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 functionis: y = loga x

An example of a natural logarithmic functionis: y = ln x

DERIVATIVES OF LOGARITHMIC FUNCTIONS

It can be proved that logarithmicfunctions are differentiable.

This is certainlyplausible from theirgraphs.

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 dy

a adx

=1 1

ln lny

dy

dx a a x a= =

If we put a = e in Formula 1, then the factoron the right side becomes ln e = 1 and we getthe formula for the derivative of the naturallogarithmic function loge x = ln x.

DERIVATIVES OF LOG FUNCTIONS

1(ln )

dx

dx x=

Formula 2

By comparing Formulas 1 and 2, we seeone of the main reasons why naturallogarithms (logarithms with base e) are usedin calculus:

The differentiation formula is simplest whena = 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

2

3 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 2with 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 )

sin

1cos cot

sin

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

dx

dx

x 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) 1

2( )(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 usingthe laws of logarithms, then the differentiationbecomes 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

xx

f x

xx x

!>""

= #" $ = <"$%

DERIVATIVES OF LOG FUNCTIONS Example 6

The result of Example 6 isworth remembering:

1ln

dx

dx x=

DERIVATIVES OF LOG FUNCTIONS Equation 4

The calculation of derivatives of complicatedfunctions involving products, quotients, orpowers can often be simplified by takinglogarithms.

The method used in the following exampleis called logarithmic differentiation.

LOGARITHMIC DIFFERENTIATION

Differentiate

We take logarithms of both sides of the equationand use the Laws of Logarithms to simplify:

3 / 4 2

5

1

(3 2)

x xy

x

+=

+

23 14 2

ln 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 explicitexpression for y, we cansubstitute 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 ofan equation y = f(x) and use the Laws ofLogarithms 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 useEquation 4.

We illustrate this procedure by proving the generalversion 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 carefullybetween:

The Power Rule [(xn)’ = nxn-1], where the baseis variable and the exponent is constant

The rule for differentiating exponential functions[(ax)’ =ax ln a], where the base is constant andthe exponent is variable

LOGARITHMIC DIFFERENTIATION

In general, there are four cases forexponents and bases:

[ ] [ ]1

( ) ( )

( )

1. ( ) 0 and areconstants

2. ( ) ( ) '( )

3. (ln ) '( )

4. To find ( / [ ( )] , logarithmic

differentiation can beused, as in thenext example.

b

b b

g x g x

g x

da a b

dx

df x b f x f x

dx

da a a g x

dx

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 1x

y x=

Another method is to write .

ln

ln

( ) ( )

( ln )

2 ln

2

x x x

x x

x

d dx e

dx dx

de x x

dx

xx

x

=

=

! "+= # $

% &

LOGARITHMIC DIFFERENTIATION E. g. 8—Solution 2ln( )x x x

x 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 eas a limit.

THE NUMBER e AS A LIMIT

From the definition of a derivative asa 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 x

xx

x x

x

! !

! !

!

+ " + "= =

+ "= = +

= +

THE NUMBER e AS A LIMIT

As f ’(1) = 1, we have

Then, by Theorem 8 in Section 2.5 andthe continuity of the exponential function,we have:

1

0lim ln(1 ) 1x

x

x!

+ =

1/1

0lim ln(1 )

1 ln(1 ) 1

0 0lim lim(1 )

xx

x

xx x

x x

e e e e x!+

+

! != = = = +

THE NUMBER e AS A LIMIT

1

0lim(1 ) x

x

e x!

= +

Formula 5

Formula 5 is illustrated by the graph ofthe function y = (1 + x)1/x here and a tableof values for small values of x.

THE NUMBER e AS A LIMIT

This illustrates the fact that, correct toseven 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

n

en!"

# $= +% &

' (

THE NUMBER e AS A LIMIT Formula 6