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In this section, we use implicit differentiation to find the derivatives of the logarithmic functions and, in particular, the natural logarithmic function.

1.
3

DIFFERENTIATION RULES

DIFFERENTIATION RULES

2.
DIFFERENTIATION RULES

3.6

Derivatives 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.

3.6

Derivatives 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.

3.
DERIVATIVES OF LOGARITHMIC FUNCTIONS

An example of a logarithmic function

is: y = loga x

An example of a natural logarithmic function

is: y = ln x

An example of a logarithmic function

is: y = loga x

An example of a natural logarithmic function

is: y = ln x

4.
DERIVATIVES OF LOG FUNCTIONS

It can be proved that logarithmic

functions are differentiable.

This is certainly

plausible from their

graphs.

It can be proved that logarithmic

functions are differentiable.

This is certainly

plausible from their

graphs.

5.
DERIVATIVES OF LOG FUNCTIONS Formula 1—Proof

d 1

(log a x)

dx x ln a

Let y = loga x.

Then, ay = x.

Differentiating this equation implicitly with respect to x,

using Formula 5 in Section 3.4, we get: y dy

a (ln a ) 1

So, dy 1 1 dx

y

dx a ln a x ln a

d 1

(log a x)

dx x ln a

Let y = loga x.

Then, ay = x.

Differentiating this equation implicitly with respect to x,

using Formula 5 in Section 3.4, we get: y dy

a (ln a ) 1

So, dy 1 1 dx

y

dx a ln a x ln a

6.
DERIVATIVES OF LOG FUNCTIONS Formula 2

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.

d 1

(ln x)

dx x

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.

d 1

(ln x)

dx x

7.
DERIVATIVES OF LOG FUNCTIONS

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.

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.

8.
DERIVATIVES OF LOG FUNCTIONS Example 1

Differentiate y = ln(x3 + 1).

To use the Chain Rule, we let u = x3 + 1.

Then y = ln u.

2

dy dy du 1 du 1 3 x

So, 3 (3 x 2 ) 3

dx du dx u dx x 1 x 1

Differentiate y = ln(x3 + 1).

To use the Chain Rule, we let u = x3 + 1.

Then y = ln u.

2

dy dy du 1 du 1 3 x

So, 3 (3 x 2 ) 3

dx du dx u dx x 1 x 1

9.
DERIVATIVES OF LOG FUNCTIONS Formula 3

In general, if we combine Formula 2

with the Chain Rule, as in Example 1,

we get:

d 1 du d g '( x )

(ln u ) or ln g ( x)

dx u dx dx g ( x)

In general, if we combine Formula 2

with the Chain Rule, as in Example 1,

we get:

d 1 du d g '( x )

(ln u ) or ln g ( x)

dx u dx dx g ( x)

10.
DERIVATIVES OF LOG FUNCTIONS Example 2

d

Find ln(sin x) .

dx

Using Formula 3, we have:

d 1 d

ln(sin x) (sin x)

dx sin x dx

1

cos x cot x

sin x

d

Find ln(sin x) .

dx

Using Formula 3, we have:

d 1 d

ln(sin x) (sin x)

dx sin x dx

1

cos x cot x

sin x

11.
DERIVATIVES OF LOG FUNCTIONS Example 3

Differentiate f ( x) ln x .

This time, the logarithm is the inner function.

So, the Chain Rule gives:

d

1 2

f '( x) (ln x)

1

2 (ln x)

dx

1 1 1

2 ln x x 2 x ln x

Differentiate f ( x) ln x .

This time, the logarithm is the inner function.

So, the Chain Rule gives:

d

1 2

f '( x) (ln x)

1

2 (ln x)

dx

1 1 1

2 ln x x 2 x ln x

12.
DERIVATIVES OF LOG FUNCTIONS Example 4

Differentiate f(x) = log10(2 + sin x).

Using Formula 1 with a = 10, we have:

d

f '( x) log10 (2 sin x)

dx

1 d

(2 sin x)

(2 sin x) ln10 dx

cos x

(2 sin x) ln10

Differentiate f(x) = log10(2 + sin x).

Using Formula 1 with a = 10, we have:

d

f '( x) log10 (2 sin x)

dx

1 d

(2 sin x)

(2 sin x) ln10 dx

cos x

(2 sin x) ln10

13.
DERIVATIVES OF LOG FUNCTIONS E. g. 5—Solution 1

d x 1

Find ln .

dx x 2

d x 1 1 d x 1

ln

dx x 2 x 1 dx x 2

x 2

x 2 x 2 1 (x 1) (x 2)

1

2

1 2

x 1 x 2

x 2 12 (x 1) x 5

(x 1)(x 2) 2(x 1)(x 2)

d x 1

Find ln .

dx x 2

d x 1 1 d x 1

ln

dx x 2 x 1 dx x 2

x 2

x 2 x 2 1 (x 1) (x 2)

1

2

1 2

x 1 x 2

x 2 12 (x 1) x 5

(x 1)(x 2) 2(x 1)(x 2)

14.
DERIVATIVES OF LOG FUNCTIONS E. g. 5—Solution 2

If we first simplify the given function using

the laws of logarithms, then the differentiation

becomes easier:

d x 1 d

ln ln( x 1) 12 ln( x 2)

dx x 2 dx

1 1 1

x 1 2 x 2

This answer can be left as written.

However, if we used a common denominator,

it would give the same answer as in Solution 1.

If we first simplify the given function using

the laws of logarithms, then the differentiation

becomes easier:

d x 1 d

ln ln( x 1) 12 ln( x 2)

dx x 2 dx

1 1 1

x 1 2 x 2

This answer can be left as written.

However, if we used a common denominator,

it would give the same answer as in Solution 1.

15.
DERIVATIVES OF LOG FUNCTIONS Example 6

Find f ’(x) if f(x) = ln |x|.

Since f ( x )

ln x if x 0

ln( x) if x 0

1

x if x 0

it follows that f ( x)

1 ( 1) 1 if x 0

x x

Thus, f ’(x) = 1/x for all x ≠ 0.

Find f ’(x) if f(x) = ln |x|.

Since f ( x )

ln x if x 0

ln( x) if x 0

1

x if x 0

it follows that f ( x)

1 ( 1) 1 if x 0

x x

Thus, f ’(x) = 1/x for all x ≠ 0.

16.
DERIVATIVES OF LOG FUNCTIONS Equation 4

The result of Example 6 is

worth remembering:

d 1

ln x

dx x

The result of Example 6 is

worth remembering:

d 1

ln x

dx x

17.
LOGARITHMIC DIFFERENTIATION

The calculation of derivatives of complicated

functions involving products, quotients, or

powers can often be simplified by taking

The method used in the following example

is called logarithmic differentiation.

The calculation of derivatives of complicated

functions involving products, quotients, or

powers can often be simplified by taking

The method used in the following example

is called logarithmic differentiation.

18.
LOGARITHMIC DIFFERENTIATION Example 7

3/ 4 2

Differentiate y x x 1

5

(3 x 2)

We take logarithms of both sides of the equation

and use the Laws of Logarithms to simplify:

2

ln y ln x ln( x 1) 5ln(3 x 2)

3

4

1

2

3/ 4 2

Differentiate y x x 1

5

(3 x 2)

We take logarithms of both sides of the equation

and use the Laws of Logarithms to simplify:

2

ln y ln x ln( x 1) 5ln(3 x 2)

3

4

1

2

19.
LOGARITHMIC DIFFERENTIATION Example 7

Differentiating implicitly with respect to x gives:

1 dy 3 1 1 2 x 3

2 5

y dx 4 x 2 x 1 3x 2

Solving for dy / dx, we get:

dy 3 x 15

y 2

dx 4 x x 1 3x 2

Differentiating implicitly with respect to x gives:

1 dy 3 1 1 2 x 3

2 5

y dx 4 x 2 x 1 3x 2

Solving for dy / dx, we get:

dy 3 x 15

y 2

dx 4 x x 1 3x 2

20.
LOGARITHMIC DIFFERENTIATION Example 7

Since we have an explicit

expression for y, we can

substitute and write:

3/ 4 2

dy x x 1 3 x 15

5 2

dx (3 x 2) 4 x x 1 3x 2

Since we have an explicit

expression for y, we can

substitute and write:

3/ 4 2

dy x x 1 3 x 15

5 2

dx (3 x 2) 4 x x 1 3x 2

21.
STEPS IN LOGARITHMIC DIFFERENTIATION

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’.

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’.

22.
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.

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.

23.
THE POWER RULE PROOF

If n is any real number and f(x) = xn,

then n 1

f '( x) nx

Let y = xn and use logarithmic differentiation:

n

ln y ln x n ln x x 0

Thus,

y ' n

y x

n

y x

Hence, y ' n n nx n 1

x x

If n is any real number and f(x) = xn,

then n 1

f '( x) nx

Let y = xn and use logarithmic differentiation:

n

ln y ln x n ln x x 0

Thus,

y ' n

y x

n

y x

Hence, y ' n n nx n 1

x x

24.
LOGARITHMIC DIFFERENTIATION

You should distinguish carefully

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

You should distinguish carefully

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

25.
LOGARITHMIC DIFFERENTIATION

In general, there are four cases for

exponents and bases:

d

1. ( a b ) 0 a and b are constants

dx

d b b 1

2. f ( x) b f ( x) f '( x)

dx

d

3. a g ( x ) a g ( x ) (ln a ) g '( x)

dx

4. To find ( d / dx[ f ( x)]g ( x ) , logarithmic

differentiation can be used, as in the next example.

In general, there are four cases for

exponents and bases:

d

1. ( a b ) 0 a and b are constants

dx

d b b 1

2. f ( x) b f ( x) f '( x)

dx

d

3. a g ( x ) a g ( x ) (ln a ) g '( x)

dx

4. To find ( d / dx[ f ( x)]g ( x ) , logarithmic

differentiation can be used, as in the next example.

26.
LOGARITHMIC DIFFERENTIATION E. g. 8—Solution 1

x

Differentiate y x .

Using logarithmic differentiation,

we have:

x

ln y ln x x ln x

y' 1 1

x (ln x)

y x 2 x

1 ln x x 2 ln x

y ' y x

x 2 x 2 x

x

Differentiate y x .

Using logarithmic differentiation,

we have:

x

ln y ln x x ln x

y' 1 1

x (ln x)

y x 2 x

1 ln x x 2 ln x

y ' y x

x 2 x 2 x

27.
LOGARITHMIC DIFFERENTIATION E. g. 8—Solution 2

x ln x x

Another method is to write x ( e ) .

d x d x ln x

( x ) (e )

dx dx

d

x ln x

e ( x ln x)

dx

x 2 ln x

x

2 x

x ln x x

Another method is to write x ( e ) .

d x d x ln x

( x ) (e )

dx dx

d

x ln x

e ( x ln x)

dx

x 2 ln x

x

2 x

28.
THE NUMBER e AS A LIMIT

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.

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.

29.
THE NUMBER e AS A LIMIT

From the definition of a derivative as

a limit, we have:

f (1 h) f (1) f (1 x) f (1)

f '(1) lim lim

h 0 h x 0 x

ln(1 x) ln1 1

lim lim ln(1 x)

x 0 x x 0 x

lim ln(1 x)1 x

x 0

From the definition of a derivative as

a limit, we have:

f (1 h) f (1) f (1 x) f (1)

f '(1) lim lim

h 0 h x 0 x

ln(1 x) ln1 1

lim lim ln(1 x)

x 0 x x 0 x

lim ln(1 x)1 x

x 0

30.
THE NUMBER e AS A LIMIT Formula 5

1x

As f ’(1) = 1, we have lim ln(1 x) 1

x 0

Then, by Theorem 8 in Section 2.5 and

the continuity of the exponential function,

we have: lim ln(1 x )1/ x 1x

1 ln(1 x ) 1x

e e e x 0

lim e lim(1 x)

x 0 x 0

1x

e lim(1 x)

x 0

1x

As f ’(1) = 1, we have lim ln(1 x) 1

x 0

Then, by Theorem 8 in Section 2.5 and

the continuity of the exponential function,

we have: lim ln(1 x )1/ x 1x

1 ln(1 x ) 1x

e e e x 0

lim e lim(1 x)

x 0 x 0

1x

e lim(1 x)

x 0

31.
THE NUMBER e AS A LIMIT

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.

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.

32.
THE NUMBER e AS A LIMIT

This illustrates the fact that, correct to

seven decimal places, e ≈ 2.7182818

This illustrates the fact that, correct to

seven decimal places, e ≈ 2.7182818

33.
THE NUMBER e AS A LIMIT Formula 6

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

as x → 0+.

So, an alternative expression for e is:

n

1

e lim 1

n

n

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

as x → 0+.

So, an alternative expression for e is:

n

1

e lim 1

n

n