Mathematics Learning Centre
Derivatives of exponential and
logarithmic functions
Christopher Thomas
c
1997 University of Sydney
Mathematics Learning Centre, University of Sydney
1
1 Derivatives of exponential and logarithmic func-
tions
If you are not familiar with exponential and logarithmic functions you may wish to consult
the booklet Exponents and Logarithms which is available from the Mathematics Learning
Centre.
Youmay have seen that there are two notations popularly used for natural logarithms,
log
e
and ln. These are just two different ways of writing exactly the same thing, so that
log
e
x ≡ ln x.Inthis booklet we will use both these notations.
The basic results are:
d
dx
e
x
= e
x
d
dx
(log
e
x)=
1
x
.
We can use these results and the rules that we have learnt already to differentiate functions
which involve exponentials or logarithms.
Example
Differentiate log
e
(x
2
+3x +1).
Solution
We solve this by using the chain rule and our knowledge of the derivative of log
e
x.
d
dx
log
e
(x
2
+3x +1) =
d
dx
(log
e
u) (where u = x
2
+3x +1)
=
d
du
(log
e
u) ×
du
dx
(by the chain rule)
=
1
u
×
du
dx
=
1
x
2
+3x +1
×
d
dx
(x
2
+3x +1)
=
1
x
2
+3x +1
× (2x +3)
=
2x +3
x
2
+3x +1
.
Example
Find
d
dx
(e
3x
2
).
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2
Solution
This is an application of the chain rule together with our knowledge of the derivative of
e
x
.
d
dx
(e
3x
2
)=
de
u
dx
where u =3x
2
=
de
u
du
×
du
dx
by the chain rule
= e
u
×
du
dx
= e
3x
2
×
d
dx
(3x
2
)
=6xe
3x
2
.
Example
Find
d
dx
(e
x
3
+2x
).
Solution
Again, we use our knowledge of the derivative of e
x
together with the chain rule.
d
dx
(e
x
3
+2x
)=
de
u
dx
(where u = x
3
+2x)
= e
u
×
du
dx
(by the chain rule)
= e
x
3
+2x
×
d
dx
(x
3
+2x)
=(3x
2
+2)× e
x
3
+2x
.
Example
Differentiate ln (2x
3
+5x
2
− 3).
Solution
We solve this by using the chain rule and our knowledge of the derivative of ln x.
d
dx
ln (2x
3
+5x
2
− 3) =
d ln u
dx
(where u =(2x
3
+5x
2
− 3)
=
d ln u
du
×
du
dx
(by the chain rule)
=
1
u
×
du
dx
=
1
2x
3
+5x
2
− 3
×
d
dx
(2x
3
+5x
2
− 3)
=
1
2x
3
+5x
2
− 3
× (6x
2
+10x)
=
6x
2
+10x
2x
3
+5x
2
− 3
.
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There are two shortcuts to differentiating functions involving exponents and logarithms.
The four examples above gave
d
dx
(log
e
(x
2
+3x + 1)) =
2x +3
x
2
+3x +1
d
dx
(e
3x
2
)=6xe
3x
2
d
dx
(e
x
3
+2x
)=(3x
2
+2)e
3x
2
d
dx
(log
e
(2x
3
+5x
2
− 3)) =
6x
2
+10x
2x
3
+5x
2
− 3
.
These examples suggest the general rules
d
dx
(e
f(x)
)=f
(x)e
f(x)
d
dx
(ln f(x)) =
f
(x)
f(x)
.
These rules arise from the chain rule and the fact that
de
x
dx
= e
x
and
d ln x
dx
=
1
x
. They can
speed up the process of differentiation but it is not necessary that you remember them.
If you forget, just use the chain rule as in the examples above.
Exercise 1
Differentiate the following functions.
a. f(x)=ln(2x
3
) b. f(x)=e
x
7
c. f(x)=ln(11x
7
)
d. f(x)=e
x
2
+x
3
e. f(x)=log
e
(7x
−2
) f. f(x)=e
−x
g. f(x)=ln(e
x
+ x
3
) h. f(x)=ln(e
x
x
3
) i. f(x)=ln
x
2
+1
x
3
− x
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Solutions to Exercise 1
a. f
(x)=
6x
2
2x
3
=
3
x
Alternatively write f (x)=ln2+3lnx so that f
(x)=3
1
x
.
b. f
(x)=7x
6
e
x
7
c. f
(x)=
7
x
d. f
(x)=(2x +3x
2
)e
x
2
+x
3
e. Write f(x)=log
e
7 − 2 log
e
x so that f
(x)=−
2
x
.
f. f
(x)=−e
−x
g. f
(x)=
e
x
+3x
2
e
x
+ x
3
h. Write f(x)=lne
x
+
3
ln x
so that f
(x)=1+
3
x
.
i. Write f(x)=ln(x
2
+1)− ln(x
3
− x)sothat f
(x)=
2x
x
2
+1
−
3x
2
− 1
x
3
− x
.