Introductory Mathematical Analysis - Chapter 12 : Additional Differentiation Topics

1. INTRODUCTORY MATHEMATICALINTRODUCTORY MATHEMATICAL
ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences
©2007 Pearson Education Asia
Chapter 12Chapter 12
Additional Differentiation TopicsAdditional Differentiation Topics

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

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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 develop a differentiation formula for y = ln u.
• To develop a differentiation formula for y = eu
.
• To give a mathematical analysis of the economic
concept of elasticity.
• To discuss the notion of a function defined implicitly.
• To show how to differentiate a function of the form uv
.
• To approximate real roots of an equation by using
calculus.
• To find higher-order derivatives both directly and
implicitly.
Chapter 12: Additional Differentiation Topics
Chapter ObjectivesChapter Objectives

5. ©2007 Pearson Education Asia
Derivatives of Logarithmic Functions
Derivatives of Exponential Functions
Elasticity of Demand
Implicit Differentiation
Logarithmic Differentiation
Newton’s Method
Higher-Order Derivatives
12.1)
12.2)
12.3)
Chapter 12: Additional Differentiation Topics
Chapter OutlineChapter Outline
12.4)
12.5)
12.6)
12.7)

6. ©2007 Pearson Education Asia
Chapter 12: Additional Differentiation Topics
12.1 Derivatives of Logarithmic Functions12.1 Derivatives of Logarithmic Functions
• The derivatives of log functions are:
( ) 













+=
→
hx
h x
h
x
x
dx
d
/
0
1limln
1
lna.
( ) 0where
1
lnb. ≠= x
x
x
dx
d
( ) 0for
1
lnc. ≠⋅= u
dx
du
u
u
dx
d

7. ©2007 Pearson Education Asia
Chapter 12: Additional Differentiation Topics
12.1 Derivatives of Logarithmic Functions
Example 1 – Differentiating Functions Involving ln x
b. Differentiate .
Solution:
2
ln
x
x
y =
( ) ( ) ( )
( )
( )
0for
ln21
2)(ln
1
lnln
'
3
4
2
22
22
>
−
=
−





=
−
=
x
x
x
x
xx
x
x
x
x
dx
d
xx
dx
d
x
y
a. Differentiate f(x) = 5 ln x.
Solution: ( ) ( ) 0for
5
ln5' >== x
x
x
dx
d
xf

8. ©2007 Pearson Education Asia
Chapter 12: Additional Differentiation Topics
12.1 Derivatives of Logarithmic Functions
Example 3 – Rewriting Logarithmic Functions before Differentiating
a. Find dy/dx if .
Solution:
b. Find f’(p) if .
Solution:
( )3
52ln += xy
( ) 2/5for
52
6
2
52
1
3 −>
+
=





+
= x
xxdx
dy
( ) ( ) ( ) ( )
3
4
2
3
1
2
1
3
1
41
2
1
31
1
1
2'
+
+
+
+
+
=






+
+





+
+





+
=
ppp
ppp
pf
( ) ( ) ( ) ( )( )432
321ln +++= ppppf

9. ©2007 Pearson Education Asia
Chapter 12: Additional Differentiation Topics
12.1 Derivatives of Logarithmic Functions
Example 5 – Differentiating a Logarithmic Function to the Base 2
Differentiate y = log2x.
Solution:
Procedure to Differentiate logbu
• Convert logbu to and then differentiate.
b
u
ln
ln
( )
( )x
x
dx
d
x
dx
dy
2ln
1
2ln
ln
log2 =





=

10. ©2007 Pearson Education Asia
Chapter 12: Additional Differentiation Topics
12.2 Derivatives of Exponential Functions12.2 Derivatives of Exponential Functions
• The derivatives of exponential functions are:
( ) dx
du
ee
dx
d uu
=a.
( ) xx
ee
dx
d
=b.
( ) ( )
dx
du
bbb
dx
d uu
lnc. =
( )( ) ( )( ) ( )( ) 0'for
'
1
d. 1
1
1
≠= −
−
−
xff
xff
xf
dx
d
dy
dx
dx
dy 1
e. =

11. ©2007 Pearson Education Asia
Chapter 12: Additional Differentiation Topics
12.2 Derivatives of Exponential Functions
Example 1 – Differentiating Functions Involving ex
a.Find .
Solution:
b. If y = , find .
Solution:
c. Find y’ when .
Solution:
x
e
x
( ) x
xx
e
x
e
dx
d
xx
dx
d
e
dx
dy −
=+= −− 1
3ln2
++= x
eey
xx
eey =++= 00'
( )x
e
dx
d
3
( ) ( ) xxx
ee
dx
d
e
dx
d
333 ==
dx
dy

12. ©2007 Pearson Education Asia
Chapter 12: Additional Differentiation Topics
12.2 Derivatives of Exponential Functions
Example 3 – The Normal-Distribution Density Function
Determine the rate of change of y with respect to x
when x = μ + σ.
( ) ( ) ( )( )2
2
1 /
2
1 σµ
σ
−−
==
x
e
x
xfy
Solution: The rate of change is
( ) ( )( )
( )
e
e
dx
dy x
x
πσ
σσ
µσµ
πσ
σµ
σµ
2
1
1
2
2
1
2
1
2
/
2
2
1
−
=

















 −+
−=
−−
+=

13. ©2007 Pearson Education Asia
Chapter 12: Additional Differentiation Topics
12.2 Derivatives of Exponential Functions
Example 5 – Differentiating Different Forms
Example 7 – Differentiating Power Functions Again
Find .
Solution:
( )xe
xe
dx
d
22
++
( ) ( )
[ ]( )
x
ex
x
eexxe
dx
d
x
e
xexe
2
2ln2
2
1
2ln2
1
2ln12
+=






+=++
−
−
Prove d/dx(xa
) = axa−1
.
Solution: ( ) ( ) 11ln −−
=== aaxaa
axaxxe
dx
d
x
dx
d

14. ©2007 Pearson Education Asia
Chapter 12: Additional Differentiation Topics
12.3 Elasticity of Demand12.3 Elasticity of Demand
Example 1 – Finding Point Elasticity of Demand
• Point elasticity of demand η is
where p is price and q is quantity.
( )
dq
dp
q
p
q ==ηη
Determine the point elasticity of the demand equation
Solution: We have
0and0where >>= qk
q
k
p
1
2
2
−=== −
q
k
q
k
dq
dp
q
p
η

15. ©2007 Pearson Education Asia
Chapter 12: Additional Differentiation Topics
12.4 Implicit Differentiation12.4 Implicit Differentiation
Implicit Differentiation Procedure
1. Differentiate both sides.
2. Collect all dy/dx terms on one side and other
terms on the other side.
3. Factor dy/dx terms.
4. Solve for dy/dx.

16. ©2007 Pearson Education Asia
Chapter 12: Additional Differentiation Topics
12.4 Implicit Differentiation
Example 1 – Implicit Differentiation
Find dy/dx by implicit differentiation if .
Solution:
73
=−+ xyy
( ) ( )
2
2
3
31
1
013
7
ydx
dy
dx
dy
y
dx
dy
dx
d
xyy
dx
d
+
=
=−+
=−+

17. ©2007 Pearson Education Asia
Chapter 12: Additional Differentiation Topics
12.4 Implicit Differentiation
Example 3 – Implicit Differentiation
Find the slope of the curve at (1,2).
Solution:
( )223
xyx −=
( ) ( )[ ]
( )
( )
( ) 2
7
2
443
223
2,1
2
32
22
223
=
−
−+
=






−−=
−=
dx
dy
xy
xxyx
dx
dy
x
dx
dy
xy
dx
dy
x
xy
dx
d
x
dx
d

18. ©2007 Pearson Education Asia
Chapter 12: Additional Differentiation Topics
12.5 Logarithmic Differentiation12.5 Logarithmic Differentiation
Logarithmic Differentiation Procedure
1. Take the natural logarithm of both sides which
gives .
2. Simplify In (f(x))by using properties of logarithms.
3. Differentiate both sides with respect to x.
4. Solve for dy/dx.
5. Express the answer in terms of x only.
( )( )xfy lnln =

19. ©2007 Pearson Education Asia
Chapter 12: Additional Differentiation Topics
12.5 Logarithmic Differentiation
Example 1 – Logarithmic Differentiation
Find y’ if .
Solution:
( )
4 22
3
1
52
+
−
=
xx
x
y
( )
( )
( ) ( )x
x
xx
xxxy
xx
x
y
2
1
1
4
1
ln252ln3
1ln52lnln
1
52
lnln
2
4 223
4 22
3






+
−−−=
+−−−=
+
−
=

20. ©2007 Pearson Education Asia
Chapter 12: Additional Differentiation Topics
12.5 Logarithmic Differentiation
Example 1 – Logarithmic Differentiation






+
−−
−+
−
=
+
−−
−
=
+
−−
−
=
)1(
2
52
6
1
)52(
'
)1(2
2
52
6
)2)(
1
1
(
4
1
)
1
(2)2)(
52
1
(3
'
24 22
3
2
2
xx
x
xxxx
x
y
x
x
xx
x
xxxy
y
Solution (continued):

21. ©2007 Pearson Education Asia
Chapter 12: Additional Differentiation Topics
12.5 Logarithmic Differentiation
Example 3 – Relative Rate of Change of a Product
Show that the relative rate of change of a product is
the sum of the relative rates of change of its factors.
Use this result to express the percentage rate of
change in revenue in terms of the percentage rate of
change in price.
Solution: Rate of change of a function r is
( ) %100
'
1%100
'
%100
'
%100
'
%100
'
'''
p
p
r
r
q
q
p
p
r
r
q
q
p
p
r
r
η+=
+=
+=

22. ©2007 Pearson Education Asia
Chapter 12: Additional Differentiation Topics
12.6 Newton’s Method12.6 Newton’s Method
Example 1 – Approximating a Root by Newton’s Method
Newton’s method:
( )
( )
,...3,2,1
'
1 =−=+ n
xf
xf
xx
n
n
nn
Approximate the root of x4
− 4x + 1 = 0 that lies
between 0 and 1. Continue the approximation
procedure until two successive approximations differ
by less than 0.0001.

23. ©2007 Pearson Education Asia
Chapter 12: Additional Differentiation Topics
12.6 Newton’s Method
Example 1 – Approximating a Root by Newton’s Method
Solution: Letting , we have
Since f (0) is closer to 0, we choose 0 to be our first x1.
Thus,
( )
( ) 44
13
' 3
4
1
−
−
=−=+
n
n
n
n
nn
x
x
xf
xf
xx
25099.0,3When
25099.0,2When
25.0,1When
0,0When
4
3
2
1
≈=
≈=
==
==
xn
xn
xn
xn
( ) 144
+−= xxxf
( )
( ) 21411
11000
−=+−=
=+−=
f
f
( )
( ) 44'
14
3
4
−=
+−=
nn
nnn
xxf
xxxf

24. ©2007 Pearson Education Asia
Chapter 12: Additional Differentiation Topics
12.7 Higher-Order Derivatives12.7 Higher-Order Derivatives
For higher-order derivatives:

25. ©2007 Pearson Education Asia
Chapter 12: Additional Differentiation Topics
12.7 Higher-Order Derivatives
Example 1 – Finding Higher-Order Derivatives
a. If , find all higher-order
derivatives.
Solution:
b. If f(x) = 7, find f(x).
Solution:
( ) 26126 23
−+−= xxxxf
( )
( )
( )
( )
( ) 0
36'''
2436''
62418'
4
2
=
=
−=
+−=
xf
xf
xxf
xxxf
( )
( ) 0''
0'
=
=
xf
xf

26. ©2007 Pearson Education Asia
Chapter 12: Additional Differentiation Topics
12.7 Higher-Order Derivatives
Example 3 – Evaluating a Second-Order Derivative
Example 5 – Higher-Order Implicit Differentiation
Solution:
( ) .4whenfind,
4
16
If 2
2
=
+
= x
dx
yd
x
xf
( )
( ) 3
2
2
2
432
416
−
−
+=
+−=
x
dx
yd
x
dx
dy
16
1
4
2
2
=
=x
dx
yd
Solution:
y
x
dx
dy
dx
dy
yx
4
082
−
=
=+
.44ifFind 22
2
2
=+ yx
dx
yd

27. ©2007 Pearson Education Asia
Chapter 12: Additional Differentiation Topics
12.7 Higher-Order Derivatives
Example 5 – Higher-Order Implicit Differentiation
Solution (continued):
32
2
3
22
2
2
4
1
16
4
getto
4
ateDifferenti
ydx
yd
y
xy
dx
yd
y
x
dx
dy
−=
+
−=
−
=