .

production costs K x, so we can write K=K(x) ∆x K(x+∆x). ∆x ∆K=K(x+∆x)- K(x).

Limit called

The concept of a derivative

The derivative of the function at the point x 0 is called the limit of the ratio of the increment of the function to the increment of the argument, provided that the increment of the argument tends to zero.

Derivative function notation:

That. a-priory:

Algorithm for finding the derivative:

Let the function y=f(x) continuous on the segment , x

1. Find the increment of the argument:

x is the new value of the argument

x0- initial value

2. Find the function increment:

f(x) is the new value of the function

f(x0)- function initial value

3. Find the ratio of the function increment to the argument increment:

4. Find the limit of the ratio found at

Find the derivative of the function based on the definition of the derivative.

Solution:

Let's give X increment Δx, then the new value of the function will be:

Let's find the increment of the function as the difference between the new and initial values ​​of the function:

Find the ratio of the function increment to the argument increment:

.

Let's find the limit of this ratio provided that:

Therefore, by the definition of the derivative: .

Finding the derivative of a function is called differentiation.

Function y=f(x) called differentiable on the interval (a;b) if it has a derivative at every point of the interval.

Theorem If the function is differentiable at a given point x 0, then it is continuous at that point.

The converse statement is not true, because there are functions that are continuous at some point but are not differentiable at that point. For example, the function at the point x 0 =0.

Find derivatives of functions

1) .

2) .

Let's perform the identical transformations of the function:

Derivatives of higher orders

Second order derivative is called the derivative of the first derivative. Denoted

n-order derivative is called the derivative of the derivative of the (n-1)-th order.

For example,

Partial derivatives

private derivative a function of several variables with respect to one of these variables is called a derivative taken with respect to this variable, provided that all other variables remain constant.

For example, for the function partial derivatives of the first order will be equal:

Maximum and minimum of a function

The value of the argument for which the function has highest value, called maximum point.

The value of the argument for which the function has smallest value, called minimum point.

The maximum point of the function is the boundary point of the transition of the function from increasing to decreasing, the minimum point of the function is the boundary point of the transition from decreasing to increasing.

Function y=f(x) has (local) maximum at the point if for all x

Function y=f(x) has (local) minimum at the point if for all X, sufficiently close to , the inequality

The maximum and minimum values ​​of the function are common name extremes, and the points at which they are reached are called extremum points.

Theorem (necessary condition the existence of an extremum) Let the function be defined on the interval and have the largest (smallest) value at the point . Then, if a derivative of this function exists at a point, then it is equal to zero, i.e. .

Proof:

Let at the point x 0 the function has the greatest value, then for any the following inequality is true: .

For any point

If x > x 0 , then , i.e.

If x< x 0 , то , т.е.

Because exists , which is possible only if they are equal to zero, therefore, .

Consequence:

If at a point the differentiable function takes on the largest (smallest) value, then at the point the tangent to the graph of this function is parallel to the Ox axis.

The points at which the first derivative is equal to zero or does not exist are called critical - these are possible extreme points.

Note that, since the equality of the first derivative to zero is only a necessary condition for an extremum, it is necessary to additionally investigate the question of the presence of an extremum at each point of a possible extremum.

Theorem (sufficient condition the existence of an extremum)

Let the function y = f(x) is continuous and differentiable in some neighborhood of the point x0. If, when passing through a point x0 from left to right, the first derivative changes sign from plus to minus (from minus to plus), then at the point x0 function y = f(x) has a maximum (minimum). If the first derivative does not change sign, then this function does not have an extremum at the point x 0 .

Algorithm for studying a function for an extremum:

1.Find the first derivative of the function.

2. Equate the first derivative to zero.

3. Solve the equation. The found roots of the equation are critical points.

4. Put the found critical points on the numerical axis. We get a number of intervals.

5. Determine the sign of the first derivative in each of the intervals and indicate the extrema of the function.

6.To build a graph:

Ø determine the function values ​​at the extremum points

Ø find the points of intersection with the coordinate axes

Ø find additional points

The tin can has the shape of a round cylinder of radius r and height h. Assuming that a clearly fixed amount of tin is used to make a can, determine at what ratio between r And h bank will have the largest volume.

The amount of tin used will be equal to the area full surface banks, i.e. . (1)

From this equality we find:

Then the volume can be calculated by the formula: . The problem will be reduced to finding the maximum of the function V(r). Find the first derivative of this function: . Equate the first derivative to zero:

. We find: . (2)

This point is the maximum point, because the first derivative is positive at and negative at .

Let us now establish at what ratio between the radius and height the bank will have the largest volume. To do this, we divide equality (1) by r2 and use relation (2) for S. We get: . Thus, the largest volume will have a jar whose height is equal to the diameter.

Sometimes it is quite difficult to study the sign of the first derivative to the left and to the right of the possible extremum point, then you can use second sufficient extremum condition:

Theorem Let the function y = f(x) has at the point x0 possible extremum, the final second derivative. Then the function y = f(x) has at the point x0 maximum if , and the minimum if .

Remark This theorem does not solve the problem of the extremum of a function at a point if the second derivative of the function at the given point is equal to zero or does not exist.

Inflection points

The points of the curve at which the convexity separates from the concavity are called inflection points.

Theorem (required inflection point condition): Let the graph of the function have an inflection at a point and the function has a continuous second derivative at the point x 0, then

Theorem (sufficient condition for the inflection point): Let the function have a second derivative in some neighborhood of the point x 0, which has different signs to the left and right of x0. then the graph of the function has an inflection at the point .

The algorithm for finding inflection points:

1. Find the second derivative of the function.

2. Equate the second derivative to zero and solve the equation: . Put the resulting roots on a number line. We get a number of intervals.

3. Find the sign of the second derivative in each of the intervals. If the signs of the second derivative in two adjacent intervals are different, then we have an inflection point at a given value of the root, if the signs are the same, then there are no inflection points.

4. Find the ordinates of the inflection points.

Examine the curve for convexity and concavity. Find inflection points.

1) find the second derivative:

2) Solve the inequality 2x<0 x<0 при x кривая выпуклая

3) Solve the inequality 2x>0 x>0 for x the curve is concave

4) Find the inflection points, for which we equate the second derivative to zero: 2x=0 x=0. Because at the point x=0 the second derivative has different signs on the left and right, then x=0 is the abscissa of the inflection point. Find the ordinate of the inflection point:

(0;0) inflection point.

Exercises to solve

No. 1 Find the derivatives of these functions, calculate the value of the derivatives for a given argument value:

1.	5.	9.
2.	6.	10.
3.	7.	11.
4.	8.	12.
	13.	14.
	15.	16.

#2 Find derivatives complex functions:

1.	2.
3.	4.
5.	6.
7.	8.
9.	10.
11.	12.
13.	14.
15.	16.
17.	18.

No. 3 Solve problems:

1. Find the slope of the tangent drawn to the parabola at the point x=3.

2. To the parabola y \u003d 3x 2 -x at the point x \u003d 1, a tangent and a normal are drawn. Write their equations.

3. Find the coordinates of the point at which the tangent to the parabola y=x 2 +3x-10 forms an angle of 135 0 with the OX axis.

4. Compose the equation of the tangent to the graph of the function y \u003d 4x-x 2 at the point of intersection with the OX axis.

5. At what values ​​of x is the tangent to the graph of the function y \u003d x 3 -x parallel to the straight line y \u003d x.

6. The point moves in a straight line according to the law S=2t 3 -3t 2 +4. find the acceleration and speed of the point at the end of the 3rd second. At what point in time will the acceleration be zero?

7. When is the speed of a point moving according to the law S=t 2 -4t+5 equal to zero?

#4 Explore functions using the derivative:

1. Investigate the function y \u003d x 2 for monotonicity

2. Find the intervals of increase and decrease of the function .

3. Find the intervals of increase and decrease of the function .

4. Explore the maximum and minimum function .

5. Explore the function for an extremum .

6. Investigate the function y \u003d x 3 for an extremum

7. Explore the function for an extremum .

8. Break the number 24 into two terms so that their product is the largest.

9. From a sheet of paper, it is necessary to cut a rectangle with an area of ​​​​100 cm 2 so that the perimeter of this rectangle is the smallest. What should be the sides of this rectangle?

10. Investigate the function y=2x 3 -9x 2 +12x-15 for an extremum and build its graph.

11. Examine the curve for concavity and convexity.

12. Find the intervals of convexity and concavity of the curve .

13. Find the inflection points of the functions: a) ; b) .

14. Explore the function and build its graph.

15. Explore the function and build its graph.

16. Explore function and plot it.

17. Find the largest and smallest value of the function y \u003d x 2 -4x + 3 on the segment

Control questions and examples

1. Define a derivative.

2. What is called the increment of the argument? function increment?

3. What is the geometric meaning of the derivative?

4. What is called differentiation?

5. List the main properties of the derivative.

6. What function is called complex? back?

7. Give the concept of a second order derivative.

8. Formulate a rule for differentiating a complex function?

9. The body moves in a straight line according to the law S=S(t). What can be said about the movement if:

5. The function is increasing on some interval. Does it follow from this that its derivative is positive on this interval?

6. What is called the extrema of the function?

7. Does the largest value of the function on a certain interval necessarily coincide with the value of the function at the maximum point?

8. The function is defined on . Can the point x=a be the point of extremum of this function?

10. The derivative of the function at the point x 0 is zero. Does it follow from here that x 0 is the extremum point of this function?

Test

1. Find derivatives of these functions:

A)	e)
b)	and)
With)	h)
e)	And)

2. Write the equations of tangents to the parabola y=x 2 -2x-15: a) at the point with the abscissa x=0; b) at the point of intersection of the parabola with the abscissa axis.

3. Determine the intervals of increase and decrease of the function

4. Explore the function and plot it

5. Find at time t=0 the speed and acceleration of a point moving according to the law s =2e 3 t

Answers to the exercises

5.

7.

9.

11.

12.

13.

14.

2.

3.

4. (the result is obtained by using the formula for the derivative of the quotient). You can solve this example in another way:

5.

8. The product will be the largest if each term is equal to 12.

9. The perimeter of the rectangle will be the smallest if the sides of the rectangle are 10 cm each, i.e. cut out a square.

17. On the segment, the function takes the largest value, equal to 3 when x=0 and the smallest value equal to –1 at x=2.

Literature

1.	Vlasov V.G. Abstract of lectures on higher mathematics, Moscow, Iris, 96
2.	Tarasov N.P. Course of higher mathematics for technical schools, M., 87
3.	I.I. Valutse, G.D. Diligul Mathematics for technical schools, M., Science, 90g
4.	I.P.Matskevich, G.P.Svirid Higher Mathematics, Minsk, Higher Mathematics. School, 93
5.	V.S.Schipachev Fundamentals of Higher Mathematics, M.Vyssh.shkola89
6.	V.S.Schipachev Higher Mathematics, M.Vyssh.shkola 85g
7.	V.P. Minorsky Collection of problems in higher mathematics, M. Nauka 67g
8.	O.N.Afanasyeva Collection of problems in mathematics for technical schools, M.Nauka 87g
9.	V.T.Lisichkin, I.L.Soloveichik Mathematics, M.Vyssh.shkola 91g
10.	N.V. Bogomolov Practical lessons in mathematics, M. Higher school 90
11.	H.E. Krynsky Mathematics for Economists, M. Statistics 70g
12.	L.G.Korsakova Higher Mathematics for Managers, Kaliningrad, KSU, 97.

KALININGRAD COMMERCE AND ECONOMIC COLLEGE

for the study of the topic

"derivative of a function"

for students of the specialty 080110 "Economics and Accounting", 080106 "Finance",
080108 "Banking", 230103 "Automated information processing and management systems"

Compiled by Fedorova E.A.

KALININGRAD

Reviewers: Gorskaya Natalya Vladimirovna, Lecturer, Kaliningrad College of Trade and Economics

In this manual, the basic concepts of differential calculus are considered: the concept of a derivative, properties of derivatives, application in analytical geometry and mechanics, basic differentiation formulas are given, examples are given that illustrate the theoretical material. The manual is supplemented with exercises for independent work, answers to them, questions and sample tasks for intermediate control of knowledge are given. Designed for students studying the discipline "Mathematics" in secondary specialized educational institutions who study full-time, part-time, evening education, as an external student or who have free attendance at classes.

KTEK
PCC of Economics and Accounting

15 copies, 2006

Introduction. 4

Requirements for knowledge and skills.. 5

The concept of a derivative. 5

The geometric meaning of the derivative. 7

The mechanical meaning of the derivative. 7

Basic rules of differentiation. 8

Formulas for differentiating basic functions. 9

Derivative inverse function. 9

Differentiation of complex functions. 10

Derivatives of higher orders. eleven

Private derivatives. eleven

Investigation of functions with the help of derivatives. eleven

Increasing and decreasing function. eleven

The maximum and minimum of a function. 13

Convexity and concavity of a curve. 15

Inflection points. 16

General scheme research functions and plotting graphs. 17

Solution exercises. 17

Test questions and examples.. 20

Test. 20

Answers to exercises.. 21

Literature. 23

Introduction

Mathematical analysis gives a number of fundamental concepts that an economist operates on - this is a function, limit, derivative, integral, differential equation. In economic research, specific terminology is often used to refer to derivatives. For example, if f(x) is a production function that expresses the dependence of the output of any product on the cost of the factor x, then called marginal product; If g(x) is a cost function, i.e. function g(x) expresses the dependence of total costs on the volume of production x, then g′(x) called marginal cost.

Marginal Analysis in Economics- a set of methods for studying the changing values ​​of costs or results when the volumes of production, consumption, etc. change. based on the analysis of their limiting values.

For example, finding productivity. Let the function u=u(t), expressing the amount of products produced u while working t. Let's calculate the amount of goods produced during the time ∆t=t 1 - t 0:

u=u(t 1)-u(t 0)=u(t 0 +∆t)-u(t 0).

Average labor productivity is the ratio of the amount of output produced to the time spent, i.e. z cf. =

Worker productivity at the moment t 0 is called the limit to which z cf. at ∆t→ 0: . The calculation of labor productivity, therefore, is reduced to the calculation of the derivative:

production costs K homogeneous products is a function of the quantity of products x, so we can write K=K(x). Let us assume that the quantity of production increases by ∆x. The quantity of production x+∆x corresponds to production costs K(x+∆x). Therefore, the increment in the amount of production ∆x corresponds to the increase in production costs ∆K=K(x+∆x)- K(x).

The average increment of production costs is ∆K/∆x. This is the increment in production costs per unit increment in the quantity of output.

Limit called marginal cost of production.