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Differentiation
Session 9
Course : MATH6135-Business
Mathematics
Effective Period: September 2020
Thank you
These slides have been adapted from:
Haeussler, Jr. E.F., Paul, R. S., Wood, R. J. (2019).
Introductory Mathematical Analysis for Business,
Economics, and the Life and Social Sciences. 14th.
Ontario: Pearson.
Chapter 11
Acknowledgement
Learning Outcomes
After studying this chapter, the students should
be able to :
• LO 1: Identify the concept of mathematics in
business decision making
• LO 2: Explain the mathematics analysis concept
properly in business decision making
• LO 3: Apply mathematics concept and critical
thinking to solve economics and business problem
in business decision making
11 - 5
Copyright © 2019 Pearson Canada Inc. All rights reserved.
Introductory Mathematical Analysis
For Business, Economics, and The Life and Social Sciences
Fourteenth Canadian Edition
Chapter 11
Differentiation
Copyright © 2019 Pearson Canada Inc. All rights reserved.
11 - 6
Copyright © 2019 Pearson Canada Inc. All rights reserved.
Chapter Outline
11.1) Differentiation Formulas
11.2) Rules for Differentiation
11.3) The Derivative as a Rate of Change
11.4) The Product Rule and the Quotient Rule
11 - 7
Copyright © 2019 Pearson Canada Inc. All rights reserved.
11.1 Differentiation Formulas
11 - 8
Copyright © 2019 Pearson Canada Inc. All rights reserved.
11.2 Rules for Differentiation (1 of 7)
Below are some rules for differentiation:
BASIC RULE 1 Derivative of a Constant:
  0
d
c
dx

BASIC RULE 2 Derivative of xn:
  1
, .
for any real number
n n
d
x nx a
dx


COMBINING RULE 1 Constant Factor Rule:
 
   , for a differentiable function and a constant.
d
cf x cf x f c
dx
 ′
COMBINING RULE 2 Sum or Difference Rule:
   
     
'
d
f x g x f x g x
dx
    ′
11 - 9
Copyright © 2019 Pearson Canada Inc. All rights reserved.
11.2 Rules for Differentiation (2 of 7)
Example 1 – Derivatives of Constant Functions
   
807.4
(3) 0 3
( ) 5, ( ) 0
4 (4) 0.
1,938,623 , 0.
a. because is a constant function.
b.If then because is a constant function.
For example, the derivative of when is
c.If then
d
dx
g x g x g
g x g
ds
s t
dt

 
 
 
′
′
11 - 10
Copyright © 2019 Pearson Canada Inc. All rights reserved.
11.2 Rules for Differentiation (3 of 7)
Example 3 – Rewriting Functions in the Form xa
     
1/2
1/2 1 1/2
1/2
3/2
3/2 1
3/2 5/2
1 1 1 1
2 2 2 2
1
( ) . ( ) ( )
3 3
2 2
a. To differentiate , we write as . Thus,
b. Let We rewrite as . We have
y x x x
dy
x x
dx x x
h x h x h x x
x x
d
h x x x x
dx
 

 
 

   
 
    
′
11 - 11
Copyright © 2019 Pearson Canada Inc. All rights reserved.
11.2 Rules for Differentiation (4 of 7)
Example 5 – Differentiating Sums and Differences of
Functions
         
 
5
5
5 1/2 5 1/2
4 1/2 4
( ) 3
3 .
3 3
1 1
3 5 15 .
2 2
Differentiate the following functions:
a.
Solution: Here is the sum of two functions, and
Therefore,
F x x x
F x x
d d d d
F x x x x x
dx dx dx dx
x x x
x

 
   
   
′
11 - 12
Copyright © 2019 Pearson Canada Inc. All rights reserved.
11.2 Rules for Differentiation (5 of 7)
Example 5 – Continued
       
4
1/3
4 1/3
1
4
4 1/3 4 1/3
3 4/3 3 4/3
5
( )
4
( ) ( ) 5 .
1 1
5 5
4 4
1 1 5
(4 ) 5
4 3 3
b.
Solution: We rewrite in the form
z
f z
z
f z f z z z
d d d d
f z z z z z
dz dz dz dz
z z z z

 
 
 
 
 
   
 
 
 
    
 
 
′
11 - 13
Copyright © 2019 Pearson Canada Inc. All rights reserved.
11.2 Rules for Differentiation (6 of 7)
Example 5 – Continued
3 2
3 2
3 2
2
2
6 2 7 8
(6 ) (2 ) (7 ) (8)
6 ( ) 2 ( ) 7 ( ) (8)
6(3 ) 2(2 ) 7(1) 0
18 4 7
c.
Solution:
y x x x
dy d d d d
x x x
dx dx dx dx dx
d d d d
x x x
dx dx dx dx
x x
x x
   
   
   
   
  
11 - 14
Copyright © 2019 Pearson Canada Inc. All rights reserved.
11.2 Rules for Differentiation (7 of 7)
Example 7 – Finding an Equation of a Tangent Line
2
2
1
2
2
1
3 2
1.
3 2
3 2
2
3 .
2
1 3 5.
1
Find an equation of the tangent line to the curve
when
Solution:
Thus, The slope of the tangent line to the curve
when is To find the -co
x
x
y
x
x
x
y x x
x x
dy
dx x
dy
x y
dx





   
 
   
2 2
3 2 3(1) 2
1. 1.
1
1 5( 1).
ordinate, we
evaluate at This gives
Therefore an equation of the tangent line is
x
y x y
x
y x
 
   
  
11 - 15
Copyright © 2019 Pearson Canada Inc. All rights reserved.
Exercise
• Differentiate the functions
11 - 16
Copyright © 2019 Pearson Canada Inc. All rights reserved.
11.3 The Derivative as a Rate of Change (1 of 3)
Applications of Rate of Change to Economics
A manufacturer's total-cost function, c = f (q), gives the total cost
c of producing and marketing q units of a product. The rate
of change of c with respect to q is called the marginal cost.
Thus, marginal cost =
dc
dq
.
Suppose r = f (q) is the total revenue function for a manufacturer.
The marginal revenue is defined as the rate of change of the total
dollar value received with respect to the total number of units
sold. Hence, marginal revenue =
dr
dq
.
11 - 17
Copyright © 2019 Pearson Canada Inc. All rights reserved.
11.3 The Derivative as a Rate of Change (2 of 3)
Example 7 – Marginal Cost
If a manufacturer's average-cost equation is c = 0.0001q2
- 0.02q + 5 +
5000
q
find the marginal-cost function.
What is the marginal cost when 50 units are produced?
Solution: c = qc = q 0.0001q2
- 0.02q + 5 +
5000
q
æ
è
ç
ö
ø
÷
c = 0.0001q3
- 0.02q2
+ 5q + 5000. Differentiating c, we have the
marginal-cost function:
dc
dq
= 0.000(3q2
)- 0.02(2q)+ 5(1)+ 0 = 0.0003q2
- 0.04q + 5.
The marginal-cost when 50 units are
produced is
dc
dq q=50
= 0.0003(50)2
- 0.04(50)+ 5 = 3.75.
If c is in dollars and production is increased by 1 unit, from q = 50
to q = 51, then the cost of the additional unit is approximately $3.75.
11 - 18
Copyright © 2019 Pearson Canada Inc. All rights reserved.
11.3 The Derivative as a Rate of Change (3 of 3)
( )
( ) .
( )
( )
( ) 100%.
( )
The relative rate of change of is
The percentage rate of change of is
f x
f x
f x
f x
f x
f x

′
′
Example 9 – Relative and Percentage Rates of Change
 
 
2
( ) 3 5 25 5.
( ) 6 5. (5) 25 (5) 75,
5
5 25
0.333.
5 75
Determine the relative and percentage rates of change of
when
Solution: Here Since and
the relative rate of change of when is
y f x x x x
f x x f f
y x
f
f
    
   

 
′ ′
′
100%
(0.333)(100) 33.3%.
Multiplying by gives the percent rate
of change: 
11 - 19
Copyright © 2019 Pearson Canada Inc. All rights reserved.
11.4 The Product Rule and the Quotient
Rule (1 of 6)
COMBINING RULE 3 The Product Rule:
   
   
     
 
( )
d d d
f x g x f x g x f x g x
dx dx dx
  
COMBINING RULE 4 The Quotient Rule:
 
 
2
2
( ) ( ) ( )
( )
( ) ( )
With the understanding about the denominator not
being zero,we can write
g x f x f x g x
d f x
dx g x g x
f gf fg
g g

 
 
 
 
  

 
 
′
′ ′
′ ′
11 - 20
Copyright © 2019 Pearson Canada Inc. All rights reserved.
11.4 The Product Rule and the Quotient
Rule (2 of 6)
Example 1 – Applying the Product Rule
    
   
 
2
2
2
3 4 5 , ( ).
( ) ( ) ( ), ( ) 3 ( ) 4 5 .
( ) ( ) ( ) ( )
If find
Solution: We consider as a product of two functions:
where and
Therefore we can apply the product rule:
F x x x x F x
F
F x f x g x f x x x g x x
F x f x g x f x g x
d
x
dx
  
    
 

′
′ ′ ′
       
     
2
2 2
3 4 5 3 4 5
2 3 4 5 3 4 12 34 15
d
x x x x x
dx
x x x x x x
   
    
   
   
       
11 - 21
Copyright © 2019 Pearson Canada Inc. All rights reserved.
11.4 The Product Rule and the Quotient
Rule (3 of 6)
Example 5 – Applying the Quotient Rule
 
       
 
     
 
  
 
2
2
2
2 2
2
2
2 2
4 3
( ) , ( ).
2 1
( )
( ) , ( ) 4 3 ( ) 2 1.
( )
( ) ( ) ( ) ( )
( ( ))
2 1 4 3 4 3 2 1
2 1
2 1 8 4 3 2 2 2 1 2 3
2 1 2 1
If find
Solution: where and
x
F x F x
x
f x
f x f x x g x x
g x
g x f x f x g x
F x
g x
d d
x x x x
dx dx
x
x x x x x
x x



    


    


    
 
 
′
′ ′
′
11 - 22
Copyright © 2019 Pearson Canada Inc. All rights reserved.
11.4 The Product Rule and the Quotient
Rule (4 of 6)
c. f (x) =
5x2
- 3x
4x
Solution: Rewriting, we have f (x) =
1
4
5x2
- 3x
x
æ
è
ç
ö
ø
÷ =
1
4
(5x - 3)
for x ¹ 0. Thus, f '(x) =
1
4
(5) =
5
4
for x ¹ 0.
Example 7 – Differentiating Quotients without Using the
Quotient Rule
11 - 23
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11.4 The Product Rule and the Quotient
Rule (5 of 6)
11 - 24
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11.4 The Product Rule and the Quotient
Rule (6 of 6)
Example 9 – Finding Marginal Propensities to Consume
and to Save
       
 
3
3/2 3
2
5(2 3)
10
100.
10 2 3 2 3 10
5
10
If the consumption function is given by determine the marginal
propensity to consume and the marginal propensity to save when
Solution:
I
C
I
I
d d
I I I I
dC dI dI
dI I





    

 



     
 
1/2 3
2
100
10 3 3 2 3 1
5 100,
10
1297
5 0.536.
12,100
100 1 0.536 0.464.
When the marginal propensity to
consume is
The marginal propensity to save when is
I
I I I
I
I
dC
dI
I






 
   
 
 
 

 
 
 
 
 
 
  
11 - 25
Copyright © 2019 Pearson Canada Inc. All rights reserved.
Home Work
1) If a manufacturer's average-cost equation is
a) Find the marginal-cost function.
b) What is the marginal cost when 15 units are
produced?
2) If the demand equation for a manufacturer's product is
where p denotes the price per unit for q units.
Find the marginal-revenue function in each case. (Recall
that revenue = p.q)
Haeussler, Jr. E.F., Paul, R. S., Wood, R. J. (2019).
Introductory Mathematical Analysis for
Business, Economics, and the Life and Social
Sciences. 14th. Ontario: Pearson.
References

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Differentiation Business Mathematics ppt

  • 1. Differentiation Session 9 Course : MATH6135-Business Mathematics Effective Period: September 2020
  • 3. These slides have been adapted from: Haeussler, Jr. E.F., Paul, R. S., Wood, R. J. (2019). Introductory Mathematical Analysis for Business, Economics, and the Life and Social Sciences. 14th. Ontario: Pearson. Chapter 11 Acknowledgement
  • 4. Learning Outcomes After studying this chapter, the students should be able to : • LO 1: Identify the concept of mathematics in business decision making • LO 2: Explain the mathematics analysis concept properly in business decision making • LO 3: Apply mathematics concept and critical thinking to solve economics and business problem in business decision making
  • 5. 11 - 5 Copyright © 2019 Pearson Canada Inc. All rights reserved. Introductory Mathematical Analysis For Business, Economics, and The Life and Social Sciences Fourteenth Canadian Edition Chapter 11 Differentiation Copyright © 2019 Pearson Canada Inc. All rights reserved.
  • 6. 11 - 6 Copyright © 2019 Pearson Canada Inc. All rights reserved. Chapter Outline 11.1) Differentiation Formulas 11.2) Rules for Differentiation 11.3) The Derivative as a Rate of Change 11.4) The Product Rule and the Quotient Rule
  • 7. 11 - 7 Copyright © 2019 Pearson Canada Inc. All rights reserved. 11.1 Differentiation Formulas
  • 8. 11 - 8 Copyright © 2019 Pearson Canada Inc. All rights reserved. 11.2 Rules for Differentiation (1 of 7) Below are some rules for differentiation: BASIC RULE 1 Derivative of a Constant:   0 d c dx  BASIC RULE 2 Derivative of xn:   1 , . for any real number n n d x nx a dx   COMBINING RULE 1 Constant Factor Rule:      , for a differentiable function and a constant. d cf x cf x f c dx  ′ COMBINING RULE 2 Sum or Difference Rule:           ' d f x g x f x g x dx     ′
  • 9. 11 - 9 Copyright © 2019 Pearson Canada Inc. All rights reserved. 11.2 Rules for Differentiation (2 of 7) Example 1 – Derivatives of Constant Functions     807.4 (3) 0 3 ( ) 5, ( ) 0 4 (4) 0. 1,938,623 , 0. a. because is a constant function. b.If then because is a constant function. For example, the derivative of when is c.If then d dx g x g x g g x g ds s t dt        ′ ′
  • 10. 11 - 10 Copyright © 2019 Pearson Canada Inc. All rights reserved. 11.2 Rules for Differentiation (3 of 7) Example 3 – Rewriting Functions in the Form xa       1/2 1/2 1 1/2 1/2 3/2 3/2 1 3/2 5/2 1 1 1 1 2 2 2 2 1 ( ) . ( ) ( ) 3 3 2 2 a. To differentiate , we write as . Thus, b. Let We rewrite as . We have y x x x dy x x dx x x h x h x h x x x x d h x x x x dx                    ′
  • 11. 11 - 11 Copyright © 2019 Pearson Canada Inc. All rights reserved. 11.2 Rules for Differentiation (4 of 7) Example 5 – Differentiating Sums and Differences of Functions             5 5 5 1/2 5 1/2 4 1/2 4 ( ) 3 3 . 3 3 1 1 3 5 15 . 2 2 Differentiate the following functions: a. Solution: Here is the sum of two functions, and Therefore, F x x x F x x d d d d F x x x x x dx dx dx dx x x x x            ′
  • 12. 11 - 12 Copyright © 2019 Pearson Canada Inc. All rights reserved. 11.2 Rules for Differentiation (5 of 7) Example 5 – Continued         4 1/3 4 1/3 1 4 4 1/3 4 1/3 3 4/3 3 4/3 5 ( ) 4 ( ) ( ) 5 . 1 1 5 5 4 4 1 1 5 (4 ) 5 4 3 3 b. Solution: We rewrite in the form z f z z f z f z z z d d d d f z z z z z dz dz dz dz z z z z                               ′
  • 13. 11 - 13 Copyright © 2019 Pearson Canada Inc. All rights reserved. 11.2 Rules for Differentiation (6 of 7) Example 5 – Continued 3 2 3 2 3 2 2 2 6 2 7 8 (6 ) (2 ) (7 ) (8) 6 ( ) 2 ( ) 7 ( ) (8) 6(3 ) 2(2 ) 7(1) 0 18 4 7 c. Solution: y x x x dy d d d d x x x dx dx dx dx dx d d d d x x x dx dx dx dx x x x x                   
  • 14. 11 - 14 Copyright © 2019 Pearson Canada Inc. All rights reserved. 11.2 Rules for Differentiation (7 of 7) Example 7 – Finding an Equation of a Tangent Line 2 2 1 2 2 1 3 2 1. 3 2 3 2 2 3 . 2 1 3 5. 1 Find an equation of the tangent line to the curve when Solution: Thus, The slope of the tangent line to the curve when is To find the -co x x y x x x y x x x x dy dx x dy x y dx                2 2 3 2 3(1) 2 1. 1. 1 1 5( 1). ordinate, we evaluate at This gives Therefore an equation of the tangent line is x y x y x y x         
  • 15. 11 - 15 Copyright © 2019 Pearson Canada Inc. All rights reserved. Exercise • Differentiate the functions
  • 16. 11 - 16 Copyright © 2019 Pearson Canada Inc. All rights reserved. 11.3 The Derivative as a Rate of Change (1 of 3) Applications of Rate of Change to Economics A manufacturer's total-cost function, c = f (q), gives the total cost c of producing and marketing q units of a product. The rate of change of c with respect to q is called the marginal cost. Thus, marginal cost = dc dq . Suppose r = f (q) is the total revenue function for a manufacturer. The marginal revenue is defined as the rate of change of the total dollar value received with respect to the total number of units sold. Hence, marginal revenue = dr dq .
  • 17. 11 - 17 Copyright © 2019 Pearson Canada Inc. All rights reserved. 11.3 The Derivative as a Rate of Change (2 of 3) Example 7 – Marginal Cost If a manufacturer's average-cost equation is c = 0.0001q2 - 0.02q + 5 + 5000 q find the marginal-cost function. What is the marginal cost when 50 units are produced? Solution: c = qc = q 0.0001q2 - 0.02q + 5 + 5000 q æ è ç ö ø ÷ c = 0.0001q3 - 0.02q2 + 5q + 5000. Differentiating c, we have the marginal-cost function: dc dq = 0.000(3q2 )- 0.02(2q)+ 5(1)+ 0 = 0.0003q2 - 0.04q + 5. The marginal-cost when 50 units are produced is dc dq q=50 = 0.0003(50)2 - 0.04(50)+ 5 = 3.75. If c is in dollars and production is increased by 1 unit, from q = 50 to q = 51, then the cost of the additional unit is approximately $3.75.
  • 18. 11 - 18 Copyright © 2019 Pearson Canada Inc. All rights reserved. 11.3 The Derivative as a Rate of Change (3 of 3) ( ) ( ) . ( ) ( ) ( ) 100%. ( ) The relative rate of change of is The percentage rate of change of is f x f x f x f x f x f x  ′ ′ Example 9 – Relative and Percentage Rates of Change     2 ( ) 3 5 25 5. ( ) 6 5. (5) 25 (5) 75, 5 5 25 0.333. 5 75 Determine the relative and percentage rates of change of when Solution: Here Since and the relative rate of change of when is y f x x x x f x x f f y x f f             ′ ′ ′ 100% (0.333)(100) 33.3%. Multiplying by gives the percent rate of change: 
  • 19. 11 - 19 Copyright © 2019 Pearson Canada Inc. All rights reserved. 11.4 The Product Rule and the Quotient Rule (1 of 6) COMBINING RULE 3 The Product Rule:                 ( ) d d d f x g x f x g x f x g x dx dx dx    COMBINING RULE 4 The Quotient Rule:     2 2 ( ) ( ) ( ) ( ) ( ) ( ) With the understanding about the denominator not being zero,we can write g x f x f x g x d f x dx g x g x f gf fg g g                  ′ ′ ′ ′ ′
  • 20. 11 - 20 Copyright © 2019 Pearson Canada Inc. All rights reserved. 11.4 The Product Rule and the Quotient Rule (2 of 6) Example 1 – Applying the Product Rule            2 2 2 3 4 5 , ( ). ( ) ( ) ( ), ( ) 3 ( ) 4 5 . ( ) ( ) ( ) ( ) If find Solution: We consider as a product of two functions: where and Therefore we can apply the product rule: F x x x x F x F F x f x g x f x x x g x x F x f x g x f x g x d x dx            ′ ′ ′ ′               2 2 2 3 4 5 3 4 5 2 3 4 5 3 4 12 34 15 d x x x x x dx x x x x x x                         
  • 21. 11 - 21 Copyright © 2019 Pearson Canada Inc. All rights reserved. 11.4 The Product Rule and the Quotient Rule (3 of 6) Example 5 – Applying the Quotient Rule                          2 2 2 2 2 2 2 2 2 4 3 ( ) , ( ). 2 1 ( ) ( ) , ( ) 4 3 ( ) 2 1. ( ) ( ) ( ) ( ) ( ) ( ( )) 2 1 4 3 4 3 2 1 2 1 2 1 8 4 3 2 2 2 1 2 3 2 1 2 1 If find Solution: where and x F x F x x f x f x f x x g x x g x g x f x f x g x F x g x d d x x x x dx dx x x x x x x x x                           ′ ′ ′ ′
  • 22. 11 - 22 Copyright © 2019 Pearson Canada Inc. All rights reserved. 11.4 The Product Rule and the Quotient Rule (4 of 6) c. f (x) = 5x2 - 3x 4x Solution: Rewriting, we have f (x) = 1 4 5x2 - 3x x æ è ç ö ø ÷ = 1 4 (5x - 3) for x ¹ 0. Thus, f '(x) = 1 4 (5) = 5 4 for x ¹ 0. Example 7 – Differentiating Quotients without Using the Quotient Rule
  • 23. 11 - 23 Copyright © 2019 Pearson Canada Inc. All rights reserved. 11.4 The Product Rule and the Quotient Rule (5 of 6)
  • 24. 11 - 24 Copyright © 2019 Pearson Canada Inc. All rights reserved. 11.4 The Product Rule and the Quotient Rule (6 of 6) Example 9 – Finding Marginal Propensities to Consume and to Save           3 3/2 3 2 5(2 3) 10 100. 10 2 3 2 3 10 5 10 If the consumption function is given by determine the marginal propensity to consume and the marginal propensity to save when Solution: I C I I d d I I I I dC dI dI dI I                         1/2 3 2 100 10 3 3 2 3 1 5 100, 10 1297 5 0.536. 12,100 100 1 0.536 0.464. When the marginal propensity to consume is The marginal propensity to save when is I I I I I I dC dI I                                  
  • 25. 11 - 25 Copyright © 2019 Pearson Canada Inc. All rights reserved. Home Work 1) If a manufacturer's average-cost equation is a) Find the marginal-cost function. b) What is the marginal cost when 15 units are produced? 2) If the demand equation for a manufacturer's product is where p denotes the price per unit for q units. Find the marginal-revenue function in each case. (Recall that revenue = p.q)
  • 26. Haeussler, Jr. E.F., Paul, R. S., Wood, R. J. (2019). Introductory Mathematical Analysis for Business, Economics, and the Life and Social Sciences. 14th. Ontario: Pearson. References

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