Chapter 3.CALCULUS-new

CHAPTER 3:
CALCULUS

At the end of this chapter students should be able to:

 Solve 7 rules of differentiation problems
 Identify the first & second derivative
 Determine the minimum point,

maximum point and inflection point
 Applications on rules of differentiation

in calculus to solve business
mathematics problems

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Rule of differentiation

 Rule 1 – Constant rule

If y(x) = c, then y’(x) = 0

Example: Find y’(x) if b) y(x) = -71
a) y(x) = 9 y’(x) = 0

y’(x) = 0

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Rule of differentiation

 Rule 2 – Constant-multiple rule

If y(x) = mx, then y’(x) = = m


Example: Find y’(x) if

a) y(x) = 13x b) y(x) = -32x

=13 = -32


x has been eliminated

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Rule of differentiation

 Rule 3 – Power rule

If y(x) = , then = n −1


Example : Find y’(x) if

a) y(x) = axn b) y(x) = 9x4

= naxn-1 = 36x3


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Rule of differentiation

 Rule 4 – Sum rule

If y(x) = a + b ,

then = na −1 + mb −1

Example : Find y’(x) if

a) y(x) = 3x3 + 9x c) y(x) = 7 + 8x2 + 3x3

= 9x2 + 9 = 16x + 9x2


b) y(x) = 3x3 + 9x d) y(x) = x2 + 5

= -12x-4 + 6x 1

= x2 + x5

= 2x + 1 −4
5 5

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Rule of differentiation

 Rule 5 – Product rule

If y(x) = uv, u and v are function of x
u(x) v(x)
then = +

Example : Find y’(x) if b) y(x) = x2(2x + 8)
a) y(x) = (2x + 7)(x – 9)

uv uv

u(x) = 2x + 7 v(x) = x – 9 u(x) = x2 v(x) = (2x + 8)

= 2x + 7 =1 = 2x =2


= (2x + 7)(1) + (x – 9)(2) = x2 (2) + (2x + 8)(2x)

= (2x + 7) + (2x – 18)
= 4x - 11 = 2x2 + 4x2 + 16x
= 6x2 + 16x

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Rule of differentiation

 Rule 6 – Quotient rule

If y(x) = , and u’(x) & v’(x) both exist and v(x)≠ 0

then = −

[ ]2

Example : Find y’(x) if

a) y(x) = 2 +1 u b) y(x) = 5 2 +6 u
v 3 +6 v

u(x) = x2 + 1 v(x) = x u(x) = 5x2 + 6 v(x) = 3 + 6

= 2x =1 = 10x = 3 2

( 3 +6) 10 − 5x2 + 6 3 2
2 −( 2 +1)(1) =
= 2 ( 3 +6)2
(10 4 15x4 + 18 2
=( 2 2 )−( 2 +1) = +60 ) −
2 ( 3 +6) ( 3 +6)
−5 4 −18 2 +60
= ( 2 −1) = 6 +12 3 + 36
2

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Rule of differentiation

 Rule 7 – Chain rule

Itfhye(nx)y’=(x[)u=(x ) ] =, =. . [u(x)] −1

Example : Find y’(x) if y(x)
a) y(x) = (7x + 1)3

u

u = 7x2 + 1 y(x) = u3

= 14x = 3u2

= 3(7x2 + 1)2
= .


= 3(7x2 + 1)2 . 14x

= 42x(7x2 + 1)(7x2 + 1)

= 42x(49x4 + 14x2 + 1)

= 2058x5 + 588x3 + 42x 9

Continued example Rule 7……

Example : Find y’(x) if

b) y(x) = (x2 – 4x)2 y(x)

u

u = x2 – 4x y(x) = u2

= 2x - 4 = 2u

= 2(x2 – 4x)

= 2x2 – 8x

= .


= (2x2 8x) . (2x – 4)

= 4x3 – 8x2 – 16x2 + 32x

= 4x3 – 24x2 + 32x

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HIGHER ORDER OF

DIFFERENTIATION
 If f(x) = 4 + 3 + 2 + x + 1

a) = 4 3+ 3 2+ 2x+ 1 1st derivative
2nd derivative
2 3rd derivative
b) 2 = 12 2+ 6x+ 2 4th derivative

c) 3 = 24x + 6
3
4
d) 4 = 24

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Application of derivatives

 First derivatives
o The first derivatives produce the

marginal functions of total revenue,
profit, cost and etc.
o The marginal function can be used to
determine the critical quantity or volume
for the function

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Example 1

 The profit function of a product is given by
P(x) = 20x - 0.02 2 - 500; x represents the
quantity. Find the critical quantity for the
function

 = 20 – o.o4x (1st derivatives)
 quantity
Critical occurs when =0

 20- 0.04x = 0

0.04x = 20
20
x = 0.04 = 500 units

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Example 2

The revenue function of a product is given by
R(x) = 18x – 0.04 2, where represents the quantity. Find
the critical quantity for the function

= 18 – 0.08x (1st derivative)


Critical quantity occurs when =0

18 – 0.08x = 0
- 0.08x = -18
18
0.08 = 0.08

= 225 units

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Example 3

The cost function of a product is given by
C(q) = 500 + 18.2q – 0.03 2 + 0.00018 3. Find the
marginal cost function.

= 0 +18.2q0 – 0.06q + 0.00054q2
= 18.2 – 0.06q + 0.00054q2 (1st derivative)

Therefore : the marginal cost function =

C’(q) = 18.2 – 0.06q + 0.00054q2

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Application of derivatives

 Second derivatives
o Can be used to determine whether a

function produces maximum point,
minimum point or inflection point for a
critical quantity

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Example

a) Find the first & second derivatives of P(x) = 20x – 0.02x2 - 500

Solution


= 20 – 0.04x (1st derivative)
2
2 = -0.04 (2nd derivative)

b) Find the first & second derivatives of R(x) = 12x – 0.02x2

Solution

= 12 – 0.04x (1st derivative)


2 = -0.04 (2nd derivative)
2

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Application calculus in economics
and business

 Extensively in business and economics
 Meet companies objectives
 Determine the best price, optimize

expenses and output

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Total cost, average cost and marginal cost

Total Cost = Fixed Costs + Variable costs

Average Cost = ̅(x) = ( )

Marginal Cost = C’(x) =



Example
A firm produces x units of a product per month and the total cost
per month in ringgit is given by C(x) = 150 + 0.02 2. Find
a) The total cost when 20 units of are produced
b) The average cost function
c) The average cost when 20 units are produced
d) Marginal cost function
e) The marginal cost when 20 units are produced.

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Solution: d) C(x) = 150 + 0.02x2
C’(x)= = 0.04x
a) C(x) = 150 + 0.02x2
C(20)= 150 + 0.02(20)2
= RM158 e) C’(x) = 0.04x
C’(20)= 0.04(20)
b) ̅(x) = 150 + 0.02x2 = RM0.80

150
= + 0.02

c) ̅(20) = 150 + 0.02(20)2
20

= RM7.90

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Total revenue, average revenue and marginal revenue

Total Revenue = R(x) = px Average Revenue � (x) = ( )

( )
Demand function = p =

Marginal Revenue R’(x) =



Example:
Total revenue per month in ringgit, R(x) is given by R(x) = 20x –
0.03 2 where x is the number of units produced and sold per
month. Find
a) The total revenue when 30 units are produced and sold
b) The average revenue function
c) The average revenue when 30 units are produced and sold
d) The marginal revenue function
e) The marginal revenue when 30 units are produced and sold

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Solution:

a) R(x) = 20x – 0.03x2 d) R(x) = 20x – 0.03x2
R(30)= 20(30) – 0.03(30)2
= RM573 � (x)= = 20- 0.06x


b) � (x) = 20 − 0.03x2

e) R’(30) = 20 – 0.06(30)
= 20 – 0.03x = RM18.20

c) � (30) = 20 – 0.03(30)
= RM19.10

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Profit, breakeven analysis and optimization

Profit Function P(x) = R(x) – C(x)
Marginal Profit P’(x) =



Example: 23
The demand for an item produced by ZuaDRIF is given by p + 0.2x =
100 where p is the price per unit and x is quantity demand.The total
cost of producing x unit, C(x) = 800 + 30x where x is the level of
output. Find
a) The total revenue function
b) The total profit function
c) The marginal profit function
d) The marginal profit when

i) 90 units are sold
ii) 300 units are sold

Prepared By: Siti Nor Bt Fauzi

Solution: d) P’(x) =

p + 0.2x = 100
p = 100 – 0.2x = 70x – 0.2x2 – 800
= 70 – 0.4x
a) R(x) = px
= (100 – 0.2x)x e) i) P’(90) = 70 – 0.4(90)
= 100x – 0.2x2 = RM34

b) P(x) = R(x) – C(x) ii) P’(300) = 70 – 0.4(300)
= (100x – 0.2x2) – (800 + 30x) = -RM50
= 70x – 0.2x2 - 800

c) P(100) = 70(100) – 0.2(100)2 – 800
= RM4,200

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