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3.6Derivatives of
Logarithmic Functions
In this section, we:
use implicit differentiation to find the derivatives of
the logarithmic functions and, in particular,
the natural logarithmic function.
DIFFERENTIATION RULES
An example of a logarithmic function
is: y = loga x
An example of a natural logarithmic function
is: y = ln x
DERIVATIVES OF LOGARITHMIC FUNCTIONS
It can be proved that logarithmic
functions are differentiable.
This is certainly plausible from their graphs.
DERIVATIVES OF LOG FUNCTIONS
1(log )
lna
dx
dx x a
DERIVATIVES OF LOG FUNCTIONS Formula 1—Proof
Let y = loga x.
Then, ay = x. Differentiating this equation implicitly with respect to x,
using Formula 5 in Section 3.4, we get: So,
(ln ) 1y dya a
dx
1 1
ln lny
dy
dx a a x a
If we put a = e in Formula 1, then the factor
on the right side becomes ln e = 1 and we get
the formula for the derivative of the natural
logarithmic function loge x = ln x.
DERIVATIVES OF LOG FUNCTIONS
1(ln )
dx
dx x
Formula 2
By comparing Formulas 1 and 2, we see
one of the main reasons why natural
logarithms (logarithms with base e) are used
in calculus:
The differentiation formula is simplest when a = e because ln e = 1.
DERIVATIVES OF LOG FUNCTIONS
Differentiate y = ln(x3 + 1).
To use the Chain Rule, we let u = x3 + 1.
Then y = ln u.
So, 2
23 3
1 1 3(3 )
1 1
dy dy du du xx
dx du dx u dx x x
DERIVATIVES OF LOG FUNCTIONS Example 1
In general, if we combine Formula 2
with the Chain Rule, as in Example 1,
we get:
1 '( )(ln ) or ln ( )
( )
d du d g xu g x
dx u dx dx g x
DERIVATIVES OF LOG FUNCTIONS Formula 3
Find .
Using Formula 3, we have:
ln(sin )d
xdx
1ln(sin ) (sin )
sin1
cos cotsin
d dx x
dx x dx
x xx
DERIVATIVES OF LOG FUNCTIONS Example 2
Differentiate .
This time, the logarithm is the inner function.
So, the Chain Rule gives:
( ) lnf x x
1 212'( ) (ln ) (ln )
1 1 1
2 ln 2 ln
df x x x
dx
xx x x
DERIVATIVES OF LOG FUNCTIONS Example 3
Differentiate f(x) = log10(2 + sin x).
Using Formula 1 with a = 10, we have:
10'( ) log (2 sin )
1(2 sin )
(2 sin ) ln10
cos
(2 sin ) ln10
df x x
dxd
xx dx
x
x
DERIVATIVES OF LOG FUNCTIONS Example 4
Find .
d
dxln
x 1
x 2
1
x 1
x 2
d
dx
x 1
x 2
x 2
x 1
x 2 1 (x 1) 12 (x 2) 1 2
x 2
x 2 1
2(x 1)
(x 1)(x 2)
x 5
2(x 1)(x 2)
1ln
2
d x
dx x
DERIVATIVES OF LOG FUNCTIONS E. g. 5—Solution 1
If we first simplify the given function using
the laws of logarithms, then the differentiation
becomes easier:
This answer can be left as written. However, if we used a common denominator,
it would give the same answer as in Solution 1.
12
1ln ln( 1) ln( 2)
2
1 1 1
1 2 2
d x dx x
dx dxx
x x
DERIVATIVES OF LOG FUNCTIONS E. g. 5—Solution 2
Find f ’(x) if f(x) = ln |x|.
Since
it follows that
Thus, f ’(x) = 1/x for all x ≠ 0.
ln if 0( )
ln( ) if 0
x xf x
x x
1if 0
( )1 1
( 1) if 0
xxf x
xx x
DERIVATIVES OF LOG FUNCTIONS Example 6
The calculation of derivatives of complicated
functions involving products, quotients, or
powers can often be simplified by taking
logarithms.
The method used in the following example is called logarithmic differentiation.
LOGARITHMIC DIFFERENTIATION
Differentiate
We take logarithms of both sides of the equation and use the Laws of Logarithms to simplify:
3 / 4 2
5
1
(3 2)
x xy
x
23 14 2ln ln ln( 1) 5ln(3 2)y x x x
LOGARITHMIC DIFFERENTIATION Example 7
Differentiating implicitly with respect to x gives:
Solving for dy / dx, we get:
2
1 3 1 1 2 35
4 2 3 21
dy x
y dx x xx
2
3 15
4 1 3 2
dy xy
dx x x x
LOGARITHMIC DIFFERENTIATION Example 7
Since we have an explicit expression for y, we can substitute and write:
3 / 4 2
5 2
1 3 15
4 3 2(3 2) 1
dy x x x
dx x xx x
LOGARITHMIC DIFFERENTIATION Example 7
1. Take natural logarithms of both sides of an equation y = f(x) and use the Laws of Logarithms to simplify.
2. Differentiate implicitly with respect to x.
3. Solve the resulting equation for y’.
STEPS IN LOGARITHMIC DIFFERENTIATION
If f(x) < 0 for some values of x, then ln f(x)
is not defined.
However, we can write |y| = |f(x)| and use
Equation 4.
We illustrate this procedure by proving the general version of the Power Rule—as promised in Section 3.1.
LOGARITHMIC DIFFERENTIATION
If n is any real number and f(x) = xn,
then
Let y = xn and use logarithmic differentiation:
Thus,
Hence,
1'( ) nf x nx
THE POWER RULE
ln ln ln 0n
y x n x x 'y n
y x
1'n
ny xy n n nx
x x
PROOF
You should distinguish carefully
between:
The Power Rule [(xn)’ = nxn-1], where the base is variable and the exponent is constant
The rule for differentiating exponential functions [(ax)’ =ax ln a], where the base is constant and the exponent is variable
LOGARITHMIC DIFFERENTIATION
In general, there are four cases for
exponents and bases:
1
( ) ( )
( )
1. ( ) 0 and areconstants
2. ( ) ( ) '( )
3. (ln ) '( )
4. To find ( / [ ( )] , logarithmic
differentiation can be used, as in the next example.
b
b b
g x g x
g x
da a b
dxd
f x b f x f xdxd
a a a g xdx
d dx f x
LOGARITHMIC DIFFERENTIATION
Differentiate .
Using logarithmic differentiation, we have:
ln ln ln
' 1 1(ln )
2
1 ln 2 ln'
2 2
x
x
y x x x
yx x
y x x
x xy y x
x x x
LOGARITHMIC DIFFERENTIATION E. g. 8—Solution 1
xy x
Another method is to write .
ln
ln
( ) ( )
( ln )
2 ln
2
x x x
x x
x
d dx e
dx dxd
e x xdx
xx
x
LOGARITHMIC DIFFERENTIATION E. g. 8—Solution 2ln( )x x xx e
We have shown that, if f(x) = ln x,
then f ’(x) = 1/x.
Thus, f ’(1) = 1.
Now, we use this fact to express the number e as a limit.
THE NUMBER e AS A LIMIT
From the definition of a derivative as
a limit, we have:
0 0
0 0
1
0
(1 ) (1) (1 ) (1)'(1) lim lim
ln(1 ) ln1 1lim lim ln(1 )
lim ln(1 )
h x
x x
x
x
f h f f x ff
h xx
xx x
x
THE NUMBER e AS A LIMIT
As f ’(1) = 1, we have
Then, by Theorem 8 in Section 2.5 and the continuity of the exponential function, we have:
1
0lim ln(1 ) 1x
xx
1/1
0lim ln(1 )1 ln(1 ) 1
0 0lim lim(1 )
xx
xx x x
x xe e e e x
THE NUMBER e AS A LIMIT
1
0lim(1 ) x
xe x
Formula 5
Formula 5 is illustrated by the graph of
the function y = (1 + x)1/x here and a table
of values for small values of x.
THE NUMBER e AS A LIMIT
This illustrates the fact that, correct to
seven decimal places, e ≈ 2.7182818
THE NUMBER e AS A LIMIT