THE INSTITUTE OF FINANCE MANAGEMENT (IFM)
Department of Mathematics
Business Mathematics
MTU 07101
Elements of Calculus in Economics
Calculus
Calculus deals with rate of change of quantity with
respect to another
Calculus has two branches
i). Differentiation
ii). Integration
Differential calculus
Is the study of
a) Derivative (rate of change of one quantity with
respect to another)
b) Motion of objects in space e.g. Velocity and
acceleration
c) Marginal concepts in economy
d) The growth or decay of Biological population
e) Curve sketching and finding Maximum and
minimum values of functions
Differentiation
The derivative f of a function f is given by
f x h f(x)
f (x) lim
h 0 h
Provided this limit exist. If f (a) exists, then is said
to be differentiable at x a.
The process of finding the derivative is called
differentiation.
Symbols for Derivatives
If y f(x), then the derivative of y ( or f) at x
symbols used for differentiation are
dy d
y , f (x), , f(x), Dy or Df(x)
dx dx
Rules of Differentiation
• There some rules or derivative formula used for
differentiation of functions
The derivative of Constant function
If f is a function such that f(x) c , then
d
(f(x)) 0
dx
derivative of constant is zero.
Example
Find the derivative of the following constant
functions
i) f(x) 5 ii). f(x) c
Solution
d d
i) f(x) 5 , dx
(5) 0 ii). f(x) c , (c) 0
dx
The derivative of x to a power
If f is a function such that f(x) x n , then
f (x) nx n 1
but n 0
The derivative of x to a power n is the exponent
times x to the next lower power
Example: Find the derivative of functions
1 1
(i) y x 5
(ii) y 5 (iii) y
x x
Solution:
Use dy
f (x) nx n 1
dx
(i). yx 5
, dy d(x 5 )
5 x4
dx dx
1
1 dy d 2
(ii). y ,
x d(x 5
) 6 1
x5 dx 5 x 5 6
dx dx x
1
1 d x 2
(iii). y 1 , dy d 3
x x 2
1
x dx dx dx 2
Derivative of sum or difference
If f is a function such that f(x) h(x) g(x) ,then
f (x) h(x) g(x)
The derivative of the sum is the sum derivative.
Example: Find the derivative of functions
1
(i). f(x) 4 x5 3 x 2 (ii). y x 3 5
5x
Solution
x
dy
(i). f(x) 4 x5 3 x 2 , 20 x 4 6 x
dx
(ii). y x 3 1 5 x , dy 1
3x 5 6 5
2
x5 dx x
Product rule
If f is a function such that f(x) h(x) g(x), then
f (x) h(x) g(x) g(x) h(x)
derivative of the product is the first factor times the
derivative of the second plus the second factor times
the derivative of the first.
Example: Differentiate the function
1
(i) f ( x) (4 x )(3x ) (ii) y (x 1)( 5 )
5 2 3
x
Use f (x) h(x) g(x) g(x) h(x)
(i). f(x) (4 x 5
)(3 x 2
)
dy
(20 x 4 )(3 x 2 ) (6 x)(4 x 5 )
dx
1
(ii). y (x3 1)( 5
)
x
dy 3 1
2 x 5 6
dx x
Derivative of quotient
If f is a function such that f(x) h(x) , then
g(x)
g(x) h(x) h(x) g(x)
f (x)
g(x) 2
The derivative of a quotient is the denominator
times the derivative of the numerator minus the
numerator times the derivative of the denominator,
all divided by the denominator squared.
Example:
2 x3 5 x1
(i) f(x) (ii) y 5
x2 x 4 x1
Solution: x1
2 x3 5
(i) f(x) (ii). y x 5 4 x 1
x2
dy x 2 (6 x 2 ) (2 x 3 5) 2 x dy (x5 4 x 4) 1 (x 1)(5x 4 4)
dx (x 2 )2 dx (x5 4 x 1)2
dy 10
2 3 dy 4 x 5 5 x 4 3
dx x dx (x5 4 x 1)2
Chain rule
Chain rule is used to find the derivative of composite
functions (function of function) fog(x) = f(g(x))
If f is a function such that f(x) = f(g(x)), then
dy dy d g
dx d g dx
Example: Differentiate
(i). y (4 3x) 5 (ii).
y x 3 2x 7
Solution:
Given y (4 3 x)
5
let g ( x) 4 3x then y (g(x))
5
dg dy
3 5 g4
dg
dx
dy dy dg
use
dx dg dx
dy
5 g 4 3 15(4 3 x) 4
dx
Derivative of Parametric Functions
If y f(t) and x f(t) when t is called a parameter,
such equations are parametric equations,
dy dy dt
dx dt dx
Example: Givenx (t 1)2 and y t 1 , Find
2
dy dt dy dx
Solution:
dy
, then 2 t and 2(t 1)
dx dt dx dt dt
1
2 t
dy
dx 2(t 1)
dy t
dx t 1
Derivative of Exponential function
For exponential function f(x) e x
, then
f (x) e x
dy
If y a , then, Iny xIna, dx
x x
yIna a Ina
Example: Find the derivative of y e
2 x3
Solution:y e 2 x3
Let y e u
and u 2 x 3
dy du
eu
dx dx
dy
6x e
2 2 x3
dx
Derivative of Logarithm of function
If f(x) Inx , then
1
f (x)
x
Example: Find the derivative of
i).y In(x 2 4 x 3) ii).y ln(2 x 5)3
Solution:
i). y In(x 2 4 x 3) ii).y ln(2x 5)3
dy 2 x 4 dy 6
2
dx x 4 x 3 dx 2 x 5
Higher Derivatives
dy d2y
If y f(x) and g(x) , then g(x) can be 2
dx
2 dx
differentiated with respect to x to give d y2 .
dx
2
d y d dy
dx 2 dx dx
3
Example: Find d y when y In x
dx 3
Solution: Given y In x
dy 1 d3 y 2
3
dx x dx 3
x
d2 y 1
dx 2 x2
Minimum and maximum
If f is a function such that y = af(x), then
2
dy d y
Maximum when 0 and 2
0
dx dx
dy d2y
Minimum when 0 and 0
dx 2
dx
Example: using second derivative find the relative
maximum or minimum points of the curve
1 3 2
f(x) x x 8 x 2
3
Solution
f (x) x 2 2 x 8
f (x) 2 x 2
When f (x) 0 2 x 8 0 x 2 or x 4
2
x
f (2) 2(2) 2 0 relative maximum
at (-2, f(-2))
1 3
f(2) (2) (2) 2 8(2) 2
3
f (4) 2(4) 2 0 , relative minimum
At (4, f(4))
1
f(4) (4) 3 4 2 8(4) 2
3
INTERGRATION
INTERGRATION AS REVERSE PROCESS OF
DIFFERENTIATION (ANTIDERIVATIVES)
Integration
Integral is the process of summing. It involve
determination of
a) Areas
b) Anti-derivatives
c) Volumes
d) Revenue, cost and profit functions.
Properties of integral
Integral of zero
If f is a function such that f(x) 0 , then
0 dx c
The integral of zero is constant
Example: Find the integral of 0 dx
Integral of constant
If f is a function such that f(x) k , then
Kdx Kx c
The integral of constant k is kX + constant
Example:
i). 2 dx ii). ydx
Solution
i). 2 dx 2 x c ii). ydx yx c
Integral of Power
If f is a function such that f(x) x n , then
n 1
x
dx n 1 c
n
x
Increase the exponent of x by one and divide by new
exponent
Examples: Evaluate the following
3
i). x dx ii). x dx
3
4
Solution: 3 7
x4 4 4
i). x dx 4 c ii). x dx 7 x c
3 4
Integral of coefficient of the function
If y is a function such that y = kf(x), then
Kf(x) dx K f( x) dx
Example: Evaluate the following functions
i) dx
3
4 x
3
ii) px 4
dx
Solution
i) 4 x 3
dx 4 dx
x 3
c
3 4
4 x dx x
3 3
ii) px 4
dx p x dx 4
3 7
4 4
px dx4
7
px C
1
The integral of x
1
If f is a function such that f(x) = , then
x
1
x dx ln x c
True only when x is positive
The integral of e
x
• If f is a function such that f(x) = e x , then
c
x x
e dx e
Integral of sum and difference
If f is a function such that f(x) h(x) g(x) , then
h(x) g(x) dx h(x) dx g(x) dx
The integral of the sum is the sum integral
Example:
i) (x 5 2 x) dx
ii) (3 x 3 2 x 2 3x) dx
iii) (4 x 2e x 3 1 x 2 ) dx
x 2
Solution
i) (x 5 2 x) dx x 5 dx 2 xdx
1 6
2 x) dx x x2 C
5
(x
6
ii)
dx 3xdx
3 2 3 2
(3 x 2 x 3 x) dx 3 x 2 x
3 4 2 3 3 2
3 x) dx x x x C
3 2
(3 x 2 x
4 3 2
iii) 3 1 2 3 1
(4 x 2 e x ) dx 4 xdx 2 e x dx dx x 2 dx
x
x 2 x 2
3 1 2 1
x ) dx 2 x 2 2e x 3 ln x x 3 C
x
( 4 x 2e
x 2 6
Definite integrals
If the function f(x) is differentiable and
continuous at the interval x = a and x = b,
then
Examples
Use the above properties to evaluate the
following integrals
1. 3xdx
3
2. 2 x 3dx
3
2
Solution
3
1. 3xdx x 3 1 12
3 3 2 3 2 3 2 27 3
1 2 1 2 2 2 2
2. 2 x 3dx x 3x 9 9 4 6 2
3 3
2
2 2
Application of Calculus
In business mathematics we experience changes of
variable with respects to others.
a) Sales revenue changes with volumes of units sold
b) Cost change with volume of unit produced or sold
c) Sales demand for a product varies with its sale price
Marginal Functions
Total cost ( C )
Total cost represents the cost of producing a specific
quantity (X) of commodity
The total cost consist of two types of costs.
i)Fixed cost is the cost independent of quantity
produced e.g. salaries, rent etc.
ii)Variable cost is the cost depend on the quantity
produced
C = Fixed cost +Variable cost
Example:
Given a total cost production function
c(x) 4 x5 3 x 2 50,000. Identify the
i). fixed cost
ii). variable cost
Solution:
i). Fixed cost = 50,000
ii). Variable cost = 4 x 5
3 x 2
Marginal cost(MC)
• MC is the cost of producing one more unit of a good.
If c(x) is the total cost of producing “x” units, the
d(C)
Marginal Cost = MC
dx
Average cost (AC)
• Average cost function (AC) given the cost per
unity is defined as
Totalcost C
AC
Quantity X
Marginal average Cost (MAC)
The marginal average cost (MAC) is defined as
d(AC)
MAC
dx
Example: Give a total cost production function
c(x) 4 x5 3 x 2 50,000. Identify the
a. fixed cost
b. variable cost
c. marginal Cost
d. average cost
e. marginal average cost
Solution: c(x) 4 x5 3 x 2 50,000.
a. fixed cost 50,000.
b. variable cost 4 x 3 x
5 2
d(C)
c. marginal Cost, MC
dx
MC 20x4 6 x
d. Average Cost
Totalcost C
AC
Quantity X
4 x 5 3 x 2 50000
AC
x
50000
AC 4 x 3 x
4
x
e. Marginal Average Cost
d(AC)
MAC
dx
d(AC) 50000
16 x 3
3
dx x2
Total Revenue functions (R).
Total revenue function is the product of price per
item by quantity.
Total revenue R Price (P) quantity (X)
Price as a function of quantity expresses the link
between the price and the quantity demanded and
referred to demand function.
Marginal Revenue Function (MR)
• The Marginal Revenue (MR)
• Marginal revenue (MR) = R
'
d(R)
MR
dx
Average Revenue Function (AR)
Average revenue function (AR) gives the revenue per
unit. It is defined as
R
AR
X
But R PX where “P” is the price per units, thus
R
P
X
Then the average revenue AR is equal to the price “P”
Marginal Average Revenue Function (MAR)
The marginal average Revenue (MAR) is defined as
d(AR)
MAR
dx
Example: Given the total revenue function
1 3
R(x) x 3 x 2 45,000.
2
Find the (i). Marginal Revenue Function
(ii). Average Revenue Function
(iii). Price when MR = 0
Solution
i)Marginal Revenue Function (MR)
d(R) 3 2
MR x 6x
dx 2
ii)Average Revenue Function(AR)
R 0.5 x3 3x 2 45000
AR
X x
45000
AR 0.5 x 2 3x
x
iii) Price when MR = 0
3 2
MR x 6 x 0
2
3x (0.5x-2) =0
x = 0 or x = 4
Since AR = P
45000
P 0.5 x 2 3x
x
when x = 4,
45000
P 0.5(4) 3(4)
2
11246
4
Hence the price when MR = 0 is 11,246
Total Profit
Profit (π) is the difference between the total
revenue and total cost. That is π = R – C
The aim of any firm is to maximize profit. This
involves finding value of which makes
dπ
0
dx
Differentiating π = R – C
dπ d(R) d(C)
dx dx dx
dπ
At maximum profit 0
dx
There fore MR MC 0 hence MR MC
Example:
The total cost and revenue function for a certain
commodity are R 50 x x 2
and c 0.01x 2
20x 100
(a) Derive the total profit function
(b) Find the profit break even point
(c) calculate the level and amount of demand that
maximize profit
Solution
a) Derive the total profit function
π R C
π 1.01x 2 30x 100
b) The break - even points are levels of demand
which makes π 0
1.01x2 30x 100 0
x = 26 or x = 4
The break even points is when the demand is
4 or 26 products
dπ d2π
(c). Profit is maximized when 0 and 0
dx dx 2
Since π 1.01x 2
30x 100
Then dπ 2.02 x 30
dx
2.02x 30 0 x 15
d2π
2
2.02 0 gives maximum profit
dx
when x 15 then π 1.01(15) 30(15) 100 123
2
Hence the maximum profit of 122.5 is raised
when the demand is 15
