 So, how is this fact useful to us? Well recall that the natural exponential function and the natural logarithm function are inverses of each other and we know what the derivative of the natural exponential function is!

So, if we have
f
(
x
)
=
e
x
and
g
(
x
)
=
ln
x
then,

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g
?
(
x
)
=
1
f
?
(
g
(
x
)
)
=
1
e
g
(
x
)
=
1
e
ln
x
=
1
x
The last step just uses the fact that the two functions are inverses of each other.

Putting this all together gives,

d
d
x
(
ln
x
)
=
1
x
x
>
0
Note that we need to require that
x
>
0
since this is required for the logarithm and so must also be required for its derivative. It can also be shown that,

d
d
x
(
ln
|
x
|
)
=
1
x
x
?
0
Using this all we need to avoid is
x
=
0
.

In this case, unlike the exponential function case, we can actually find the derivative of the general logarithm function. All that we need is the derivative of the natural logarithm, which we just found, and the change of base formula. Using the change of base formula we can write a general logarithm as,

log
a
x
=
ln
x
ln
a
Differentiation is then fairly simple.

d
d
x
(
log
a
x
)
=
d
d
x
(
ln
x
ln
a
)
=
1
ln
a
d
d
x
(
ln
x
)
=
1
x
ln
a
We took advantage of the fact that
a
was a constant and so
ln
a
is also a constant and can be factored out of the derivative. Putting all this together gives,

d
d
x
(
log
a
x
)
=
1
x
ln
a
Here is a summary of the derivatives in this section.

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