Derivatives of Logarithmic and Exponential Functions of Logarithmic and Exponential Functions We...

Derivatives of Logarithmic and Exponential Functions

We begin by finding the derivative of a logarithmic function directly from the definition and the properties of logarithms. We remind you of the following facts:

1. log log loga a cb b bc

= −

2. log ( ) logra r ab b

=

3. If f is a continuous function, and then lim ( ( )) ( ( ))f g x f g a

x a=

→

lim ( ) ( )g x g ax a

=→

( )11 14. lim 1 lim 1 where

0

xe v vvx xx v

= + = + = →∞ →

Proof. log ( ) loglog lim

0

x h xd b bxbdx hh

+ −= →

( )1lim log 10

vbvxv

= +→

1lim log0

x hb xhh

+ = →

1lim log 10

hb xhh

= + →

(Definition)

(Fact 1)

(algebra)

(Substitute v = h/x)

Theorem. The derivative of f(x) = logb(x) is

And more generally

1(log )eb x

log ( ) 1log ( )d u dub e

bdx u dx

=

( )11 lim log 1

0v vbx v

= + →

1 log ( )ebx

=

( )1 1lim log 10

vbx vv

= + →

( )11 log lim 1

0v vbx v

= + →

(Constant 1/x comes outside limit)

(Fact 2)

(Fact 3)

(Fact 4)

Example. Find the derivative of

(a) (b) (c)

As a special case, when b = e, we have

Theorem. The derivative of f(x) = ln(x) is

It follows that

1 1(log )ee x x

=

This simple formula explains in part the use of e as a base for logarithms in mathematical work.

[ ]ln( ) 1d u dudx u dx

=

log( )1 log( )

xx+

ln(ln( ))x ln(2 )x+

[ ]

[ ]

1 11 log( ) log( ) log( ) log( )log( )

21 log( ) 1 log( )

x e x ex x xx x

′ + − = + +

1 [ln( )] 1ln(ln( ))ln( ) ln( )

d xxx dx x x

′= =

(a)

(b)

[2 ]1ln(2 )2

d xxdxx

′ + + = + (c)

[ ] [ ]

1log( ) log( )

2 21 log( ) 1 log( )

e exx x x

= =+ +

1 1 12 2 4 2x x x x

= = + +

Slope of tangent line is 1 at x = 1

Derivative of ln(x) is 1x

Slope of tangent line at x = 2 is 1/2.

Derivative of ln(x) is 1x

Slope of tangent line at x = 4 is 1/4.

Derivative of ln(x) is 1x

If you are taking the derivative of the logarithm of a complicated expression involving primarily products, quotients, and powers , it is usually best to simplify the logarithm first before taking derivatives.

Example. Compute the derivative of the function3( 1) cos( )

( ) ln2

x xf x

x x

+=

−

We can write this derivative as but the computation

of the derivative of u will be very long and involved.

1 duu dx

1( ) 3ln( 1) ln(cos( )) ln( ) ln(2 )2

f x x x x x= + + − − −So

Then( )

1 sin( ) 1 1 1( ) 31 cos( ) 2 2

xf xx x x x

− −′ = + − −+ −

3 1 1tan( )1 4 2

xx x x

= − − ++ −

Better Solution: First we simplify the function.

3( 1) cos( ) 3ln ln ( 1) ln(cos( )) ln( ) ln( 2 )2

x xx x x x

x x

+ = + + − − − −

Logarithmic Differentiation

Even if a function does not involve a logarithm, it’s differentiation can often be simplified by using a technique called logarithmic differentiation. This technique should be used on functions that are primarily composed of products, quotients, and roots.

Example. Find the derivative of the function 3sin( )cos( ) tan ( )x x xy

x=

1. We begin by taking the natural logarithm of both sides of the equation 3sin( )cos( ) tan ( )x x xy

x=

3sin( )cos( ) tan ( )ln( ) ln

x x xy

x

=

1ln(sin( )) ln(cos( )) 3ln(tan( )) ln( )2

x x x x= + + −

2. Now take the derivative of both sides, treating y as an implicit function of x on the left, getting

21 cos( ) sin( ) sec ( ) 113sin( ) cos( ) tan( ) 2

dy x x xy dx x x x x

−= + + −

Thus

2sec ( ) 1cot( ) tan( ) 3

tan( ) 2x

y x xx x

= − + −

3 2sin( )cos( ) tan ( ) sec ( ) 1cot( ) tan( ) 3

tan( ) 2x x x x

x xx xx

= − + −

2cos( ) sin( ) sec ( ) 1 13

sin( ) cos( ) tan( ) 2x x xdy yx x x xdx

−= + + −

Derivatives of arbitrary powers of x

Theorem. For any real number r, rational or irrational, we have

ln( ) ln( ) ln( )ry x r x= =Proof. We use logarithmic differentiation. Let y = xr. Then

1d r rx rxdx

−=

, therefore

Examples. 1d x x

dxπ ππ −=

5 5 15x x

′ −=

1 dy ry dx x

= 1rdy ry rx rrx

dx x x−= = =and so

Derivatives of Exponential FunctionsWe can also use logarithmic differentiation to find the derivatives of the functions f(x) = bx.

Theorem.

Proof. Let y = bx. Then ln(y) = ln(bx) = xln(b). Therefore1 ln( )dy by dx

= and ln( ) ln( )xdy y b b b

dx= = . The second formula

Theorem.

follows by the chain rule. If b = e,we have:

and, more generally, d dux ux ue e e edx dx

′ = =

ln( ) and, more generally, ln( )x d dux u ub b b b b b

dx dx

′ = =

Height is 1 at x = 1.

Slope of tangent line is 1 at x = 1

Derivative of exp(x) is exp(x)

Slope of tangent lineHeight here

Derivative of exp(x) is exp(x)

Height

Slope of tangent line

Derivative of exp(x) is exp(x)

Example. Find the derivative of with respect to x.cos( )xe

Example. Find the derivative of with respect to x. 34

2x

Solution:

Solution:

[ ] [ ]cos( ) cos( ) cos( )cos( ) sin( )x x xe e x e x′ ′= =−

( )3 3 34 4 3 2 42 ln(2)2 4 ln(2) 12 2d x x xx xdx

′ = =

We have now shown the following rules for differentiation.

Basic Rule Generalized Rule

11. log logd x eb bdx x=

1log logd duu eb bdx u dx=

12. ln( )d xdx x

= 1ln( )d duu

dx u dx=

3. ln( )d x xb b bdx

=

ln( )d duu ub b bdx dx

=

4. d x xe edx

=

d duu ue edx dx

=

15.

rd rx rxdx

− =

1rd duru rudx dx

− =

We have now shown the following rules for differentiation.

Basic Rule Generalized Rule

11. log logd x eb bdx x=

1log logd duu eb bdx u dx=

12. ln( )d xdx x

= 1ln( )d duu

dx u dx=

3. ln( )d x xb b bdx

=

ln( )d duu ub b bdx dx

=

4. d x xe edx

=

d duu ue edx dx

=

15.

rd rx rxdx

− =

1rd duru rudx dx

− =

WARNING

Do not confuse 3 and 4 with 5

Example. Find the derivative of with respect to x.3ln( )y x=

Solution1: 21 3 33

3 3dy d xxdx dx xx x

= = =

3ln( )y x=Solution2: so 1 33dydx x x

= =

Example. Find the derivative of with respect to x.3

ln( )y x=

Solution: 22 3ln( )3ln( ) ln( )dy d xx x

dx dx x= =

Miscellaneous Examples

Example. Find the derivative of with respect to x.2sin (ln( ))y x=

Solution: 2sin(ln( ))cos(ln( ))2sin(ln( ))cos(ln( )) ln( )dy d x xx x x

dx dx x= =

Example. Find the derivative of with respect to x.2ln(sin )y x=

Solution: 2ln(sin( ), so2 cos( )sin( ) 2 2cot( )

sin( ) sin( )

y xdy d xx xdx x dx x

=

= = =

Example. Find the derivative of with respect to x.

2 2ln 3( 3) ( 1)

x xy

x x

−=

+ −

Solution: First simplify to get

( )2 3ln( ) ln( 2 ) ln ( 3) ln( 1)y x x x x= + − − + − −12ln( ) ln(2 ) 3ln( 3) ln( 1)2

x x x x= + − − + − −

Then

12 1 3 122 3 1

yxx x x

− ′= + − − − + −

2 1 3 14 2 3 1x x x x

= − − −− + −

Example. Find the derivative of with respect to x.3( )xx ey e −=

Solution: 3 33 3( ) ( ) 1 3x xdy d x xx e x ee x e e e

dx dx − −= − = −

Example. Find the derivative of with respect to x.( ) 22 1y x= +

Solution: ( )

2 1 2 12 2 22 1 1 2 1 2dy dx x x xdx dx

− − = + + = + 2 122 2 1x x

− = +

Example. Find the derivative of with respect to x by using logarithmic differentiation.

Solution:

151

xx

−+

1Let 5 .1

xyx

−=+

1 11 1 15Then ln( ) ln ln ln( 1) ln( 1)1 15 5 5

x xy x xx x

− − = = = − − + + +

1 11 1 1 2so 1 1y 5 5 5( 1)( 1)

dyx xdx x x

= − = − + − +

2 1 255( 1)( 1) 1 5( 1)( 1)

xdy yx x x x xdx

− = = − + + − +

Example. Find the derivative of with respect to xby using logarithmic differentiation.

Solution:

tan( )ln( ) xx

tan( )Let ln( ) xy x=

tan( )Then ln( ) ln ln( ) tan( )ln(ln( ))xy x x x = = 1 1 12so sec ( )ln(ln( )) tan( )y ln( )

dy x x xdx x x

= +

1 12sec ( ) ln(ln( )) tan( )ln( )

dy y x x xx xdx

= +

tan( )tan( ) 2ln( ) sec ( ) ln(ln( ))ln( )

xxx x xx x

= +