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DERIVATIVE OF A FUNCTION AND ITS APPLICATIONS IN ECONOMICS
Sh.B. Akhmedov
Kokand university Andijan branch
ANNOTATSIYA:
Ushbu maqolada hosila tushunchasining iqtisodiy muammolarni hal
qilishdagi ahamiyati yoritilgan. Xususan, funksiyaga tangens chizig‘i tenglamasini tuzish,
tezlik va tezlanishni hisoblash, ishlab chiqarish xarajatlari, funksiyaning elastikligi, daromad
va narxga nisbatan talab elastikligi kabi iqtisodiy masalalarga hosilaning qo‘llanilishi tahlil
qilinadi. Maqolada oliy ta'limda matematikani iqtisodiy yo‘nalishda o‘qitish uslubiy
yondashuvlar bilan boyitilishi lozimligi ko‘rsatib o‘tilgan.
KALIT SO‘ZLAR:
Hosila, iqtisodiy masalalar, funksiyaning elastikligi, talab elastikligi,
narx, daromad, matematik tahlil, o‘qitish metodikasi
АННОТАЦИЯ:
В данной статье рассматривается значение понятия производной
функции при решении экономических задач. В частности, анализируется применение
производной при построении уравнения касательной к графику функции, расчёте
скорости и ускорения, затрат на производство, эластичности функции, эластичности
спроса по доходу и цене. В статье подчёркивается необходимость методического
обогащения преподавания математики в экономических направлениях высшего
образования.
КЛЮЧЕВЫЕ СЛОВА:
Производная, экономические задачи, эластичность функции,
эластичность спроса, цена, доход, математический анализ, методика преподавания
ANNOTATION.
This article highlights the significance of the derivative concept in solving
economic problems. Specifically, it analyzes the use of derivatives in constructing tangent
line equations, calculating speed and acceleration, production costs, function elasticity, and
the elasticity of demand with respect to income and price. The article emphasizes the need for
methodological enrichment of teaching mathematics in economics-oriented higher education
programs.
KEYWORDS.
derivative, economic problems, function elasticity, demand elasticity, price,
income, mathematical analysis, teaching methodology
INTRODUCTION
. The introduction of new standards of higher education in the
Republic of Uzbekistan is a necessity dictated by the current situation and the rapid
development of science and technology, as well as the integrated development of all sectors
of the national economy. It is the new standards that set the goal of forming a personality
with a set of qualities that allow a person to be successful in the 21st century.
The purpose of higher education is to create conditions for student self-realization in
the educational process, to develop the ability to pose practical problems in one’s specialty,
the ability to choose methods for solving a given problem, to solve, analyze and apply
solutions to practical problems, using the latest achievements of science and technology. To
achieve these goals, we consider it advisable to use professionally oriented modern
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pedagogical technologies in teaching all subjects at universities, taking into account the
specifics of the subject and specialty.
The subject of mathematics has its own special place in this process. Here you should
definitely take into account how much attention the government and state of Uzbekistan pays
to the issue of education in general, as well as the teaching of mathematics at all levels of
education and the integration of science and production.
The emergence of students' interest in mathematics depends to a large extent on the
methodology of its teaching, on the chosen style of communication with students, and on the
extent to which the applied problems of this particular specialty are demonstrated.
When solving many problems in science, technology, and economics, the concept of a
derivative function is encountered. For example, when drawing up an equation for a tangent
to a curve, when calculating speed and acceleration, when studying and plotting functions,
when calculating production costs, elasticity of a function, elasticity of demand relative to
income, relative to price, and others. Therefore, studying the concept of derivative is one of
the most important issues in mathematics.
METHODS.
Definition 1. Let
f x
be a function defined outside some neighborhood
of the point
x
0
. The derivative of the function
f x
at the point
x
0
is the number equal to
lim
x→x
0
f x −f(x
0
)
x−x
0
(1)
If this limit exists and is denoted by
f'(x
0
)
.
To find the derivative of a given function, you must perform the following work:
1)
Let’s assign an increment
∆x
to the free variable x, then the function y takes on an
increment
∆y
, i.e.
y + ∆y = f(x + ∆x
).
2) Let’s find the increment of the function
∆y
:
∆y = f x + ∆x − y = f x + ∆x − f(x)
.
3) Find the ratio of the increment of the function to the increment of the argument:
∆y
∆x
=
f x+∆x −f(x)
∆x
.
4) Let us calculate the limit of the ratio of the increment of the function to the increment of the
argument as
∆x → 0
, i.e.
lim
∆x→0
∆y
∆x
= lim
∆x→0
f x+∆x −f(x)
∆x
.
If this limit exists, then it is called the derivative of the function
f(x)
at the point
x
0
and is denoted by
y'
or
f'(x
0
)
, i.e.
lim
∆x→0
∆y
∆x
= lim
∆x→0
f x+∆x −f(x)
∆x
= f
'
x
0
= y'
(2)
If limit (2) exists, then
∆y → 0
as
∆x → 0
. This means that
f x − f(x
0
) → 0
or
f x →
f(x
0
)
. Consequently, the derivative at a point exists only for functions that are continuous at
the point
x
0
(and even then not for all).
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Limit (2) may not exist at all or may exist for some values of x and not exist for other
values. At those points x where limit (2) does not exist, the derivative of the function is not
defined.
For example, find the derivative of the function
y = f x = x
2
Solution.
1) Let’s set x the increment
∆x
, then
y = f x
takes the increment
∆y
, i.e.
y + ∆y = f x + ∆x =
(x + ∆x)
2
.
2) Let’s find the increment of the function
∆y
:
∆y = x + ∆x
2
− y = x + ∆x
2
− x
2
= x
2
+ 2x∆x + ∆x
2
− x
2
= 2x∆x + ∆x
2
3) Let's find the ratio
∆y
∆x
:
∆y
∆x =
2x∆x + (∆x)
2
∆x
= 2x + ∆x.
4) Let's calculate
∆y
∆x
:
lim
∆x→0
∆y
∆x
= lim
∆x→0
(2x + ∆x) = 2x
.
This means
lim
∆x→0
∆y
∆x
exists and therefore the derivative of the function
y = x
2
will be:
y' = f' x = 2x
.
2. Geometric meaning of the derivative.
As we mentioned above, the concept of derivative is one of the most important concepts in
mathematical analysis. Many theoretical and applied questions, being similar to each other in
their content, come down to the same mathematical problem - finding the limit of the ratio of
the increment of a function to the increment of the argument, provided that the latter tends to
zero, i.e. to finding the derivative.
Let us consider the most important of these problems—the problems of the tangent to
a curve and the speed of alternating motion.
Problem about the tangent to the graph of a curve
Consider the graph of the function
y = f(x)
, defined in the circle of the point
x
0
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At point B we bring the tangent to the curve
y = f(x)
. It forms an angle α with the
positive direction of the ox axis. On the curve
y = f(x)
we take an arbitrary point D and give
a secant BD. It forms an angle φ with the positive direction of the ox axis, i.e.
∠BDE = φ и ∠TFO = α
A-priory
f
'
x
0
= lim
x→x
0
f x −f(x
0
)
x−x
0
From the drawing we can write:
OA = x
0
, OC = x AC = OC − OA = x − x
0
= ∆x.
B = f x
0
, CD = f x DE = CD − CE = CD − AB = y − y
0
= f x − f x
0
= ∆y, BE = AC
= ∆x.
From
ED
BD
= tgφ,
∆y
∆x
= tgφ
then
∆y
∆x
=
f x −f(x
0
)
x−x
0
is the ratio of the legs of the right triangle
BDE , i.e. tangent of the angle φ. If
x → x
0
, then point D tends to point B along the curve,
and straight line BD tends to take the position of the tangent BT to the curve at point B.
Next, we have
f
'
x
0
= lim
x→x
0
f x − f(x
0
)
x − x
0
= lim
D→B
tgφ = tgα
So, the geometric meaning of the derivative is as follows:
The derivative of the function at the point
x
0
is equal to the tangent of the angle
formed by the tangent to the curve
y = f(x)
at the point
x
0
, f(x
0
)
and the positive direction
of the ox axis.
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DISCUSSION.
The most important functions found in economics
When solving various economic problems, the following functions are encountered.
1) Linear function
� = �� + �
The graph of this function is a straight line and it is defined in the interval
( − ∞; ∞)
.
Example: production costs at an industrial enterprise where homogeneous products are
manufactured can be divided into two groups:
a) variable costs proportional to the volume of production, for example material costs.
b) fixed costs, i.e. those that mainly do not depend on the volume of production, for
example, the costs of maintaining administrative buildings, their heating, etc. If fixed costs
are denoted by b, and variable costs per unit of production are denoted by a, then with a
production volume of x units, the total production costs in a given period will be
y = ax + b
.
If linear is expressed by the formula
y = ax
, then it is said that it determines direct
proportionality between y and x.
2) Power function
� � = �
�
.
a) If α is a natural number, then the interval of definition is
−∞; ∞ .
b) If α is a negative number, then substituting
α =− n
, we get:
f x = x
α
= x
−n
=
1
x
n
.
Here the function is defined for all values of
x ≠ 0
.
c) If α is a reverse natural number, then substituting
α =
1
n
, then
f x = x
α
= x
1
n
=
n
x
is defined
( − ∞; ∞)
.
3) Exponential function
� � = �
�
.
If
a > 0
, defined on
( − ∞; ∞)
. If
0 < a < 1
– decreases,
a > 1
, increases.
Example: If 1 zloty is invested at compound interest at a rate of 5%, then its value after X
years will be:
y = 1,05
x
. If
a = e
, then
y = e
x
or
y = exp x
.
4) Logarithmic function
� � = ���
�
� � > � .
It is defined only by
x > 0
. Logarithmic functions with base
0 < a < 1
are not used in
practice. If
a = e
, then
f x = log
a
x = lnx
.
Example: The Italian economist Pareto formulated a theorem about the distribution of income
in any society. If y denotes the number of persons with income not less than x, then
y =
a
x
m
, where a and m are constants.
For low incomes, Pareto’s law does not apply. Let the distribution of income in some
society be determined by the equation
y =
2 000 000 000
x
1,5
a = 2 000 000 000, m = 1,5
.
Find:
a) the number of persons who have an income exceeding 100 000;
b) the lowest income among the 100 richest individuals.
Solution.
a) We have,
x = 100 000
y =
2 000 000 000
100 000
1,5
;
lgy = lg
2 000 000 000
100 000
1,5
= lg2 000 000 000 − 1,5lg100 000 = 9,3010 − 1,5 ∙ 5 = 1,8010
lgy = 1,8010; y = 63,2
.
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Therefore, 63 people have an income exceeding 100,000.
b) We have,
100 =
2 000 000 000
x
1,5
;
x
1,5
= 20 000 000; 1,5 lg x = lg20 000 000 = 7.3010
lg x = 4,8673; x = 73700
.
That. the lowest income among the 100 richest persons (that is, the income of the
hundredth person, counting the richest) is 73,700.
Production costs for homogeneous products are a function of the quantity of
production x. Therefore we can write:
K = k(x)
Let us assume that the quantity of production increases by ∆x. Product x+∆x
corresponds to production costs
K(x + ∆x)
Consequently, the increment in the quantity of products ∆x corresponds to the
increment in production costs
∆K = K x + ∆x − K(x)
The average increase in production costs is:
∆K
∆x =
K x + ∆x − K(x)
∆x
This is an increase in production costs per unit increase in the quantity of production.
Limit
lim
∆x→0
∆K
∆x = K'(x)
called the marginal cost of production.
Similarly, if we denote by
U x
the revenue from the sale of x units of goods, then the
limit
lim
∆x→0
∆U(x)
∆x = U'(x)
we will call it marginal revenue.
Example 1. Production costs K depend on the volume of production x according to the
formula:
K = 100x −
1
30 x
3
Determine marginal costs if the production volume is: a) 5 units; b) 10 units of
production.
Determine marginal costs if the production volume is: a) 5 units; b) 10 units of
production.
We have:
K
'
= 100 −
1
10
x
2
.
K
'
5 = 100 −
1
10
∙ 25 = 97,5
K
'
10 = 100 −
1
10
∙ 100 = 90.
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This means that with a production volume of 5 units of production, the cost of
producing the next (sixth) unit of production is 97.5; with a production volume of 10 units
they will be 90.
Elasticity of function. In many problems, it is more convenient to calculate the
percentage increase (relative increment) of the dependent variable corresponding to the
percentage increase of the independent variable. This brings us to the concept of elasticity of
a function (sometimes called relative derivative).
If the function
y = f(x)
is given, then
lim
∆x→0
(
∆y
y
;
∆x
x ) = lim
∆x→0
x
y ∙
∆y
∆x =
x
y lim
∆x→0
∆y
∆x =
x
y f
'
x =
x
y ∙
dy
dx
It is called the elasticity of the function
y = f x
with respect to the variable x and
denotes
E
x
y
E
x
y =
x
y ∙
dy
dx
Elasticity with respect to x is an approximate percentage increase in a function
(increase or decrease), corresponding to an increase in the independent variable by 1%.
Elasticity of demand relative to price.
The functional relationship between demand
for a given product and its price (provided that the price of other goods, consumer income
and the structure of needs are constant values) allows the price to be brought into line with
demand, properly defined. However, in many economic studies it is necessary to determine
not the quantity of demand, but the change in demand caused by a certain change in price, i.e.
The elasticity of demand relative to price is determined.
Let us assume that demand q depends on price p:
q = f(p)
Let
∆p
be the price increment and
∆q
the corresponding demand increment.
∆p
p
– relative price change,
∆q
q
- relative change in demand.
The elasticity of demand relative to price is called the limit
E
p
q = E
c
= lim
∆p→0
(
∆q
q ;
∆p
p ) =
p
q lim
∆p→0
(
∆q
∆p ) =
p
q ∙
dq
dp
E
c
=
p
q ∙
dq
dp
The price elasticity of demand approximately determines how the demand for a given
good will change if its yen increases by 1%.
a) If E_
E
c
> 1
, i.e. if an increase in price by 1% corresponds to a decrease in demand
by more than 1%, they say that demand is elastic, and if
E
c
= 1
, i.e. a decrease in demand by
1%, then demand is neutral.
b) If E_
E
c
< 1
, then demand is inelastic.
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Because the demand function in most cases is a decreasing function, then
dq
dp
< 0
and
therefore the elasticity of demand should be written as follows:
E
c
=−
p
q ∙
dq
dp
Example. If the demand function is
q = 10 − p
, then the elasticity of demand is:
E
c
=−
p
q ∙
dq
dp =−
p
10 − p −1 =
p
10 − p
If, for example,
p = 2
, then
E
c
=
2
10−2
=
1
4
.
This means that at price 2, a 1% increase in price will cause a decrease in demand by
1
4
%. Therefore, the elasticity of demand in this case is inelastic. It is also possible to show the
elasticity of total and average costs.
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