4.1 Implicit Differentiation

4.1.1 Definition. We will say that a given equation in x and y defines the function f implicitly if the graph of y = f(x) coincides with a portion of the graph of the equation.

Example:

•The equation implicitly defines functions

2 2 1x y 2 2

1 2( ) 1 and ( ) 1f x x f x x

2x y

1 2( ) and ( )f x x f x x

•The equation implicitly defines the functions

There are two methods to differentiate the functions defined implicitly by the equation.

For example: Find / if 1dy dx xy

One way is to rewrite this equation as , from which it

follows that

1y

x

2

1dy

dx x

Two differentiable methods

With this approach we obtain [ ] [1]

[ ] [ ] 0

0

d dxy

dx dxd d

x y y xdx dxdyx ydxdy y

dx x

The other method is to differentiate both sides of the equation before solving for y in terms of x, treating y as a differentiable function of x. The method is called implicit differentiation.

Since , 1

yx

2

1dy

dx x

Two differentiable methods

Example: Use implicit differentiation to find dy / dx if

Solution: 2 2

2 2

[ ] [3 ]

[ ] [ ] 3

2 2 3

2 3 2

3 2

2

d dx y x

dx dxd dx y

dx dxdy

x ydx

dyy xdx

dy x

dx y

2 2 3x y x

Example

Example: Find dy / dx if 3 3 11 0y x

Solution: 3

3

2

2

2

[ 3 11] [0]

[ ] 3 [ ] [11] [0]

2 3 0

2 3

3

2

d dy x

dx dxd d d dy x

dx dx dx dxdy

ydxdy

ydx

dy

dx y

Example

By implicit differentiation, we can show that if r is a rational number, then

1[ ]r rdx rx

dx

Example: 2/5 2/5 1 3/52 2[ ]

5 5

dx x x

dx

1/3 1/3 1 2/33 1 1[ ] [ ]

3 3

d dx x x x

dx dx

Example

In general, 1[ ]r rd du

u rudx dx

Example 3 2 4/5 3 2 1/5 3 2

3 2 1/5 2

4[ 2 4] ( 2 4) [ 2 4]

54

( 2 4) (3 4 )5

d dx x x x x x

dx dx

x x x x

2/3 5/3

5/3

5/3

2[(sin 4 ) ] (sin 4 ) [sin 4 ]

32

(sin 4 ) (cos4 4)38

(sin 4 ) (cos4 )3

d dx x x

dx dx

x x

x x

Example

4.2 Derivatives of Logarithmic Functions

1 1[ln ] , 0 [log ] , 0

lnb

d dx x x x

dx x dx x b

Generalized derivative formulas

1 1[ln ] , [log ]

lnb

d du d duu u

dx u dx dx u b dx

3Find [ln( 4)]d

xdx

Example

Solution: 3Let 4, we obtainu x

3 33

23

2

3

1[ln( 4)] [ 4]

41

= (3 )4

3 =

4

d dx x

dx x dx

xx

x

x

Example

1[ln | |] if 0

dx x

dx x

Example:

1 1[ln | cos |] [cos ] ( sin ) tan

cos cos

d dx x x x

dx x dx x

Find [ln | cos |]d

xdx

Solution:

Example

From section 4.1, we know that the differentiation formula

holds for rational values of r.

1[ ]r rdx rx

dx

In fact, we can use logarithmic differentiation to show that holds for any real number (rational or irrational).

1[ ]r rdx rx

dx

Example: 3 3 1[ ] 3dx x

dx

1[ ]dx x

dx

1[ ]e edx ex

dx

4.3 Derivatives of Exponential and Inverse Trigonometric Functions

Differentiability of Exponential Functions

Example:2 2 22[ ] [ ] 2x x xd d

e e x xedx dx

sin sin sin[3 ] 3 ln3 [sin ] ln3cos 3x x xd dx x

dx dx

Derivatives of the Inverse Trigonometric Functions

Example

Example: Find dy/dx if

Solution:

1tan ( )xy e

1 2sin ( )y x

2 2 2 2

1 2(2 )

1 ( ) 1 ( )

dy xx

dx x x

Example: Find dy/dx if

Solution:2 2

1( )

1 ( ) 1 ( )

xx

x x

dy ee

dx e e

4.4 L’Hopital’s Rule; Indeterminate Forms

Applying L’hopital’s Rule

Example: Find the limit using L’Hopital’s rule, and check the result by factoring.

2

3

9lim

3x

x

x

Solution: The numerator and denominator have a limit of 0, so the limit is an indeterminate form of type 0/0.

Applying L’Hopital’s rule yields

22

3 3 3

[ 9]9 2lim lim lim 6

3 1[ 3]x x x

dxx xdx

dx xdx

This agrees with the computation

2

3 3 3

9 ( 3)( 3)lim lim lim( 3) 6

3 3x x x

x x xx

x x

Example

Example: Find 0

sin3lim

5x

x

x

Solution: The limit is a indeterminate form of type 0/0.

Applying L’Hopital’s rule yields

0 0 0

[sin3 ]sin3 3cos3 3lim lim lim

5 5 5[5 ]x x x

dxx xdx

dx xdx

Example

Example: Find 3

30

1lim

x

x

e

x

Solution: The limit is a indeterminate form of type 0/0.

Applying L’Hopital’s rule yields

33 3

3 20 0 03

[ 1]1 3lim lim lim

3[ ]

xx x

x x x

dee edxdx xxdx

Example

Indeterminate Forms of Type /

Example: Find 2

5lim

xx

x

e

2

5 5 5

2 2lim lim lim 0

5 25x x xx x x

x x

e e e

Solution: The limit is a indeterminate form of type /

Applying L’Hopital’s rule yields

In fact, we can use LHopital’s rule to show that

lim 0 and limn x

x nx x

x e

e x

Example

Example: Find ln

limx

x

x

1ln 1

lim lim lim 01x x x

x xx x

Solution: The limit is a indeterminate form of type /

Applying L’Hopital’s rule yields

Similar methods can be used to find the limit of f(x)/g(x) is an Indeterminate form of the types: 0 00 , ,0 , ,1

Example