DERIVATIVES OF EXPONENTIAL
AND LOGARITHMIC FUNCTIONS
Section 3.9
If you recall, the number e is important in many
instances of exponential growth:
 1
e  lim 1  
x 
 x
x
Find the following important limit using graphs
and/or tables:
h
e 1
lim
h 0
h
1
Derivative of e
xh
d x
e e
e   lim

h 0
dx
h
x
h
x
e e e
 lim
h 0
h
 x eh  1 
 lim  e

h 0
h 

 e 1 
 e lim 

h 0
 h 
h
x
x
x
Definition of the
derivative!!!
The limit we just figured!
x
e 1
d x
x
e e

dx
The derivative of this
function is itself!!!
Derivative of a
x
Given a positive base that is not one, we can use a property
x
x
of logarithms to write a in terms of e :
a e
x
 
ln a
x
e
x ln a
d x
d x ln a
x ln a d
a  e
e
 x ln a 

dx
dx
dx
e
x ln a
ln a  a ln a
x
Derivative of ln x
y  ln x
e x
d y
d
e    x

dx
dx
Imp. Diff.
dy
e
1
dx
y
dy 1
 y
dx e
dy 1

dx x
Substitution!
y
Derivative of log a x
First off, how am I able to express
following way???
log a x
in the
ln x
log a x 
COB Formula!
ln a
d
d  ln x 
1 d
log a x  
ln x

dx
dx  ln a  ln a dx
1
1 1


ln a x x ln a
Summary of the New Rules
(keeping in mind the Chain Rule and any variable restrictions)
d u
u du
e e

dx
dx
d u
du
u
a   a ln a

dx
dx
d
1 du
ln u 
dx
u dx
d
1 du
log a u 
dx
u ln a dx
u  0
 a  0, a  1
 a  0, a  1
Now we can realize the FULL POWER
of the Power Rule……………observe:
Start by writing x with any real power as a power of e…
x e
n
n ln x
d n d n ln x
n ln x d
x  e   e
 n ln x 
dx
dx
dx
e
n ln x
n
n 1
n
1
 x nx  nx
x
Power Rule for Arbitrary Real Powers
If u is a positive differentiable function of x and n is
n
any real number, then u is a differentiable function
of x, and
d n
n 1 du
u  nu
dx
dx
The power rule works for not only
integers, not only rational numbers, but
any real numbers!!!
Quality Practice Problems
dy
Find
:
dx
yx
3 3
dy
Find
:
dx
y  4e
3x
dy
Find
:
dx
4 x 1
y 5
dy

dx


3 3 x
dy
3x
 12e
dx
dy
4 x 1
 4 5 ln 5
dx
3 4
Quality Practice Problems
 
dy
3
Find
: y  ln x
dx
dy
Find
: y  log5
dx
x
3
dy 1
2
 3  3x   , x  0
x
dx x
dy

dx
1
1
x ln 5 2 x
1

,x 0
2 x ln 5
Quality Practice Problems
How do we differentiate a function
dy
x
Find
: y  x when both the base and exponent
dx
contain the variable???
Use Logarithmic Differentiation:
1. Take the natural logarithm of both sides of the
equation
2. Use the properties of logarithms to simplify the
equation
3. Differentiate (sometimes implicitly!) the
simplified equation
Quality Practice Problems
dy
x
Find
: yx
dx
ln y  ln x
x
ln y  x ln x
d
d
 ln y    x ln x 
dx
dx
1 dy
1
 x  1 ln x
y dx
x
dy
 y  ln x  1
dx
dy
x
 x  ln x  1
dx
Quality Practice Problems
dy
Find
using logarithmic differentiation:
dx
2x 2

y
x
x 1
2
2x 2

 ln y  ln
x
x 1
2
1
2
 ln y  ln  2 x   x ln 2  ln  x  1
2
1 dy 1
1 1
Differentiate:

 2   ln 2  2  2 x 
y dx 2 x
2 x 1
Quality Practice Problems
dy
Find
using logarithmic differentiation:
dx
1 dy 1
1 1

 2   ln 2  2  2 x 
y dx 2 x
2 x 1
dy
x 
1

 y   ln 2  2 
dx
x 1
x
Substitute:
2x 2


x
x 
1
 ln 2  2


2
x 1
x 1  x
Quality Practice Problems
A line with slope m passes through the origin and is tangent
to the graph of y  ln x . What is the value of m?
 What does the graph look like?
The slope of the curve:
 a,ln a 
 0, 0 
y  ln x
1
m
a
1
m
a
The slope of the line:
ln a  0 ln a
m

a0
a
Now, let’s set them equal…
Quality Practice Problems
A line with slope m passes through the origin and is tangent
to the graph of y  ln x . What is the value of m?
 What does the graph look like?
ln a 1

a
a
 a,ln a 
 0, 0 
y  ln x
ln a  1
1
m
a
ae
1
So, our slope:
1
m   0.368
e