§3.5 Derivatives of Logarithmic
and Exponential Functions.
The student will learn about:
the derivative of ln x and
the ln f (x), the derivative
of e x and e f (x) and,
applications.
1
Derivative Formula for ln x.
d
1
ln x 
dx
x
The above derivative can be combined with the
power rule, product rule, quotient rule, and
chain rule to find more complicated derivatives
2
Examples.
f (x) = 5 ln x.
f (x) = x5 ln x.
f ‘ (x) = (5)(1/x) = 5/x
Note: We need the product rule.
f ‘ (x) = (x 5 )(1/x) + (ln x)(5x 4 )
= x 4 + (ln x)(5x 4)
3
Derivative Formula for ln f (x).
We just learned that
d
1
ln(x) 
dx
x
What if instead of x we had an ugly function?
d
1 d
1
OR dx ln(u)  u dx u  u u'
The above derivative can be combined with the
power rule, product rule, quotient rule, and
chain rule to find more complicated derivatives
4
Examples.
f (x) = ln (x 4 + 5)
1
d
4
(
x
 5)
f ‘ (x) = 4
x  5 dx
3
1
4
x
3

4
x
 4
f ‘ (x) = 4
x 5
x 5
f (x) = 4 ln √x = 4 ln x 1/2
1
f ' ( x)  4 1 2
x
1 1 2
x
2
 2 ln x
2

x
5
Examples.
f (x) = (5 – 3 ln x) 4 .
f ‘ (x) = 4 (5 – 3 ln x) 3
= 4 (5 – 3 ln x)
3
d
(5  3 ln x)
dx
  3
 x 


 12 (5  3 ln x) 3

x
6
Derivative Formulas for
x
e.
d x
x
e e
dx
The above derivative can be combined with the
power rule, product rule, quotient rule, and
chain rule to find more complicated derivatives
7
Examples.
Find derivatives for
f (x) = 3 e x.
f (x) = x 4 e x
f ‘ (x) = 3 e x .
Hint, use the product rule.
f ‘ (x) = x 4 e x + ex 4x 3
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Derivative Formulas for
We just learned that
f
(x)
e .
d x
e  ex
dx
What if instead of x we had an ugly function?
d u
u d
u
OR
e e
u  e u'
dx
dx
The above derivative can be combined with the
power rule, product rule, quotient rule, and
chain rule to find more complicated derivatives
9
Example.

x
f ( x)  e
3
 3x2  1


x
f ' (x)  e
3
x

f '(x)  e
 3x2  1
3
 3x2  1
 d (x 3  3x 2  1)
dx
 (3x2  6x)
10
General Derivative Rules
Power Rule
d n
x  nxn1
dx
Exponential Rule
d x
e  ex
dx
Log Rule
d
1
ln x 
dx
x
General Power Rule
d n
u  n u n  1 u'
dx
General Exponential Derivative Rule
d u
u
e  e u'
dx
General Log Derivative Rule
d
1
ln u  u'
dx
u
11
Maximizing Consumer Expenditure
The amount of a commodity that consumers will
buy depends on the price of the commodity.
For a commodity whose price is p, let the
consumer demand be given by a function D(p).
Multiplying the number of units D(p) by the
price p gives the total consumer expenditure for
the commodity.
Example
Consumer Demand and Expenditure.
The consumer expenditure, is E (p) = p · D (p), where
D is the demand function.
Let consumer demand be
D (p) = 8000 e – 0.05 p
Graph this on your calculator
and see if it makes sense.
0 ≤ x ≤ 15 and 0 ≤ y ≤ 6,000
13
Consumer Demand and Expenditure.
Continued
The consumer expenditure, is E (p) = p · D (p), where
D is the demand function.
Let consumer demand be D (p) = 8000 e – 0.05 p
Maximize the consumer expenditure.
Consumer expenditure
E (p) = p 8000 e – 0.05 p
Use your calculator
to maximize this.
E (20) = $58,860.71
0 ≤ x ≤ 30 and 0 ≤ y ≤ 65,000
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Summary.
• The derivative of f (x) = ln x is f ' (x) = 1/x.
• The derivative of f (x) = ln u is f ' (x) = (1/u) u'.
• The derivative of f (x) = ex is f ' (x) = ex.
• The derivative of f (x) = eu is f ' (x) = eu u'.
• We did an application involving consumer
expenditure.
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ASSIGNMENT
§3.5 on my website.
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