product & quotient rules higher order derivatives lesson 2.3
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Product & Quotient RulesHigher Order Derivatives
Lesson 2.3
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 trig function
• Find a higher-order derivative of a function
Basic Rules
• Product Rule
( ) ( ) ' ( ) '( ) '( ) ( )f x g x f x g x f x g x
2
( ) ( ) '( ) ( ) '( )'
( ) ( )
f x g x f x f x g x
g x g x
How would you put this rule into words?
How would you put this rule into words?
The derivative of a product of 2 functions is equal to the first
times the derivative of the second plus the second
function times the derivative of the first.
Theorem 2.7 The Product Rule
Try Some
• Use additional rules to determine the derivatives of the following function
3 2) ( ) 2 3 6 3b p x x x
) ( ) 3 cos 4sina h x x x x
Try Some More
• Use additional rules to determine the derivatives of the following function
• a)
'
( ) 3 cos 4sin
( ) 3cos 3 sin 4cos
h x x x x
h x x x x x
Try Some More
• Use additional rules to determine the derivatives of the following function
• b)
3 2
' 2 3
( ) 2 3 6 3
( ) 6 6 3 2 3 (6 )
p x x x
p x x x x x
Try Some More
• Use additional rules to determine the derivatives of the following function
• b)
3 2
' 2 3 2 3
( ) 2 3 6 3
( ) 6 6 3 2 3 (6 ) 6 (6 3 2 3)
p x x x
p x x x x x x x x
Try Some More
• Use additional rules to determine the derivatives of the following function
• b)
3 2
' 2 3 2 3
3 2
( ) 2 3 6 3
( ) 6 6 3 2 3 (6 ) 6 (6 3 2 3)
6 (2 3 3)
p x x x
p x x x x x x x x
x x x
NOTE!!! YOU MUST FACTOR COMPLETELY!!!!!!!!!!
GET IT DOWN TO THE POINT WHERE YOU COULD SOLVE FOR THE ZEROS OF THE EQUATION!!!!!
Basic Rule
• Quotient Rule
2
( ) ( ) '( ) ( ) '( )'
( ) ( )
f x g x f x f x g x
g x g x
How would you put this rule into words?
How would you put this rule into words?
The derivative of a quotient of 2 functions is equal to the
denominator times the derivative of the numerator
minus the numerator times the derivative of the denominator
all over the square of the denominator.
Theorem 2.8 The Quotient Rule
Just Checking . . .
• Find the derivatives of the given functions
sin xy
x
4 2( ) 1
1f x x
x
2
7 4( )
5
xq x
x
Just Checking . . .
• Find the derivatives of the given functions
2'
2 2 2
7 4 7(5 ) ( 2 )(7 4)( ) ( )
5 (5 )
x x x xq x q x
x x
Just Checking . . .
• Find the derivatives of the given functions
2'
2 2 2
2 2
2 2
7 4 7(5 ) ( 2 )(7 4)( ) ( )
5 (5 )
35 7 14 8
(5 )
x x x xq x q x
x x
x x x
x
Just Checking . . .
• Find the derivatives of the given functions
2'
2 2 2
2 2 2
2 2 2 2
7 4 7(5 ) ( 2 )(7 4)( ) ( )
5 (5 )
35 7 14 8 7 8 35
(5 ) (5 )
x x x xq x q x
x x
x x x x x
x x
4 2( ) 1
1f x x
x
4
' 3 42
2( ) 1
1
2 ( 0)( 1) ( 2)(1)( ) 4 1
1 ( 1)
f x xx
xf x x x
x x
4
' 3 42
3 42
2( ) 1
1
2 ( 0)( 1) ( 2)(1)( ) 4 1
1 ( 1)
2 1 2 (2)4
1 ( 1)
f x xx
xf x x x
x x
xx x
x x
4
' 3 42
3 42
3 42
2( ) 1
1
2 ( 0)( 1) ( 2)(1)( ) 4 1
1 ( 1)
2 1 2 (2)4
1 ( 1)
8 4 (2)
1 ( 1)
f x xx
xf x x x
x x
xx x
x x
xx x
x x
4
' 3 42
3 42
33 4
2
2( ) 1
1
2 ( 0)( 1) ( 2)(1)( ) 4 1
1 ( 1)
2 1 2 (2)4
1 ( 1)
8 4 (2) 8 4 2
1 ( 1) 1 1
f x xx
xf x x x
x x
xx x
x x
x x x xx x
x x x x
4
' 3 42
3 42
33 4
2
3 3
2( ) 1
1
2 ( 0)( 1) ( 2)(1)( ) 4 1
1 ( 1)
2 1 2 (2)4
1 ( 1)
8 4 (2) 8 4 2
1 ( 1) 1 1
10 4 2 (5 2)
1 1 1 1
f x xx
xf x x x
x x
xx x
x x
x x x xx x
x x x x
x x x x
x x x x
4
' 3 42
3 42
33 4
2
3 3 3
2( ) 1
1
2 ( 0)( 1) ( 2)(1)( ) 4 1
1 ( 1)
2 1 2 (2)4
1 ( 1)
8 4 (2) 8 4 2
1 ( 1) 1 1
10 4 2 (5 2) 2 (5 2
1 1 1 1
f x xx
xf x x x
x x
xx x
x x
x x x xx x
x x x x
x x x x x x
x x x x
2
)
( 1)x
sin xy
x
'2
2
sin cos ( sin )
cos sin
x x x xy y
x xx x x
x
Other Trig Derivatives
• Now try it out
2 2tan sec cot csc
sec sec tan csc csc cot
d dx x x x
dx dxd d
x x x x x xdx dx
) 4 tan ' ?a f f
) sec tand
b x xdx
Theorem 2.9 Derivatives of Trigonometric Function
Other Trig Derivatives
2
4 tan
' 4(tan ) 4 (sec )
f
f
2
2
4 tan ' 4(tan ) 4 (sec )
4(tan sec )
f f
2
22
2
4 tan ' 4(tan ) 4 (sec )
sin4(tan sec ) 4
cos cos
cos sin4
cos
f f
2
22
2 2 2
4 tan ' 4(tan ) 4 (sec )
sin4(tan sec ) 4
cos cos
cos sin 1/ 2(sin 2 ) sin 2 24 4 4
cos cos 2cos
f f
2
22
2 2 2
2
4 tan ' 4(tan ) 4 (sec )
sin4(tan sec ) 4
cos cos
cos sin 1/ 2(sin 2 ) sin 2 24 4 4
cos cos 2cos
sin 2 22
cos
f f
JUST AN EXAMPLE OF HOW COMPLICATED THEY CAN GET!!!! CAN WE SIMPLIFY FURTHER???
Derivatives of the remaining trig functions can be determined the same way.
sin cosd
x xdx
cos sind
x xdx
2tan secd
x xdx
2cot cscd
x xdx
sec sec tand
x x xdx
csc csc cotd
x x xdx
Find Those High Orders
• Find the requested derivatives
2
2 32
) 4 1 ?d y
a y x xdx
4 3 2) ( ) 2 9 6 5 '''( ) ?b p x x x x p x
Find each derivative.
2 21. ( ) 3 2 5 4f x x x x
2. ( ) sinf x x x
'( )f x
'( )f x
2 23 2 8 5 4 3 4x x x x x 2 3 2 324 16 15 20 12 16x x x x x
3 232 36 20 15x x x
cos sin 1x x xcos sinx x x
Find the derivative.
33 4 4 13. ( )
2 3
x xg x
x
'( )g x
3 2 4 3
2
2 3 48 48 3 12 16 3 4 2
2 3
x x x x x x
x
4 312 16 3 4
2 3
x x x
x
4 3 3 2 4 3
2
96 96 6 144 144 9 24 32 6 8
2 3
x x x x x x x x
x
4 3 2
2
72 80 144 17
2 3
x x x
x
Find the derivative.
1 cos1. ( )
sin
xf x
x
'( )f x
2
sin sin 1 cos cos
sin
x x x x
x
2 2
2
sin cos cos
sin
x x x
x
2
1 cos
sin
x
x
2
1 cos
1 cos
x
x
1 cos
1 cos 1 cos
x
x x
1
1 cos x
2 2
2
sin cos cos
sin
x x x
x
( )( )
( )
g xf x
h x
'( )f x
2
' '
( )
h x g x g x h x
h x
'(2) 2, '(2) 4, (2) 4, (2) 3g h g h
2. Find '(2).f
'(2)f
2
2 ' 2 2 ' 2
(2)
h g g h
h
6 16
9
10
9
Equation of a Horizontal Tangent Line
Find the equations of the horizontal tangent lines for 2( )
1
xf x
x
Equation of a Horizontal Tangent Line
Find the equations of the horizontal tangent lines for 2( )
1
xf x
x
2 2'2 2 2 2
(1)( 1) (2 )( ) 1( )( 1) ( 1)
which is 0, when x = 1. And when x = 1, in the original equation, we have (1, 1/2) and (-1, -1/2). So, we have two tangent lines:
x x x xf xx x
1 1 and 2 2
y y
Example: ( ) tanf x x xWrite the equation of the tangent line at
4x
Equation of the Tangent Line
Example: ( ) tanf x x xWrite the equation of the tangent line at
4x
Equation of the Tangent Line
' 2 2 1 3( ) sec tan x = , sec ( ) tan( ) 1
4 4 4 2 2
, from the original equation, at , have ( , ), so the equation of the4 4 4
tangent line is:
3 3
4 2 4 2 8
f x x x x at
So we
b y x
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