2.3 the product and quotient rules and higher order derivatives

25
2.3 The Product and Quotient Rules and Higher Order Derivatives

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Page 1: 2.3 The Product and Quotient Rules and Higher Order Derivatives

2.3 The Product and Quotient Rules and Higher Order Derivatives

Page 2: 2.3 The Product and Quotient Rules and Higher Order Derivatives

After this lesson, you should be able to:

Find the derivative of a function using the Product Rule

Find the derivative of a function using the Quotient Rule

Find the derivative of a trigonometric function

Find a higher-order derivative of a function

Page 3: 2.3 The Product and Quotient Rules and Higher Order Derivatives

Theorem 2.7 The Product Rule

uvd

dx '' vuuv

dx

d uv

Example Find dy/dx

2( 6 )(3 2)y x x x

dx

dy )6( 2 xx )3( )23( x )62( x

( )u f x ( )v g x

vud

dx

uvd

dx

The product of two differentiable functions f and g is itself differentiable. The derivative is

where

12329 2 xx

Page 4: 2.3 The Product and Quotient Rules and Higher Order Derivatives

Proof of Theorem 2.7

x

xfxxfx

)()(lim

0)(lim

0xxg

x

)(lim0xf

x x

xgxxgx

)()(lim

0

and

and

exist.

)(

)()(lim

0xxg

x

xfxxfx

x

xgxxgxf

x

)()()(lim

0

exist.

exist.

exist.

Since f (x) and g(x) are differentiable, so g(x) is continuous.

Therefore

Therefore

Page 5: 2.3 The Product and Quotient Rules and Higher Order Derivatives

Proof of Theorem 2.7

x

xgxfxxgxxfxgxf

dx

dx

)()()()(lim)]()([

0

x

xgxfxfxxgxfxxgxxgxxfx

)()()()()()()()(lim

0

x

xgxxgxfxxg

x

xfxxfxx

)()()(lim)(

)()(lim

00

x

xgxxgxfxxg

x

xfxxfx

)()()()(

)()(lim

0

x

xgxxgxfxxg

x

xfxxfxxxx

)()(lim)(lim)(lim

)()(lim

0000

)(')()()(' xgxfxgxf

)(

)()(lim

0xxg

x

xfxxfx

x

xgxxgxf

x

)()()(lim

0

Page 6: 2.3 The Product and Quotient Rules and Higher Order Derivatives

The Product Rule Example

Example Find dy / dx :

xxxxy cossin23

dx

dy 23x (2 xx cos )sin x )sin( x23x xx cos2 2sin x sin x

23 2 cos sinx x x x

Page 7: 2.3 The Product and Quotient Rules and Higher Order Derivatives

The Product Rule Example

2/1

2

1x

Example Find f ’(x) )1)(32()( xxxf

)1)(32()( 21

xxxf 2)(' xf )1( 2

1

x )32( x

xx

2

332

Page 8: 2.3 The Product and Quotient Rules and Higher Order Derivatives

The Product Rule Example

Example Find y’ )9)(43( 23 xxxy

278365

)2)(43()9)(33('24

322

xxx

xxxxxy

Ans:

Page 9: 2.3 The Product and Quotient Rules and Higher Order Derivatives

The Quotient Rule

u

vd

dx

dx

d )vu

(

2

''

v

uvvu

Example Find y ’ 52

13

x

xy

2

u vv ud ddx dxv

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

2

6 15 6 2

(2 5)

x x

x

2

17

(2 5)x

( )u f x ( )v g xwhere

The quotient of two differentiable functions f and g is itself differentiable at all values of x for which g(x) ≠ 0. The derivative is:

Page 10: 2.3 The Product and Quotient Rules and Higher Order Derivatives

x

xfxxfx

)()(lim

0

x

xgxxgx

)()(lim

0

and

and

exist.

)(

)()(

lim0 xxg

xxfxxf

xexist.

exist.

exist.

Since f (x) and g(x) are differentiable, so g(x) is continuous.

Therefore

Therefore

Proof of Theorem 2.8, The Quotient Rule

)()(

)()()(

lim0 xxgxg

xxgxxg

xf

x

)(

1lim

0 xxgx

)(

)(lim

0 xg

xfx

Page 11: 2.3 The Product and Quotient Rules and Higher Order Derivatives

Proof of Theorem 2.8, The Quotient Rule

xxgxxg

xxgxfxgxxfx

)()(

)()()()(lim

0

)()(

)()()(

lim)()(

)()()(

lim00 xxgxg

xxgxxg

xf

xxgxgx

xfxxfxg

xx

)(

)(

xg

xf

dx

d

xxgxf

xxgxxf

x

)()(

)()(

lim0

xxgxxg

xxgxfxgxfxgxfxgxxfx

)()(

)()()()()()()()(lim

0

xxxgxg

xgxxgxf

xxxgxg

xfxxfxg

x )()(

)()()(

)()(

)()()(lim

0

Page 12: 2.3 The Product and Quotient Rules and Higher Order Derivatives

Proof of Theorem 2.8, The Quotient Rule

)()(

)()()(

lim)()(

)()()(

lim00 xxgxg

xxgxxg

xf

xxgxgx

xfxxfxg

xx

)]()([lim

)()()(lim

)]()([lim

)()()(lim

0

0

0

0

xxgxgx

xgxxgxf

xxgxgx

xfxxfxg

x

x

x

x

)(lim)(lim

)()(lim)(lim

)(lim)(lim

)()(lim)(lim

00

00

00

00

xxgxgx

xgxxgxf

xxgxgx

xfxxfxg

xx

xx

xx

xx

2)]([

)(')()()('

xg

xgxfxgxf

Page 13: 2.3 The Product and Quotient Rules and Higher Order Derivatives

The Quotient Rule Example

Example Find f ’(x): 1

)(2

x

xxf

'( )f x 2( 1)x (1) ( )x (2 )x

2 2( 1)x

2 2

2 2

1 2

( 1)

x x

x

2

2 2

1

( 1)

x

x

Page 14: 2.3 The Product and Quotient Rules and Higher Order Derivatives

Equation of a Horizontal Tangent LineFind the equations of the horizontal tangent lines for

2( )

1

xf x

x

Since we are asked to find horizontal tangent lines, we know the slopes of these lines are 0. So, set the 1st derivative equal to 0, then solve for x. This gives us the x-values of points on the graph where there are horizontal tangent lines. Then, we can find the points and write the equations.

2

2 2

1'( )

( 1)

xf x

x

2

2 2

10

( 1)

x

x

20 1 x

2 1x

1x

Therefore, the points where there are horizontal tangent lines are

12(1, ); 1

2( 1, ) Since the tangent lines are horizontal, the equations are 1

;2

y 1

2y

Page 15: 2.3 The Product and Quotient Rules and Higher Order Derivatives

The Quotient Rule Example

Example32

)4)(4( 32

x

xxxy

Page 16: 2.3 The Product and Quotient Rules and Higher Order Derivatives

More Quotient Rule Examples

xx

xxf

23

4)(

2

3

22

322

)23(

)26)(4()23(3)('

xx

xxxxxxf

Sometimes, you need rewrite or simplify the function before you try to take the derivative(s).Example

23

/4)(

2

x

xxxf

22

34

)23(

82443

xx

xxx

Rewrite

Then

Page 17: 2.3 The Product and Quotient Rules and Higher Order Derivatives

More Quotient Rule Examples

12

1

kk

k

kxx

kx

In Section 2.2, the Power Rule was provided only for the case where the exponent n is a positive integer greater then 1. By using the Quotient Rule, we can generalize the Power Rule to n is an integer.

If n is a negative integer, there exists a positive integer k such that n = –k So, by the quotient rule 2

1

)(

))(1()0(]

1[][

k

kk

kn

x

kxx

xdx

dx

dx

d

1 nnx

Page 18: 2.3 The Product and Quotient Rules and Higher Order Derivatives

xdx

xdcos

sin x

dx

xdsin

cos

dx

xd tan

xdx

xd 2sectan

The Derivative of Tangent

Page 19: 2.3 The Product and Quotient Rules and Higher Order Derivatives

dx

xd sec

xxdx

xdtansec

sec

The Derivative of Secant

Page 20: 2.3 The Product and Quotient Rules and Higher Order Derivatives

xxdx

xdcotcsc

csc x

dx

xd 2csccot

The Derivatives of Cosecant and Cotangent

Page 21: 2.3 The Product and Quotient Rules and Higher Order Derivatives

Example Given( ) tanf x x x

Write the equation of the tangent line at 4

x

Equation of the Tangent Line

2

2)2(

41

4sec

44tan)

4(' 2

f

xxxxxdx

dxf 2sectan]tan[)('

44tan

4)

4(

f

We can find the derivative of f (x) by using the rules.

The equation of the tangent line at is

4

x

)4

(2

2

4

xy

Page 22: 2.3 The Product and Quotient Rules and Higher Order Derivatives

Note

Because of the trigonometric identities, the derivative of a trigonometric function may have many forms. This presents a challenge when you are trying to match your answers to those given ones.

Page 23: 2.3 The Product and Quotient Rules and Higher Order Derivatives

Higher Order Derivatives

')(' ydx

dyxf Ex:

83)( 24 xxxf

xxxf 64)(' 3

'')(''2

2

ydx

ydxf Ex: 612)('' 2 xxf

''')('''3

3

ydx

ydxf xxf 24)(''' Ex:

)4(4

4)4( )( y

dx

ydxf Ex: 24)()4( xf

First DerivativeSecond Derivative

Third Derivative

Fourth Derivative

)()( )( nn

nn y

dx

ydxf Ex: nn xnnPxg ),2()()( n-th

Derivative

nxxg 2)(

Page 24: 2.3 The Product and Quotient Rules and Higher Order Derivatives

Higher Order Derivatives of sin(x)

xy sin

xdx

dycos

xdx

ydsin

2

2

xdx

ydcos

3

3

xdx

ydsin

4

4

xdx

ydcos

5

5

Page 25: 2.3 The Product and Quotient Rules and Higher Order Derivatives

Homework

Section 2.3 page 124 #1-23odd, 27, 41, 43, 51, 89, 91, 101, 103