This chapter discusses differentiation, including: - Defining the derivative using the limit definition of the slope of a tangent line. - Basic differentiation rules for constants, polynomials, sums and differences. - Interpreting the derivative as an instantaneous rate of change. - Applying the product rule and quotient rule to differentiate products and quotients. - Using differentiation to find equations of tangent lines, velocities, marginal costs, and other rates of change.

1. INTRODUCTORY MATHEMATICALINTRODUCTORY MATHEMATICAL
ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences
©2007 Pearson Education Asia
Chapter 11Chapter 11
DifferentiationDifferentiation

2. ©2007 Pearson Education Asia
INTRODUCTORY MATHEMATICAL
ANALYSIS
0. Review of Algebra
1. Applications and More Algebra
2. Functions and Graphs
3. Lines, Parabolas, and Systems
4. Exponential and Logarithmic Functions
5. Mathematics of Finance
6. Matrix Algebra
7. Linear Programming
8. Introduction to Probability and Statistics

3. ©2007 Pearson Education Asia
9. Additional Topics in Probability
10. Limits and Continuity
11. Differentiation
12. Additional Differentiation Topics
13. Curve Sketching
14. Integration
15. Methods and Applications of Integration
16. Continuous Random Variables
17. Multivariable Calculus
INTRODUCTORY MATHEMATICAL
ANALYSIS

4. ©2007 Pearson Education Asia
• To compute derivatives by using the limit
definition.
• To develop basic differentiation rules.
• To interpret the derivative as an instantaneous
rate of change.
• To apply the product and quotient rules.
• To apply the chain rule.
Chapter 11: Differentiation
Chapter ObjectivesChapter Objectives

5. ©2007 Pearson Education Asia
The Derivative
Rules for Differentiation
The Derivative as a Rate of Change
The Product Rule and the Quotient Rule
The Chain Rule and the Power Rule
11.1)
11.2)
11.3)
Chapter 11: Differentiation
Chapter OutlineChapter Outline
11.4)
11.5)

6. ©2007 Pearson Education Asia
Chapter 11: Differentiation
11.1 The Derivative11.1 The Derivative
• Tangent line at a point:
• The slope of a curve at P is the slope of the
tangent line at P.
• The slope of the tangent line at (a, f(a)) is
( ) ( ) ( ) ( )
h
afhaf
az
afzf
m
haz
−+
=
−
−
=
→→ 0
tan limlim

7. ©2007 Pearson Education Asia
Chapter 11: Differentiation
11.1 The Derivative
Example 1 – Finding the Slope of a Tangent Line
Find the slope of the tangent line to the curve y = f(x)
= x2
at the point (1, 1).
Solution: Slope = ( ) ( ) ( ) ( ) 2
11
lim
11
lim
22
00
=
−+
=
−+
→→ h
h
h
fhf
hh
• The derivative of a function f is the function
denoted f’ and defined by
( ) ( ) ( ) ( ) ( )
h
xfhxf
xz
xfzf
xf
hxz
−+
=
−
−
=
→→ 0
limlim'

8. ©2007 Pearson Education Asia
Chapter 11: Differentiation
11.1 The Derivative
Example 3 – Finding an Equation of a Tangent Line
If f (x) = 2x2
+ 2x + 3, find an equation of the tangent
line to the graph of f at (1, 7).
Solution:
Slope 
Equation 
( ) ( ) ( ) ( ) ( )( ) ( ) 24
322322
limlim'
22
00
+=
++−++++
=
−+
=
→→
x
h
xxhxhx
h
xfhxf
xf
hh
( )
16
167
+=
−=−
xy
xy
( ) ( ) 62141' =+=f

9. ©2007 Pearson Education Asia
Chapter 11: Differentiation
11.1 The Derivative
Example 5 – A Function with a Vertical Tangent Line
Example 7 – Continuity and Differentiability
Find .
Solution:
( )x
dx
d
( ) xh
xhx
x
dx
d
h 2
1
lim
0
=
−+
=
→
a. For f(x) = x2
, it must be continuous for all x.
b. For f(p) =(1/2)p, it is not continuous at p = 0, thus
the derivative does not exist at p = 0.

10. ©2007 Pearson Education Asia
Chapter 11: Differentiation
11.2 Rules for Differentiation11.2 Rules for Differentiation
• Rules for Differentiation:
RULE 1 Derivative of a Constant:
RULE 2 Derivative of xn
:
RULE 3 Constant Factor Rule:
RULE 4 Sum or Difference Rule
( ) 0=c
dx
d
( ) 1−
= nn
nxx
dx
d
( )( ) ( )xcfxcf
dx
d
'=
( ) ( )( ) ( ) ( )xgxfxgxf
dx
d
'' ±=±=

11. ©2007 Pearson Education Asia
Chapter 11: Differentiation
11.2 Rules for Differentiation
Example 1 – Derivatives of Constant Functions
a.
b. If , then .
c. If , then .
( ) 03 =
dx
d
( ) 5=xg
( ) ( ) 4.807
623,938,1=ts
( ) 0' =xg
0=dt
ds

12. ©2007 Pearson Education Asia
Chapter 11: Differentiation
11.2 Rules for Differentiation
Example 3 – Rewriting Functions in the Form xn
Differentiate the following functions:
Solution:
a.
b.
xy =
( )
x
x
dx
dy
2
1
2
1 12/1
== −
( )
xx
xh
1
=
( ) ( ) ( ) 2/512/32/3
2
3
2
3
' −−−−
−=−== xxx
dx
d
xh

13. ©2007 Pearson Education Asia
Chapter 11: Differentiation
11.2 Rules for Differentiation
Example 5 – Differentiating Sums and Differences of Functions
Differentiate the following functions:
( ) xxxF += 5
3a.
( ) ( ) ( )
( ) ( )
x
xxx
x
dx
d
x
dx
d
xF
2
1
15
2
1
53
3'
42/14
2/15
+=+=
+=
−
( ) 3/1
4
5
4
b.
z
z
zf −=
( )
( ) 3/433/43
3/1
4
3
5
3
1
54
4
1
5
4
'
−−
+=





−−=






−





=
zzzz
zdz
dz
dz
d
zf

14. ©2007 Pearson Education Asia
Chapter 11: Differentiation
11.2 Rules for Differentiation
Example 5 – Differentiating Sums and Differences of Functions
8726c. 23
−+−= xxxy
7418
)8()(7)(2)(6
2
23
+−=
−+−=
xx
dx
d
x
dx
d
x
dx
d
x
dx
d
dx
dy

15. ©2007 Pearson Education Asia
Chapter 11: Differentiation
11.2 Rules for Differentiation
Example 7 – Finding an Equation of a Tangent Line
Find an equation of the tangent line to the curve
when x = 1.
Solution: The slope equation is
When x = 1,
The equation is
x
x
y
23 2
−
=
2
1
2
23
23
23
−
−
+=
−=−=
x
dx
dy
xx
xx
x
y
( ) 5123
2
1
=+=
−
=xdx
dy
( )
45
151
−=
−=−
xy
xy

16. ©2007 Pearson Education Asia
Chapter 11: Differentiation
11.3 The Derivative as a Rate of Change11.3 The Derivative as a Rate of Change
Example 1 – Finding Average Velocity and Velocity
• Average velocity is given by
• Velocity at time t is given by
( ) ( )
t
tfttf
t
s
vave
∆
−∆+
=
∆
∆
=
( ) ( )
t
tfttf
v
t ∆
−∆+
=
→∆ 0
lim
Suppose the position function of an object moving
along a number line is given by s = f(t) = 3t2
+ 5,
where t is in seconds and s is in meters.
a.Find the average velocity over the interval [10,
10.1].
b. Find the velocity when t = 10.

17. ©2007 Pearson Education Asia
Chapter 11: Differentiation
11.3 The Derivative as a Rate of Change
Example 1 – Finding Average Velocity and Velocity
Solution:
a. When t = 10,
b. Velocity at time t is given by
When t = 10, the velocity is
( ) ( )
( ) ( ) ( ) ( ) m/s3.60
1.0
30503.311
1.0
101.10
1.0
101.010
=
=
=
−
=
−+
=
∆
−∆+
=
∆
∆
=
ffff
t
tfttf
t
s
vave
t
dt
ds
v 6==
( ) m/s60106
10
==
=tdt
ds

18. ©2007 Pearson Education Asia
Chapter 11: Differentiation
11.3 The Derivative as a Rate of Change
Example 3 – Finding a Rate of Change
• If y = f(x),
then
( ) ( )





∆+
=
∆
−∆+
=
∆
∆
xxx
x
xfxxf
x
y
tofrominterval
theoverxtorespectwith
yofchangeofrateaverage



=
∆
∆
=
→∆ xrespect toy with
ofchangeofrateousinstantane
lim
0 x
y
dx
dy
x
Find the rate of change of y = x4
with respect to x,
and evaluate it when x = 2 and when x = −1.
Solution:
The rate of change is .
3
4x
dx
dy
=

19. ©2007 Pearson Education Asia
Chapter 11: Differentiation
11.3 The Derivative as a Rate of Change
Example 5 – Rate of Change of Volume
A spherical balloon is being filled with air. Find the
rate of change of the volume of air in the balloon with
respect to its radius. Evaluate this rate of change
when the radius is 2 ft.
Solution: Rate of change of V with respect to r is
When r = 2 ft,
( ) 22
43
3
4
rr
dr
dV
ππ ==
( )
ft
ft
1624
3
2
2
ππ ==
=rdr
dV

20. ©2007 Pearson Education Asia
Chapter 11: Differentiation
11.3 The Derivative as a Rate of Change
Applications of Rate of Change to Economics
• Total-cost function is c = f(q).
• Marginal cost is defined as .
• Total-revenue function is r = f(q).
• Marginal revenue is defined as .
dq
dc
dq
dr
Relative and Percentage Rates of Change
• The relative rate of change of f(x) is .
• The percentage rate of change of f (x) is
( )
( )xf
xf '
( )
( )
( )%100
'
xf
xf

21. ©2007 Pearson Education Asia
Chapter 11: Differentiation
11.3 The Derivative as a Rate of Change
Example 7 – Marginal Cost
If a manufacturer’s average-cost equation is
find the marginal-cost function. What is the marginal
cost when 50 units are produced?
Solution: The cost is
Marginal cost when q = 50,
q
qqc
5000
502.00001.0 2
++−=
5000502.00001.0
5000
502.00001.0
23
2
++−=






++−==
qqq
q
qqqcqc
504.00003.0 2
+−= qq
dq
dc
( ) ( ) 75.355004.0500003.0
2
50
=+−=
=q
dq
dc

22. ©2007 Pearson Education Asia
Chapter 11: Differentiation
11.3 The Derivative as a Rate of Change
Example 9 – Relative and Percentage Rates of Change
11.4 The Product Rule and the Quotient Rule11.4 The Product Rule and the Quotient Rule
Determine the relative and percentage rates of
change of
when x = 5.
Solution:
( ) 2553 2
+−== xxxfy
( ) 56' −= xxf
( ) ( )
( )
( )
%3.33333.0
75
25
5
5'
change%
255565'
=≈==
=−=
f
f
f
The Product Rule
( ) ( )( ) ( ) ( ) ( ) ( )xgxfxgxfxgxf
dx
d
'' +=

23. ©2007 Pearson Education Asia
Chapter 11: Differentiation
11.4 The Product and Quotient Rule
Example 1 – Applying the Product Rule
Example 3 – Differentiating a Product of Three Factors
Find F’(x).
( ) ( )( )
( ) ( ) ( ) ( ) ( )
( )( ) ( )( ) 153412435432
543543'
543
22
22
2
++=++++=






++++





+=
++=
xxxxxx
x
dx
d
xxxxx
dx
d
xF
xxxxF
Find y’.
( )( ) ( ) ( )( )( ) ( )
26183
432432'
)4)(3)(2(
2
++=






+++++





++=
+++=
xx
x
dx
d
xxxxx
dx
d
y
xxxy

24. ©2007 Pearson Education Asia
Chapter 11: Differentiation
11.4 The Product and Quotient Rule
Example 5 – Applying the Quotient Rule
If , find F’(x).
Solution:
The Quotient Rule
( )
( )
( ) ( ) ( ) ( )
( )( )2
''
xg
xgxfxfxg
xg
xf
dx
d −
=





( )
12
34 2
−
+
=
x
x
xF
( )
( ) ( ) ( ) ( )
( )
( )( ) ( )( )
( )
( )( )
( )22
2
2
22
12
32122
12
234812
12
12343412
'
−
−+
=
−
+−−
=
−
−+−+−
=
x
xx
x
xxx
x
x
dx
d
xx
dx
d
x
xF

25. ©2007 Pearson Education Asia
Chapter 11: Differentiation
11.4 The Product and Quotient Rule
Example 7 – Differentiating Quotients without Using the Quotient Rule
Differentiate the following functions.
( )
( ) ( ) 5
6
3
5
2
'
5
2
a.
2
2
3
x
xxf
x
xf
==
=
( ) ( )
( ) ( ) 4
4
3
3
7
12
3
7
4
'
7
4
7
4
b.
x
xxf
x
x
xf
−=−=
==
−
−
( ) ( )
( ) ( )
4
5
5
4
1
'
35
4
1
4
35
c.
2
==
−=
−
=
xf
x
x
xx
xf

26. ©2007 Pearson Education Asia
Chapter 11: Differentiation
11.4 The Product and Quotient Rule
Example 9 – Finding Marginal Propensities to Consume and to Save
If the consumption function is given by
determine the marginal propensity to consume and
the marginal propensity to save when I = 100.
Solution:
Consumption Function
dI
dC
consumetopropensityMarginal =
consumetopropensityMarginal-1savetopropensityMarginal =
( )
10
325 3
+
+
=
I
I
C
( ) ( ) ( ) ( )
( )
( )( ) ( )( )
( ) 







+
+−++
=












+
++−++
= 2
32/1
2
32/3
10
1323310
5
10
10323210
5
I
III
I
I
dI
d
II
dI
d
I
dI
dC
536.0
12100
1297
5
100
≈





=
=IdI
dC

27. ©2007 Pearson Education Asia
Chapter 11: Differentiation
11.5 The Chain Rule and the Power Rule11.5 The Chain Rule and the Power Rule
Example 1 – Using the Chain Rule
a. If y = 2u2
− 3u − 2 and u = x2
+ 4, find dy/dx.
Solution:
Chain Rule:
Power Rule:
dx
du
du
dy
dx
dy
⋅=
( ) dx
du
nuu
dx
d nn 1−
=
( ) ( )
( )( )xu
x
dx
d
uu
du
d
dx
du
du
dy
dx
dy
234
4232 22
−=
+⋅−−=⋅=
( )[ ]( ) ( )[ ]( ) xxxxxx
dx
dy
26821342344 322
+=+=−+=

28. ©2007 Pearson Education Asia
Chapter 11: Differentiation
11.5 The Chain Rule and the Power Rule
Example 1 – Using the Chain Rule
Example 3 – Using the Power Rule
b. If y = √w and w = 7 − t3
, find dy/dt.
Solution: ( ) ( )
3
22
32/1
72
3
2
3
7
t
t
w
t
t
dt
d
w
dw
d
dt
dy
−
−=−=
−⋅=
If y = (x3
− 1)7
, find y’.
Solution: ( ) ( )
( ) ( ) ( )622263
3173
121317
117'
−=−=
−−=
−
xxxx
x
dx
d
xy

29. ©2007 Pearson Education Asia
Chapter 11: Differentiation
11.5 The Chain Rule and the Power Rule
Example 5 – Using the Power Rule
Example 7 – Differentiating a Product of Powers
If , find dy/dx.
Solution:
2
1
2
−
=
x
y
( )( ) ( )
( )22
2112
2
2
221
−
−=−−−=
−
x
x
x
dx
d
x
dx
dy
If , find y’.
Solution:
( ) ( )452
534 +−= xxy
( ) ( )( ) ( ) ( )( )
( ) ( ) ( ) ( )
( ) ( ) ( )2425215342
4531053412
453534'
2342
424352
524452
−++−=
−+++−=
−+++−=
xxxx
xxxxx
x
dx
d
xx
dx
d
xy