DIFFERENTIATION RULES 3. 3.6 Derivatives of Logarithmic Functions In this section, we: use implicit differentiation to find the derivatives of the logarithmic.

DIFFERENTIATION RULESDIFFERENTIATION RULES

3

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 result of Example 6 is

worth remembering:

1ln

dx

dx x

DERIVATIVES OF LOG FUNCTIONS Equation 4

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

If we put n = 1/x in Formula 5, then n → ∞

as x → 0+.

So, an alternative expression for e is:

1lim 1

n

ne

n

THE NUMBER e AS A LIMIT Formula 6