3.9: Derivatives of Exponential and Logarithmic Functions Mt. Rushmore, South Dakota

-1

0

1

2

3

-3 -2 -1 1 2 3x

Look at the graph of xy e=

The slope at x=0 appears to be 1.

If we assume this to be true, then:

0 0

0lim 1

h

h

e e

h

+

→

−=

definition of derivative

→

Now we attempt to find a general formula for the derivative of using the definition.

xy e=

( )0

limx h x

x

h

d e ee

dx h

+

→

−=

0lim

x h x

h

e e e

h→

⋅ −=

0

1lim

hx

h

ee

h→

⎛ ⎞−= ⋅⎜ ⎟

⎝ ⎠

0

1lim

hx

h

ee

h→

⎛ ⎞−= ⋅ ⎜ ⎟

⎝ ⎠

1xe= ⋅xe=

This is the slope at x=0, which we have assumed to be 1.

→

( )x xde e

dx=

→

xe is its own derivative!

If we incorporate the chain rule:

u ud due e

dx dx=

We can now use this formula to find the derivative ofxa

→

( )xda

dx

( )ln xade

dx( and are inverse functions.)

xe ln x

( )lnx ade

dx

( )ln lnx a de x a

dx⋅ (chain rule)

→

( is a constant.)ln a

( )xda

dx

( )ln xade

dx

( )lnx ade

dx

( )ln lnx a de x a

dx⋅

ln lnx ae a⋅

lnxa a⋅

Incorporating the chain rule:

( ) lnu ud dua a a

dx dx=

→

So far today we have:

u ud due e

dx dx= ( ) lnu ud du

a a adx dx

=

Now it is relatively easy to find the derivative of .ln x

→

lny x=

ye x=

( ) ( )yd de x

dx dx=

1y dyedx

=

1y

dy

dx e=

1ln

dx

dx x=

1ln

d duu

dx u dx=

→

To find the derivative of a common log function, you could just use the change of base rule for logs:

logd

xdx

ln

ln10

d x

dx=

1ln

ln10

dx

dx=

1 1

ln10 x= ⋅

The formula for the derivative of a log of any base other than e is:

1log

lna

d duu

dx u a dx=

→

u ud due e

dx dx= ( ) lnu ud du

a a adx dx

=

1log

lna

d duu

dx u a dx=

1ln

d duu

dx u dx=