2.3 The Product and Quotient Rules and Higher Order Derivatives.
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Transcript of 2.3 The Product and Quotient Rules and Higher Order Derivatives.
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
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
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
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
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
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
The Product Rule Example
Example Find y’ )9)(43( 23 xxxy
278365
)2)(43()9)(33('24
322
xxx
xxxxxy
Ans:
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:
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
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
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
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
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
The Quotient Rule Example
Example32
)4)(4( 32
x
xxxy
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
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
xdx
xdcos
sin x
dx
xdsin
cos
dx
xd tan
xdx
xd 2sectan
The Derivative of Tangent
dx
xd sec
xxdx
xdtansec
sec
The Derivative of Secant
xxdx
xdcotcsc
csc x
dx
xd 2csccot
The Derivatives of Cosecant and Cotangent
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
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.
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)(
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
Homework
Section 2.3 page 124 #1-23odd, 27, 41, 43, 51, 89, 91, 101, 103