1.2.2 Derivatives of Exponentials and Logarithms.
In this section we establish the following two formulas and use them to find derivatives of various functions
involving exponentials and logarithms.
Proposition 1.
(1)
d
1
[ ln x ] =
dx
x
(2)
d
[ ex ] = ex
dx
Proof. Let y = ln x = f(x). Then
dy
f(x+h) - f(x)
ln(x+h) - ln x
1 x+h
1 x
h
= lim
= lim
= lim ln
= lim     ln1 + 
dx
h
h
h
x
x
h
x







h0
h0
h0
h0
=
(x/h)
1 lim ln1 + h
x
x  h0 
x
Let n = . Then
h
dy
=
dx
n
1 lim ln1 + 1 = 1 ln e = 1
n
x
x  n 
x 
This proves (1). To prove (2), let y = ex. Then x = ln y. We use implicit differentiation. Take the
1 dy
dy
derivative of both sides with respect to x. This gives 1 =   . So = y = ex which proves (2). //
y
dx
dx
 
Example 1. Find
dy
if y = esin x.
dx
Solution. This has the form y = f(u(x)) where f(u) = eu and u(x) = sin x. The chain rule says
dy
du
du
= f '(u(x)) . By Proposition 1 one has f '(u) = eu. So f '(u(x)) = esin x. Also
= cos x. Therefore,
dx
dx
dx
dy
du
= f '(u(x))
= (esin x)(cos x) = (cos x)esin x.
dx
dx
This example illustrates the following rules that are useful for calculating derivatives of functions involving
exponentials and logarithms.
Proposition 2. If u = u(x) is some formula involving x then
(3)
d
1 du
( ln[u(x)] ) =
dx
u(x) dx
(4)
d
du
[ eu(x) ] = eu(x)
dx
dx
1.2.2 - 1
Proof. Let y = ln[u(x)] = f(u(x)) where f(u) = ln u. The chain rule says
dy
du
= f '(u(x)) . By Proposition 1
dx
dx
1
dy
1 du
one has f '(u) = . So =
which proves (3). The proof of (4) is similar. //
u
dx u(x) dx
dy
if y = ln(sin x).
dx
Example 2. Find
Solution. Using (3) with u(x) = sin x, we have
Problem 1. Find
d
( ln(sin x) ) =
dx
 1   d sin x = cos x = cot x.
sin x dx
 sin x
dy
for each of the following.
dx
a.
y = e7x
c.
y = e-(x
e.
y = x + ln x
g.
y = ln
i.
y = e2x(2cos 3x + 3sin 3x)
2)


3x + 2
3x - 2 
x + x2 )
b.
y = ln(
d.
y = tan(ex)
f.
y = sec(etan(x ))
h.
y = x[sin(ln x) + cos(ln x)]
i.
y =
2
ln x
 sin(et) dt

0
_
j.
y = ex ln(x2 + 1)
l.
y =
k.
y =
tan(ln(x+1))
2
sin(ex )
ln(tan x)
ecos x
If we have to take the derivative of something like y = 2x where the base is something other than e, then it is
convenient to use the following proposition.
Proposition 3. If a > 0 is a constant and u = u(x) is some formula involving x then
(5)
d
[ ax ] = (ln a) ax
dx
(6)
d
du
[ au(x) ] = (ln a) au(x)
dx
dx
Proof. Write a = eln a so that ax = (eln a)x = e(ln a) x. Then
d
d
[ ax ] = [ e(ln a) x ] = e(ln a) x (ln a) = (ln a) ax.
dx
dx
This proves (5). (6) follows from (5) by use of the chain rule. //
1.1 - 2
dy
if y = 10x.
dx
Example 3. Find
Solution. Using (5) one has
dy
= (ln 10) 10x  2.303  10x.
dx
dy
if y = 2cos x.
dx
Example 4. Find
Solution. Using (6) one has
dy
d
= (ln 2) 2cos x  cos x = (ln 2) 2cos x (- sin x) = - (ln 2) (sin x) 2cos x 
dx
dx

- 0.693 (sin x) 2cos x.
Example 5. A population of bacteria doubles every 2 hours. At time t = 0 there are 1000 bacteria. Find
a.
A formula for the number N of bacteria at time t
b.
A formula for the rate of growth at time t.
c.
The number of bacteria at time t = 10 and the rate of
t
0
2
4
6
8
t
growth at time t = 10.
Solution. At the right is a table of value of t and N.
Problem 2. Find
a.
y = 3x
dy
for each of the following.
dx
2
b.
y =
N
1000
(1000)(2)
(1000)(2)(2) = 1000  22
(1000)(2)(2)(2) = 1000  23
(1000)(2)(2)(2)(2) = 1000  24
1000  2t/2
esec x + 21/x
x ln x
If a is a positive number the y = loga x is the inverse function of the function x = ax. In other words,
y = loga x is that number such that ay = x.
1
1
Examples 4. log10 1000 = 3 since 103 = 1000. log2  = - 5 since 2-5 = .
32
32
loga x can be converted to ln x by using the following proposition.
Proposition 4. If a > 0 and b > 0 then
(7)
loga x =
ln x
ln a
(8)
loga x =
logb x
logb a
Proof. Let y = loga x. The ay = x. Take ln of both sides and use the fact that ln ay = y ln a. This gives
y ln a = ln x. Dividing by ln a gives (7). The //
From (7) we get the following formulas for taking the derivative of loga x.
Proposition 5. If a > 0 is a constant and u = u(x) is some formula involving x then
1.1 - 3
(8)
d
1
[ loga x ] =
dx
(ln a) x
(9)
d
1
du
[ loga u(x) ] =
dx
(ln a) u(x) dx
Proof. (8) follows simply by taking the derivative of (7). (9) follows from (8) by using the chain rule. //
Example 5. Find
dy
if y = log10 x.
dx
Solution. Using (8) one has
dy
1
=
 2.303  10x.
dx (ln 10) x
1.1 - 4