UNIT ONE INTRODUCTION

Dear colleague! In this unit we shall revise what you have studied in Quantitative method for
economists I and II such as basic concepts from calculus which incorporates differential calculus,
integral calculus, differential and difference equations. In addition to these, we will revise matrix
algebra which includes matrix representation, determinants, inverse of a matrix, solutions of
simultaneous linear equation systems, Leontief input - output model and linear programming.
Therefore, you should read your module of the above two courses to understand this unit easily.

Unite Objective
Up on the completion of this unit, you should be able to
- explain the meaning of differential calculus.
- describe the rules of differentiation
- explain the difference between differential calculus and integral calculus.
- describe basin concepts of a matrix.
- discuss the methods of solving simultaneous linear equations using matrix algebra.

Section 1.1 Basic Concepts from Calculus

After completing this section, you are expected to

- describe the meaning of differentiation.

- discuss the different rules of differentiation
- explain the meaning and methods of integration
- describe the concept of differentials.
- apply the techniques of differentiation and integration for
solving economic problems.

1.1.1 Differential Calculus - Differentiation

Given a function y = ƒ(x), where x and y are independent and dependent variables
respectively, if x is an increment in the value of x and y is an increment in the value of y,
y
then x is referred to as the average rate of change or the rate of change of y with respect to
the given change in x. This limiting value is also known as derivative which is the instantaneous
dy
rate of change of y due to a very small change in x represented by dx .

This means
y f ( x  x)  f ( x) dy
Lim x = Lim x = dx
x 0 x 0

1
Thus, differentiation is the process of finding the rate of change of the dependent variable (y)
with respect to a given change in the independent variable (x). In other words, it is a process of
determining the slope of the function y= ƒ(x) at any point of x in the domain of the function.

What is the difference between the instantaneous rate of change and the average rate of
change of the function y= f(x)?
---------------------------------------------------------------------------------------------------------------------
----------------------------------------------------

Rules OF Differentiation
Dear colleague! Do you remember the different rules of differentiation from the discussion of
your previous courses? What are they?
---------------------------------------------------------------------------------------------------------------------
-----------------------------------
Have you answered this question? Ok good. Now try to relate your answer with the following
analysis.
Given the function y= ƒ(x) and other functions such as g(x) and h(x) which are differentiable, the
following are the rules of differentiation.

1.Constant function Rule

Given a function y = ƒ(x) = k, where k is constant
dy dk
 0,
dx dx That is the derivative of the constant function is zero.
dy
0 0 1
Example If ƒ (x) = 4 = 4x , dx = 0 (4x ) = 0

2.Linear function Rule

Given a function y = ax + b, where a and b are constants
dy
a
dx

Example If ƒ (x) = 3x + 5, then ƒ' (x) = 3

If f(x) = 5x, then ƒ' (x) = 5

3.Power function rule

n
Given a function ƒ(x) =x where x¿ 0 and n is any real number
n 1
ƒ '(x) = n x
5 1
5
Example If ƒ(x) = 2x , ƒ ' (x) = 5 (2x ) = 10x 4

3
1 x  3  3 1 4
If ƒ(x) = x , ƒ' (x) = -3 (x ) = -3x
3
ƒ '(x) = x4

2
 −1
1 1 2
−1 x
If ƒ (x) = x  x ½, ƒ’ (x) = ½ x
2
=2
4.Sum and Difference Rule
When y = g(x)  h(x), then ƒ' (x) = g' (x)  h' (x)
2
Example If g (x) = 2x+3 and h (x) = 5x + 2x, then
ƒ ' (x) =g'(x)  h' (x)
= 2 10x +2
g' (x) +h' (x) = 10x +4
g' (x)- h' ( x) = 2 - (10x+2)
= -10x

5.Product rule
This rule enables us to differentiate two functions which are multiplied together.
If ƒ' (x) =g' (x) h(x) + h' (x) g(x)
Example
dy
2 1
1. If ƒ(x) = ( 7.5+ 0.2 x ) ( 4+8x ) , What is ƒ' (x) = dx ?
This function can be multiplied out and differentiated without using the product rule. But, let us
first use the product rule and then compare the answers obtained by the two methods.
2 1
Let g (x) = 7.5 + 0.2x and h(x) = 4+8x
2 2
Therefore, g' (x) = 0.4x and h' (x) = -1 (8x ) = -8x
Using the product rule
1 2 2
ƒ' (x) = 0.4x ( 4+8x ) + - 8x ( 7.5 + 0.2x )
2
= 1.6x + 3.2 - 60x - 1.6
2
ƒ' (x) = - 60x + 1.6x +1.6---------------------------------(1)

If we multiply the original function, we get

2 1
ƒ(x) = ( 7.5 + 0.2x ) ( 4+8x )
1 2
= 30+ 60x + 0.8x + 1.6 x
Therefore
2
ƒ' (x) = - 60x + 1.6x +1.6-----------------------------------(2)
These two derivatives, i.e. (1) and (2) are the same. But in some cases, it is not possible to
multiply out the different components of a function and then we must use the product rule to
differentiate.

Example
2.A firm faces anon - linear demand function P = ( 650 - 0.25q) 1.5
Determine the marginal revenue (MR) function of this firm.
In this case first we should determine the total revenue function and then differentiate it with
respect to the level of output using the product rule.
1 .5
TR = p x q = q (650 - 0.25q)
To differentiate TR with respect to q using the product rule,
1 .5
Let u = (650 - 0.25 q) and v=q

3
 du
Then dq = 1.5 (650 - 0.25q)
0 .5

du
dq = 1
dTR
Therefore, dq = 1 (650 - 0.25q)
1 .5 0 .5
+ (q) (- 0.375) (650 - 0.25q)
0 .5
MR = (650 - 0.25q) (650 - 0.625q)
6.Quotient rule
Dear colleague! This rule allows as differentiating two functions where one function is divided by
the other function.
If ƒ'(x) = h(x).g' (x) – g(x).h' (x)
( h (x) )2
Example
4x2
1.Determine ƒ' (x) if f ( x) = 8  0.2 x
2
Let g (x) = 4x and h(x) = 8+0.2x,
g' (x) = 8x and h' ( x) = 0.2
According to the quotient rule,
ƒ'(x) = 8x(8+0.2x) – 0.2(4x2)
(8+0.2x) 2
= 64x + 1.6x2 – 0.8x2
(8+0.2x) 2
64 x  0.8 x 2
2
ƒ' (x) = (8  0.2 x)

2. If a monopoly faces a non- linear demand function

252
p= (4  q )0.5
Derive the marginal revenue function (the rate of change of TR with respect to output).
As we know, TR = p x q,
252q
0.5
TR = (4  q )
0 .5
If u = 252q and v = (4+q) , then
du dv
 252 
dq And dq 0.5 (4+q)-0.5
Now we can find the marginal revenue function using the quotient rule as follows.
252( 4+q )0 . 5−252 q (0 .5 )(4 +q )−0 .5
MR = [(4+q )0. 5 ] 2
= 252(4+q) 0.5 – 252q (0.5) (4+q) 0.5
(4+q)

4
 (4  q )252  125q
= (4  q )1.5
1,008  252q  126q
= (4  q )1.5
1,008  126q
MR = (4  q )1.5
7. Chain Rule
Dear colleague! The chain rule enables us to differentiate' functions within functions ', for
instance, given a function y = ƒ (z) and z = g(x), then y= ƒ [g (x)].
dy  dy dz
dx dz dx
To differentiate y with respect to x, we use chain rule which states that
Example
0 .5
1. Given a non- linear demand function, p = (1500 -0.2q) , find its slope.
0 .5
Let z = 150 - 0.2q as a result p = z
dz dp dz
 0 .5
dq = dz dq = (0.5z ) (-0.2)
 0 .5
= -0.1(150-0.2q)
= - 0.1
( 150 – 0.2q) 0.5
2. If the present value of a one birr due in 8 years time is given by the formula

1
8
PV = (1  r )
Where r is the given interest rate, find the rate of change of PV with respect to r.
1
Let (1+r) = z, then PV= z8 = z  8
dP   9
-8z
dz Vdz
This means dr = 1 and
Thus, using the chain rule
dpv dpv dz

dr dz ( dr ) = - 8 (1+r)  9 (1) = -8 (1+r)  9
8
=
(1  r )9
Dear colleague! As you know in your microeconomics course the marginal revenge productivity
theory of demand for labor states that profit is maximized when MRP L = MC L . Where MRP L
= marginal revenue product of labor
MC L = Marginal cost of labor.

dTR DTC L
MRP L = dL 5MC L = DL
 and

Where TR is total revenue and TC L is total cost of labor.

3. A firm is a monopoly seller of good q and faces the demand schedule P = 200 - 2q, where p is
0.5
price in birr and the short run production function is given as q = 4L Determine the marginal
revenue productivity of labor (MRP L ) function.

dTR dq
,
MRP L = dq dL

Thus, first we need TR, given P = 200 - 2q

2
TR = (200 - 2q) q = 200q - 2q
Therefore,
dTR dq
dq = 200 - 4 q, dL = 2L  0 .5

 0 .5
As a result, MRP L = (200 - 4q) (2L )
 0 .5
= [200-4(4L0.5)] 2L
0 .5  0 .5
= (200- 16 L ) 2 L
MRPL = 400 – 32L 0.5
L 0.5
7.Implicit function Rule
Functions of the form y = ƒ(x) express the dependent variable y explicitly in terms of the
independent variable x and are called explicit functions. Where as functions of the form ƒ (x, y)
=0 do not express y in terms of x. These functions are referred to as implicit functions. If the
implicit function ƒ(x, y) = 0 exists, and the first order partial derivative of the function on with
respect to y is different from zero (ƒ y ¿ 0), then the total differential
f x dx +f y dy=d 0
f x dx =−f y dy
Rearranging these terms,
dy  fx ( f ( x , y )
dx fy x
= , where ƒx =
f ( x, y )
ƒy =
y
Example

6
 dy
2 2
1. Given the function ƒ (x, y) = x + y dx we should partially differentiate the
- 9 = 0, find
dy
function with respect to x and with respect to y, and then determine dx .
f x =2 x and f y =2 y
 2x
dy  x
Thus, dx = 2 y = y
dy
3 2 2
2. Given the implicit function ƒ (x, y) = x - 2x y + 3xy - 22 = 0, find dx .
2 2 2
ƒ x = 3x - 4xy + 3y and ƒ y = - 2x +6xy
dy
2
Therefore, dx = - (3x – 4xy + 3y2)
2
- (- 2x +6xy)
9. Inverse function rule
Dear colleague! Given the function y= ƒ(x), it has an inverse if and only if each value of y
corresponds with one and only one value of x. According to the inverse function rule, the
derivative of the inverse function is the reciprocal of the derivative of the original function. The
dy
derivative of the given original function is dx . Thus, the derivative of the inverse function x =
1
dy dx
1
ƒ (y) which is dx =
dy
3 2
Example If Q = P + 2P +7p, find the derivative of P with respect to Q
dQ dp 1
dp = 3 P 2 + 4P + 7, Thus, dQ = 3 p 2  4 P  7

10. Logarithm and Exponential Function Rules

a. Logarithmic function rule
2
If we have two numbers 8 and 64 that can be related with each other by the equation 8 = 64 the
exponent 2 can be expressed as the logarithm of 64 to the base 8. This means.

x
Log648 = 2. In general, if y = a , then log y a = x where a >0 and a ¿ 1
In logarithmic application two numbers are mainly selected as abase. These numbers are 10 and
e. If the base is 10, then the logarithmic is referred to as common logarithm. But if we use e as
abase, then the logarithm is known as natural logarithm.
Natural logarithmic function rule
Given the function ƒ(x) = n 9( x )
= loge
9( x )
where ƒ(x) and g (x) are differentiable, and
1
g (x)¿ 0, f ( x ) = g ( x) [ g ( x ) ]
' '

dy 6
Example If y =  n
6 x 7
, then dx = 6 x  7

7
 Logarithmic function of base a
g( x )
Given the function f(x) = loga , when f(x) and g (x) are differentiable, and g(x) ¿ 0,
dy 1 1
dx = ƒ' (x) = g ( x ) [g' (x)] n
a

Example
1 1 1
x
og 6 ln 6
, then ƒ' (x) = 3x  og6 + x x ln = x ( 3
3 x 2 x 6 2 3
1. If f(x) = x log6 + )
2
x
x
2 og 6
ƒ' (x)= 3x 6 + ln

( 3 x2 +2 x )
2. If ƒ(x) =, Log3 then ƒ' (x) = 6x+2
(3x2+2x) ln3
b) Exponential Function Rule
x
The function y = a where a is constant and a > 0, a ¿ 1 is known as exponential function to
base a. If the base is e, the function is referred to as natural exponential function.
x
y = e , where e= 2.71828
In economics, natural exponential functions are particularly useful for analyzing the growth rates
of different variables.
Natural exponential function rule
g ( x) g ( x)
If ƒ(x) = e where ƒ (x) and g(x) are differentiable, then ƒ' (x) = e g' (x).

dy
Given the function y =ex, then dx = ex.
Example
1−x2 1−x2
1. If ƒ(x) = 5e , then ƒ' (x) = -10x e
e5 x 5e5 x (e5 x  1)  5e5 x (e5 x  1)
5x
2. If ƒ (x) = e  1 , then ƒ' (x) = (e5 x  1) 2
10e5 x
ƒ' (x) = (e5 x  1) 2

Exponential Function Rule for Base a

Given the function f(x) - a g(x) Where g(x) and ƒ(x) are differentiable, a>0, a ¿ 1,
ƒ' (x) = a g(x) n
a

a
3 n
3x 3x
Example.1. ƒ(x) = a , then ƒ' (x) = a
ƒ' (x) = 3a
3x
n a
1.1.2 Integration

Integration is the reverse of differentiation. Thus integrating a function means finding another
function when it is differentiated gives the primitive function.

8
Indefinite Integrals
Given the derivative of the function ƒ' (x), we can find the primitive function f(x) using the
method of integration if we have appropriate information to definitive the arbitrary constant. The
standard notation which is used to denote the integration of ƒ' (x) is

ƒ( x) = ∫ f ( x )dx
'

Where ƒ (x) is the integral of a function ƒ'(x), the sign ∫ is referred to as the sign of integration.
The dx ‘indicates that ƒ(x) shall be equal to ƒ' (x) when it is differentiated with respect to x. The
function ƒ' (x) is known as the integrand, i.e. the function to be integrated.
Dear colleague! As you remember from the process of differentiation constant numbers disappear
when a function is differentiated. As a result, we cannot know the constant that should appear in
an integrated function if additional information is not available. Thus, we ought to incorporate a
constant of integration represented by C. The integral
∫ f ( x )dx is known as indefinite integral of ƒ'(x) as it does not have definite numerical value.
'

This value varies with the value of the independent variable x.

Rules of Integration
Dear colleague! As you know integration is the reverse of differentiation. As a result, the rules of
integration are the reverse of that of differentiation. They are explained as follows

1. The power rule

n
Given the derivative function ƒ'(x) = x , according of the power function rule the primitive
function
'
ƒ( x) = f ( x )dx ax n 1
n
ƒ(x) = ax dx = n 1 + C

When a and n are parameters and n ¿ -1

Example
3 x 4 1  C
1. If ƒ'(x) = 3x , ƒ(x) = 3x dx =
4 4
4 1
5
3x
C
= 5
5 x 0 1
2. If ƒ’(x) = 5, then ƒ(x) = 5dx = 0  1 + C = 5x+C
2. Exponential Rule
x
Given the function ƒ'(x) = e , the primitive function ƒ(x) is
x x
ƒ( x) = e dx e  C
3. Logarithmic function Rule
1
Given the derivative of the function ƒ' (x) = x , the original function

9
 1
dx 
ƒ(x) =  x n /x/ + c , x ¿ 0 .
4. The Integral of sum
The integral of sum of functions is the sum of the integral of those functions. Given the two
derivative functions ƒ' (x) and g '(x), then
ƒ(x) = [ ƒ'(x) + g'(x) ] dx =  ƒ' (x) dx + g'(x) dx
= ƒ (x) + g(x) + C
Example
2
Find (2 x + 3x+2) dx
(2 x 2 +3x+2) dx = 2 x 2 dx  3xdx  2dx
2 x 3  c1 3 x 2  c2 ) 2 x  c3
= 3 + 2 + 1
2 3 3 2
x + x 2 x+ C
=3 2
In the final answer the arbitrary constants of integration can always be combined in single
arbitrary constants.

5.The Integral of a multiple

The integral of the product of a constant and a function is equal to the product of the constant and
the integral of the function, that is

∫ ƒ' (x) dx = c∫ ƒ'(x) dx where c is constant.

3 x 4 1
Example ∫ 3 x dx=3 ∫ x dx=
4 4
4 1 + C
3
= x 5+ C
5
Techniques of Integration
Dear colleague! As you know from your Quantitative method for economists I study, there are
different derivative functions whose primitive functions cannot be obtained using one of the
above rules of integration. Thus, certain techniques of integration have been developed for
integrating these functions. Some of these are explained below.

a) Integration by Substitution
Given the derivative function ƒ' (x) and x = g (u) be a differentiable function then
∫ f ' ( x)dx=∫ f ' [ g(u)] g' (u)du
Example
2 3
a. 2 x( x  5) dx
du u 31
Let u= x +5, then dx = 2x,du=2 xdx .Therefore u du = 3  1 + c
2 3

1
=4 u + C
4

10
 1
= 4 (x +5) + C
2 4

2 x−3
∫( )dx
b. √2 x 2−6 x +1 =
du
2
Let u= 2x - 6x +1 so that dx = 4x – 6, u'=2(2x-3)
du=2(2 x−3)dx
du
2 = (2x - 3)dx
2x  3 du 1 du
( )dx  1 
2
2x  6x 1 2 u= 2 u
Therefore,
 1 1
1 U 2

1
2 C
1 1
= 2 U du =
2
2

= U (1/2) +C
2
Substituting 2x -6x+1 in place of u gives as
2x  3
(
2 x 2  6 x  1 ) dx = √ 2 x 2−6 x +1 + C
b. Integration by Parts
It is a method that enables us to integrate certain products ƒ'(x) g' (x). It is a restatement of the
product rule for differentiation. The formula for this method is
f (x) .g (x) = ƒ(x) G(x) -  ƒ'(x) G(x)dx , where G(x)= g(x)dx .

Example
1
1. Evaluate x( x+1) dx
2

3 3
x( x  1) 2
( x  1) 2
dx
1 3 3
x( x  1) 2 dx = 2 -  2
5
( x  1) 2
2 3
( x  1) 2
2 5
= 3 x- 3 2 +C
3 2 2 5
2 ( x  1) 2 ( )( x  1) 2
= 3 - 3 5 +C
3 5
2 x( x  1) 2 4 ( x  1) 2
= 3 - 15 +C
x x x
2. Evaluate (1  x)e dx (1  x)e  ( 1)e dx
x x
= (1-x) e + e +C

11
 x x x
= e - x e +e + C
x x
= 2e - x e + C
= (2 -x) ex + C

Economic Applications
Dear colleague! You know that integration of marginal functions shall give us the corresponding
total functions. Thus let us discuss some applications of integration.
Total Costs
It is clear that the total cost is a summation of total fixed cost and total variable cost. Thus, the
integration of the marginal cost with zero constant of integration will give total variable cost.
Example
If a firm spends 650 Birr on fixed costs and its marginal cost is given as
2
MC= 82-16Q + 1.8Q Where Q is quantity produced, determine the total cost
function of the firm.
Using the method of integration
M .dQ
TVC = C

2 16 1 .8
TVC = (82  16Q  1.8Q )dQ = 82Q - 2 Q2 + 3 Q3
2 3
TVC = 82Q - 8Q +0.6Q
3 2
We know that TC = TVC + TFC. Therefore, TC = 0.6Q - 8Q + 82Q + 650

Total Revenue
Dear colleague! Similarly we can determine total revenue function provided that we know the
marginal revenue functions.

0 .5
Given the marginal revenue function MR = 520 - 3Q , find the total revenue function and the
corresponding demand function.

TR = ∫ MRdQ
=∫
[520−(3) √Q ]dQ
1 .5
TR = 520Q - 2Q
Dividing the total revenue (TR) function by the quantity Q gives us the corresponding demand
function. Thus,
520Q  2Q1.5
P=
Q
0 .5
P = 520 - 2Q
Dear colleague! What will be the total revenue if the firm charges a price of P= 120 Birr?
0 .5
P = 520 - 2 Q
0 .5
Rearranging this equation, 2Q = 520 - P
520  P
0 .5
Q = 2

12
 Q = [(520 – P)/2]2

Thus, Q = [(520 – P)/2] 2= (200)2=40,000 units.

But we know that Total Revenue (TR) = PQ, where P is price and Q is quantity sold. Thus,

TR = 120x40, 000 = 4, 800,000 Birr

Consumption and Saving Functions

Dear colleague! Applying the same technique we can drive the consumption and saving functions
provided that marginal propensity to consume and autonomous consumption are given.

Example
Suppose the marginal propensity to consume (MPC) out of income for the economy as a whole is
4
given as 5 and it is known that when income is zero, consumption is equal to 12 billion Birr.
Find the function which relates aggregate consumption to national income. Find the aggregate
saving function of the economy.

Dear colleague! As you remember from your macroeconomics study

Income = consumption + saving
dc 4

dy 5
Here MPC = where MPC is Marginal Propensity to Consume C is consumption and y
,
is income.
Integrating MPC With respect to income gives us the aggregate consumption function.
MP .d  4
5 4
C= C Y dy = 5 Y + K, where k is an arbitrary constant.
However, when income is zero, consumption is 12 billion birr.
C= 12, when y = 0,
4 (0)  K  K 12
This means, 12= 5
Therefore, the consumption function is

C= (4/5) Y +12

And the aggregate saving function is S = y- [(4/5) Y+12]

S= y- [(4/5) Y +12]
= y-(4/5) Y - 12

S = (1/5) Y -12

Definite Integrals
Dear colleague! Until now we have discussed about indefinite integrals. However, there is also
another form of integral i.e. definite integral, which is specified with two values of the
independent variables and defined as the value of the integral at one value minus the value of the
integral at another value . Given the derivative function ƒ’(x),

13
 b
∫a f ' (x )dx =f (b )−f (a )
From this difference we can get a specific numerical value which is free of x and the arbitrary
constant, C. This value is referred to as the definite integral of f(x) from x=a to x=b . In this
case, a is the lower limit of integration and b is the upper limit of integration.

Example
6

1. Given 
2 ( 3x2 + 2x) dx , find the value of the primitive function from x = 2 to x = 6

First we ought to determine the original function ƒ(x),

(3x
2
 2 x)dx
ƒ( x) =
3x3 2x2
 c  x3  x 2  c
= 3 + 2
6

Therefore 
2
(3 x  2 x)dx  f (6)  f (2)
2
3 2 3 2
= (6 + 6 + C) - (2 + 2 + C)
= 216 + 36 + C - 8 -4 - C
= 252 - 12 = 240
Economic Applications

An important feature of the definite integrals is that they are equal to the area between a function
and the horizontal axis, and between the two specified values of the independent variables. As a
result, we can calculate the consumer's surplus and producer's surplus using the method of
definite integrals.
Examples
2
1. Given the non - linear demand function P = 1, 800 - 0.6Q and the corresponding marginal
2
revenue function MR = 1,800 -1.8Q
Using the technique of definite integral, find
a) Total revenue ( TR) when Q = 10
b) Change in total revenue when Q increases from 10 to 20,
c) Consumer's surplus when Q = 10.

Solution
a). When Q = 10 , TR shall be
10 10
∫0 MRdQ=∫0 (1,800−1.8Q 2 )dQ

1. 8Q 3 10
]
= [1,800Q - 3 0
2 3
= [1,800Q - 0.6 (10) ] - [1,800(0) -0.6(0 )]

14
 = 18,000 - 600
TR = 17,400 Birr

b).The change in total revenue when Q increases from 10 to 20 units is

20 20
∫10 ( MR )dQ=∫10 (1 , 800−1 . 80 Q2 )dQ
20
3
= [1,800Q - 0.6Q ]10
3
= [1,800(20) - 0.6(203)] - [1,800(10) -0.6(10) ]
= (36,000 - 4,800) - (18, 000 -600)
= 13, 800 Birr
c).The consumer's surplus when Q = 10 shall be the definite integral of the demand function
minus the total revenue actually spent.
10
Consumer's
∫
= 0
(1 , 800−0 . 6 Q2 )dQ−TR
Surplus
10 10
∫0 (1 , 800−0 . 6 Q2 )dQ=[1 .800 Q−0 .2 Q3 ] =18 , 000−200=17 ,800 birr
0
But we know that TR = PQ = (1,800 - 0.6Q2)Q
TR = 1,800 Q - 0.6Q3
3
When Q = 10, TR = 1,800(10) - 0.6(10)
= 18,000 -600 = 17,400 Birr
Therefore, consumer's surplus = 17,800 - 17,400= 400 Birr

2. Suppose a company whose annual sales are currently 500,000 Birr has been experiencing sales
increase by 20% per year. Assuming this rate of growth continues, what will be the total sales of
the company in five years time?
Here the rate of sales is
0 . 2t
S' (t) = 500,000 e
Thus, total sales (at the end of year)
5 5
=
∫0
S'(t ) dt ∫0 500 ,000 e dt
=
0 . 2t

500 , 000 0 .2 t 5 500 , 000( e−1)

|e | =
= 0 .2 0 0.2
Total sales = 4,295, 694 Birr

3. The market demand and market supply functions under perfect competition are given as P= 16
2 2
- Q and P = 2 Q + 4 respectively. Find the producer's surplus.

Dear colleague! It is clear that in perfectly competitive market equilibrium price and quantities
are determined when market demand and market supply are equal. Thus,
2 2 2 2
16-Q = 2Q + 4 ⇒ Q = 4, Q = 2 as a result P=16 = 2 = 12

2
∫ (2 Q2+4 )dQ
Thus, Producer’s surplus = PQ - 0

15
 2
2
Q 3 + 4Q ]
= 12 (2) - [ 3 0
2 3
( 2 ) + 4( 2 )]
= 24-[ 3
16 32
units
= 24 - 3- 8 = 3

1.1.3 Differentials
Given the function y = ƒ(x) if we know the rate at which x changes, we can find the change in y,
Δy Δy dy
Δy=( )Δx Δx →0 , ≈
Δx . As Δx dx
Denoting dy as small change in y and dx as small change in x,
dy dy '
(
)dx , where =f ( x )
dy = dx dx

'
dy =f ( x ) dx It is the differential.

Example
Given y = (5x3 + 2x2+ x)dy = (15x2 + 4x+1) dx
Suppose x changes from 2 to 2.01 (i.e. dx = 0.01), by what value does the dependent variable y
changes?
dy = (15 x2 + 4x +1)( 0.01)
2
= [15 (2 ) + 4 (2) + 1] (0.01)
= (60+ 8+1) (0.01)
dy
= (69)0.01 = 0.69
From the original function, the actual change in y is
Δy = ƒ (2.01) - ƒ (2)
3 2
ƒ (2) = 5(2 ) + 2(2 ) +2
= 5 (8) + 2(4) +2
ƒ (2) = 40+8+2 = 50
3 2
ƒ(2.01) = 5(2.01) + 2(2.01) + 2.01
= 40.606020+8.0802+2.01
=50.696220
Δy=f (2. 01 )−f (20=50 . 696220−50=0 .696220
Δy−dy=0 .696220−0 . 69
= 0.006220
The value 0.006220 is the error of calculation.
As x changes from 2 to 3, that is Δx = 1, what is the change in y?
dy = [15 (2) 2+ 4(2) + 1] 1
= 60+8 +1 = 69

16
 Δy= ƒ (3) - ƒ(2)
But we have determined that
ƒ (2)=50
ƒ (3) = 5 ( 33 ) + 2(32) + 3
= 5 (27) + 2(9) +3
ƒ (3) = 135 + 18 + 3 = 156
Thus, Δ y = 156 - 50 = 116. The error term is 116-69 = 47
Note. Dear colleague! By now we have realized that the smaller the change in x the smaller error
term and the greater the change in x the larger will be the error term.

Total differentials
The concept of differential can be extended to a function of two or more independent variables.
Given the utility function U = U ( x1, x2)
Supposing U as continuous and differentiable
∂u
∂x
Ux1 = 1 which is marginal utility of x1.
∂u
Ux =
∂ x 2 which is marginal utility of x
2 2.

∂u
The change in total utility that results from small change in x 1 is given by
∂ x 1 .dx1 and from
∂u
small change in x
∂ x 2 dx2
2 is

Then the total change in utility is

∂u ∂u
dx 1 +
du = ∂ x 1 ∂ x 2 dx
2

Example Find the total differential, given

3
1. Z = 3x2 + x y-2 y
∂z ∂z
dx+ dy
dz = ∂ x ∂y
dz = (6x + y) dx + (x-6y2) dy
2 2
x +6 x 1 x2 +4 x
2. U = 1 2
du = (4x1 + 6x2) dx1+ (6x1 + 8x2) dx2
Dear colleague! By now you have completed the first section of this unit. Therefore, try to do
the following self - test questions in order to examine how you have understood this section.

Self - Test 1.1

17
 Do the following questions based on the above information .
Part I
1. What is differentiation?
---------------------------------------------------------------------------------------------------------------------
---------------------------------------------------------------------------------
2. Explain the different rules of differentiation.
---------------------------------------------------------------------------------------------------------------------
------------------------------------------------------------
3. What is the difference between differentiation and integration? -------------------------------------
---------------------------------------------------------------------------------------------------------------------
4. Describe the difference between definite and indefinite integrates. ---------------------------------
-------------------------------------------------------------------------------------------------------------------

Part II workout
1. Differentiate the following functions
a) ƒ(x) = 3x3 + 4x2 + 6x+2
b) ƒ(x) = x4e3x c) f ( x )=3( x 3 +7 x2 −5 x )12
2. Solve the following integrals
4 5
a)
∫ 2
6 x 2 dx
b)
∫ 3
(3 x 2 +2 x +2 )dx
3 3
c)
∫ 0
4 e2 x dx
d)
∫1
( x 3 +x +6 )dx

e)∫ (120x 4−60 x3 )dx f) ∫ (42−18 x−2 )dx

720
0. 5
,
3. Given the demand function P = (25+Q ) derive the marginal revenue function.
4. If the marginal cost function of a firm is given as MC = 40 - 18 Q+4.5Q2
What would be the increase in total cost if output were increased from 30 to 40 units?
---------------------------------------------------------------------------------------------------------------------
--------------
5. Given the non - linear demand function P= 600 – 6Q 0.5 and the corresponding marginal
revenue function MR= 600 – 9Q0.5. Using the definite integral, find
a).the total revenue (TR ) when Q = 2, 500 units
b).the change in total revenue when q increases from 2, 025 to 2,500
a) Consumer’s surplus when Q = 2, 500 and price = 300 birr.
b) the change in consumer's surplus when Q increases from 2, 025 to 2,500 due to
decrease in price from 330 birr to 300
birr.-------------------------------------------------------------------------------------------------
-----------------------------------
Dear colleague! Have you answered these questions? If you answer is no, please reread this
section and try to answer these question. If your answer is yes, go to the following section.

18
Section 1.2 Basic Concepts from Linear Algebra

After completing this section, you are expected to

- discuss about the definition of matrix

- explain the use of a determinant
- describe the various operation regarding matrix
- describe the concept of rank of a matrix
- discuss about Leontief input - output model
- describe the methods of solving simultaneous
linear equations

Dear colleague! You are expected to remember and reread what you have studied in quantitative
method for economists II to understand this section properly. Now we will carry out a revision on
basic concepts of this course.

What is matrix?
---------------------------------------------------------------------------------------------------------------------
-------------------------------------------------------------
Matrix is a rectangular array of numbers, parameters, and variables arranged in rows and columns
subject to certain rules of operation. These numbers, parameters, or variables are known as
elements of a matrix. The study of matrix is a concern of linear algebra .The reason for taking this
name is that the variables and numbers involved are of first degree, and the operations applied are
addition, subtraction and multiplication.
Given the following three linear equations
2x+3y+z=5
6x +2y+3z = 12
8x +3y+5z = 16

The coefficients of the variables of these equations make up a matrix. Matrices are usually
identified by capital letters A, B, C, etc. Thus, the matrix which represents the above linear
equations is that

A= 2 3 1
6 2 3
8 3 5 3x3

In matrix the elements are enclosed in parenthesis is or brackets. The numbers in the horizontal
line are called rows and the numbers in the vertical line are referred to as columns.

Determinant of a matrix

19
 Dear colleague! It is clear that every square matrix has it's own determinant. The determinant of
a matrix is written by enclosing the elements of the matrix by vertical bars. Therefore, the third
order determinant of the matrix

a 11 a 12 a 13 is
a 11 a 12 a 13
a 21 a 22 a 23 a 21 a 22 a 23
a 31 a 32 a 33 3x3 a 31 a 32 a 33

The value of the determinant can be expressed as a single numbers. For instance, the value of
the following second order determinant is

a 11 a 12 = a 11 a 22 - a 21 a 12
a 21 a 22

The third order determinant of the matrix is

a 11 a 12 a 13 a 22 a 23 a 21 a 23 a 21 a 22
a 21 a 22 a 23 =
a 11 a 32 a 33 -
a 12 a 31 a 33 +
a 13 a 31 a 32
a 31 a 32 a 33

=
a 11 (a 22 a 33 -a 32 a 23 ) - a 12 (a 21 a 33 -a 31 a 23 ) + a 13 (a 32 a 21 -a 31 a 22 )

Example Given A= 2 3 1
6 2 3
8 3 5

/ A/ = (2) 2 3 - (3) 6 3 + (1) 6 2

3 5 8 5 8 3

= 2 (10 -9) -3 (30-24) +1 (18-16)

/A/ = 2 - 18 + 2 = - 14
From the above matrix we have seen that
a 11 is multiplied by a lower order determinant
a 22 a 23
a 32 a 33 Obtained by deleting the column and

20
the row which contains
a 11 . Therefore, a 22 a 23 is referred to as
a 32 a 33

the minor of
a 11 in the original matrix. The minor of the element multiplied by
i+ j
( -1) where the (i)th row and j th column have been deleted since they contain the element is
known as cofactor of an element.

Inverse of a matrix
Given a square matrix A, A-1 is said to be the inverse of a square matrix if and only if
−1
AA = I where I is identity matrix.
This means, if the inverse of a matrix exists, this matrix is the only matrix which gives identity
matrix when multiplied by the original matrix. The concept of inverse of a matrix is crucial in
economies in solving simultaneous equations, input - output analysis and regression analysis.
There are two methods of finding the inverse of a matrix. These are
- Gauss elimination method
- Adjoint matrix - co factor method

Dear colleague! You should refer what you have studied from quantitative method for economists
II and try to understand these two methods.

Linear dependency and Rank of a matrix

A matrix is said to be linearly independent (non - singular) provided that none of its row or
column is a linear combination of other rows (columns). We can identify whether a matrix is
linearly dependent or independent using its determinant. If the determinant of a square matrix is
zero (vanish), then the matrix is referred to as linearly dependent (singular). Other wise, it is said
to be linearly independent (non - singular)
Given a linear equation system Ax = d, where a and n x n coefficient matrix,
|A| ¿ 0 implies that
- There is row or column independence
- A is non - singular
−1
- A exists
- Unique solution x = A
−1
d exists.

Rank of a matrix is the maximum number of linearly independent rows or columns. The rank of
an m x n matrix is at most m or n which ever is smaller. This means,
r (A) ¿ min ( m, n)

Solutions of Linear systems of Equations

There are three methods of solving simultaneous linear equations. These are
- The inverse method
- Gauss - Jordan elimination method
- Cramer’s rule

Dear colleague! Let us discuss each of them separately.

21
1. The Inverse Method

It is one of the ways of solving simultaneous linear equations

Adj. A
¿
= ¿ A /¿ pre multiplying both
−1 −1
Given A ( mxn) X( nx1 ) = dmx 1 , if it is possible to get A where A
−1
sides of the equation AX = d by A results in
−1 −1
A AX= A d
−1
IX= A d
−1
X= A d , where IX= X

Example
The ABC Company produces products X and Y. Each product is first processed in a machine M 1
and then sent to another machine M2 for finishing. Each unit of X requires 20 minutes time in M 1
and 10 minutes time in M2 where as each unit of Y requires 10 minutes of time in M 1 and 20
minutes in M2 . The total time available in each machine is 600 minutes. Calculate the amount of
these two types of products produced using matrix inversion.
The data is summarized as

X Y Total time
20 10 600
M1
10 20 600
M2

It can be expressed in matrix form AX = d as

20 10 X = 600
10 20 Y 600

First we should find the determinant to get the inverse of the coefficient matrix A.
−1
/ A/ = 20 (20) - 10(10) = 400 -100 = 300 ¿ 0. Thus, A exists so that there is a unique solution
adj. A
¿
d = ¿ A /¿ d, where adj. A is ad joint of
−1 −1
of AX = d which is given by X = A d. X = A
matrix A.
The ad joint of matrix is A the transpose of the cofactor of matrix A. Thus, we should find the
cofactor of matrix A.

C = 20 -10 20 -10
-10 20 which implies that adj. A = -10 20

22
 1
−1
This means, A = 300 20 -10
-10 20

1 −1
15 30
[ X ¿] ¿¿¿ −1 1 600 40−20 20
Therefore, ¿ = 30 15 600 = −20+40 = 20
X= 20 and Y = 20

2. Gauss - Jordan Elimination Method

To solve simultaneous linear equations using this method, it is necessary to express a system of
linear equations in augmented matrix and apply different row operations on this matrix until the
coefficient matrix becomes identity matrix. Finally, we can read the solution from the remaining
elements of the column vector d.

Example

Solve the following linear equations using the Gauss – Jordan elimination method.
2X+12Y = 40
8X+4Y = 28
The augmented matrix of this equation is
2 12 40
8 4 28
Applying row operation on this matrix

½ R1 2 12 40 1 6 20
8 4 28 8 4 28
-8R1 + R2 1 6 20 1 6 20
8 4 28 0 -44 -132

6
44 R2 + R1 1 6 20 1 0 2
0 -44 132 0 -44 -132

−1
44 R2 1 0 9 1 0 2
0 -44 -132 0 1 3

23
Now the coefficient matrix is reduced to an identify matrix. Thus,
X = 2 and Y = 3

3. Cramer's Rule

Dear colleague! It is another method of solving simultaneous linear equations through the use of
the determinant.
According to the Cramer rule
Ai/¿
¿¿ ¿
Xi = ¿A/¿ where Xi = the ith column variable in a series of equation. / A i/ represents the
determinant of a special matrix formed from the original coefficient matrix by replacing the
coefficient of Xi with the column of the constant. /A/ represents the determinant of matrix A.
Given the simultaneous linear equation systems
a11 X1 + a12X2 = d1
a21X1 + a22 X2 = d2
The matrix representation of these equations in the form AX = d
a11 a12 X1 = d1
a21 a22 X2 d2

¿
A 1/¿
X1 = ¿A/¿
¿¿
=
d1 a lignl¿12¿¿¿d2 a21¿ A 2/¿
X 2= ¿ A/¿ ¿¿
¿
=
a 11 d1
a 21 d
/A / /A /
Example
Solve the following simultaneous linear equations using the Cramer's rule.

A company produces three products every day. Their total production on a certain day is 45
tones. It is found that the production of the third product exceeds the production of the first
product by 8 tones while the total production of the first and the third product is twice of the
production of the second product. Determine the production level of each product using the
Cramer's rule.
Let X1 , X2 , X3 be the daily production of the three products. Therefore,
X1+X2 + X3 = 45
X3 = X1 +8
X1 + X3 = 2X2
Rearranging these equations gives us
X1+X2 + X3 = 45
-X1 + 0X2 + X3 = 8
X1 - 2X2 + X3 = 0

The matrix representation of these equations is

1 1 1 X1 45
-1 0 1 X2 = 8
1 -2 1 X3 0

24
 Where 1 1 1
-1 0 1 = A that is a 3x3 coefficient matrix
1 -2 1

Dear colleague! Now we should determine the determinant of this coefficient matrix A in order to
identify the solution of the equations. Thus,

/A/ = 1 0 1 - 1 -1 1 +1 -1 0
-2 1 1 1 1 -2

= 1 (2) - 1 (-2) +1 (2)

/A/ = 2+2+2 = 6

The determinant of matrix a, i.e. / A/ = 6 ¿ 0. Therefore, this system of the equations has unique
solutions.

45 1 1
A 1/¿ 8 0 1
¿ ¿¿
X1= ¿A/¿ , / A1/ = 0 −2 1
/A1/ = 45 (2) - 1 (8) + 1 ( -16

= 90 - 8 -16 = 66

66
|A 1|=66 , Thus X 6 11
1= =

1 45 1
A 2/¿ −1 8 1
¿ ¿¿
X2 = ¿ A/¿ , / A2/ = 1 0 1
/ A2/ = 1 (8) - 45 (-2) + 1 (- 8)
90
/A2/ = 8+ 90 - 8 = 90 Therefore, X2 = 6 = 15
+1 1 45
A 3/¿ −1 0 8
¿ ¿¿
X3 = ¿ A/¿ , /A3 / = 1 −2 0
/A3/ = 1 (16) - 1(-8) +45(2)
114
/A3/ = 16 + 8+90 = 114, Thus, X3 = 6 = 19
Dear colleague! As we have seen this unit is a revision of basic concept of Quantitative method
for economists I and II. Therefore, please read these materials and understand the remaining

25
concepts in linear algebra such as Leontief input - out put analysis and linear programming. Unit
three of quantitative method for economist II is mainly emphasized on these two areas.

By now you have completed the second section of this unit. Thus, try to do the following self -
test questions to examine how you have understood the main concepts in this section.

Self - Test 1.2

Do the following questions based on the information given in this section.
1. What is matrix?
----------------------------------------------------------------------------------------------------------------
------------------------------------------------------------------------
2. Define rank of a matrix
----------------------------------------------------------------------------------------------------------------
-------------------------------------------------------------
3. Describe properties of the determinant ------------------------------------------------------
----------------------------------------------------------------------------------------------------
4. Given the matrix
2 1 5
4 6 3
A= 3 2 7 , find the determinant of matrix A, i.e. /A/.

5. Given the following two related markets A and B, find the equilibrium quantity and
equilibrium prices in each market using inverse method.
In market A In market B
Qd = 82- 3PA + PB Qd = 92 + 2PA - 4PB
Qs = - 5+15 PA Qs = -6 + 32 PB

6. Find the solution of these simultaneous linear equations using the Cramer’s rule
2x - y + 3z = 9
x+ y + z = 6
x- y+ z = 2
7. What is the difference between minor and cofactor of an element of a matrix?
------------------------------------------------------------------------------------------------------
8. Discuss the methods of solving simultaneous linear equations briefly.
----------------------------------------------------------------------------------------------------
9. Describe the Leontief input - output model -------------------------------------------------
--------------------------------------------------------------------------------------------------

Check List
Write √ inside the box which corresponds to problem that you can solve.
1. Define what differentiation is-------------------------------------------------
2 Explain the rules of differentiation-----------------------------------------
3 What is integration of a function? -------------------------------------------
4 Describe the difference between differentiation and integration -------
5 Explain the difference between definite and indefinite integrals-----
6 What is matrix? --------------------------------------------------------------

26
 7 Describe the various operations regarding matrix-----------------------
8 Explain properties of the determinant ---------------------------------------
9 Describe Leontief Input - output model---------------------------------
10 Discuss the methods of solving linear programming problem---------
11 Discuss the methods of solving simultaneous linear equation systems-

Unit Summary

Differentiation is the process of finding the rate of change of the dependent variable with respect
to a given change in the independent variable. There are different rules of differentiation. Given
the function Y = ƒ (X), the derivative of this function with respect X represents the slope of the
function.
Using the technique of differentiation one can determine the marginal functions from their
corresponding total functions. For instance, it is possible to derive the marginal revenue and
marginal cost function from the total revenue and total cost functions respectively.
Integration is the reverse of differentiation. Given the derivative of the function, we can easily
find the primitive function using the method of integration if we have enough information
concerning the arbitrary constant. The rules of integration are the reverse of rules of
differentiation.

The method of definite integral enables us to calculate the consumer's surplus and producer's
surplus.

Matrix is the rectangular array of numbers, parameters, or variables arranged in rows and
columns subject to certain rules of operation. In matrix the elements in the horizontal line are
rows and those in the vertical line are columns of a matrix.

Every square matrix has determinant and its determinant is written by enclosing the elements of
the matrix by vertical bars. The value of the determinant is explained by a single number.

−1
For a given square matrix A, A is referred to as the inverse of a square matrix A if and only if
−1
AA = I, where I is an identity matrix.
Rank of a matrix represents the maximum number of linearly independent rows or columns of a
matrix. The rank of an m x n matrix is at most m or n whichever is smaller. This means
R (A) ¿ min. (m, n )

Given the linear equation system AX = d where a is an n x n coefficient matrix, / A/ ¿ 0 implies

that
- there is row or column independence.
- A is non singular.
−1
- A exists.
- a unique solution x =A
−1
d exists .

There are three methods of solving simultaneous linear equation systems. These are the inverse
method, Cramer's rule and the gauss - Jordan elimination method.

27
 Important Points

Differentiation
Chain rule linear programming
Determinant Simplex method
Logarithmic functions Duality theorem
Differential Exponential functions
Cramer’s rule Rate of change
Integration Inverse of a matrix
Definite Integral Matrix
Input- out put model Indefinite integral
Integrand
Ad joint matrix Consumer’s surplus
Cofactor matrix Producer’s surplus

Answers for Self Test Questions

Self Test 1.1
Part II
2
1) a. 9 x +8 x +6
b. x 3 e 3 x ( 4+3 x 4 )
3 2 11 2
c. 36( x +7 x −5 x ) (3 x +14 x −5)
2) a. 112 b. 118 c. 804.8 d. 36
5 4 −1
e. f ( x )=24 x −15 x +c f. f ( x )=42 x +18 x +c
−1 . 5
3) MR=(18 , 000+360 Q )(25+Q)
4) 49,600
5) a. 750,000 b. 81,750 c. 1,000,000 d. 149,500
Self Test 1.2
4) |A|=33 5) P A =5 , P B =3 , Q A =70 , Q B =90 6) x=1 , y=2 , z=3

28