the product rule f ' ( x ) = v ' ( x ) · u ( x ) + u ' ( x ) · v ( x ) is probably...
TRANSCRIPT
The Product Rule f ' ( x ) = v ' ( x ) · u ( x ) + u ' ( x ) · v ( x ) is probably the rule we
will use the most in conjunction with the other types of derivatives.
For example, finding u ' ( x ) may involve one rule while finding v ' ( x ) may
involve another rule.
Example: y = 8 x 2 · ln ( 4 x 2 – 3 x + 2 )
Solution: We first determine that the main rule to follow is the product rule.
Therefore
u ( x ) = 8 x 2 and v ( x ) = ln ( 4 x 2 – 3 x + 2 )
We determine to use the power rule to find u ' and Dx [ ln | g ( x ) | ] to find v '.
u ' ( x ) = 16 x g ( x ) = 4 x 2 – 3 x + 2, theng ' ( x ) = 8 x – 3
)(
)('])(ln[
xg
xgxgDx
234
38)('
2
xx
xxv
Now substitute these answers into the product rule
f ' ( x ) = v ' ( x ) u ( x ) + u ' ( x ) v ( x ).
)234(ln168234
)38(' 22
2
xxxx
xx
xy
Let us work a problem that looks similar but uses a different set of rules.
)234(ln8 22 xxy x
Solution: We first determine that the main rule to follow is the product rule.
Therefore
and v ( x ) = ln ( 4 x 2 – 3 x + 2 ) 2
8)( xxu
Dx [ ln | g ( x ) | ] to find v '.
Then use the derivative of a g (x) which is
D x [ a g (x) ] = ( ln a) [ g ' ( x ) ] a g (x) to find u ' ( x ).and
a = 8
g ( x ) = x 2
g ' ( x ) = 2 x
g ( x ) = 4 x 2 – 3 x + 2, theng ' ( x ) = 8 x – 3
)(
)('])(ln[
xg
xgxgDx
234
38)('
2
xx
xxv
28)2)(8ln()(' xxxu
You can not multiply these numbers together.
Now substitute these answers into the product rule
f ' ( x ) = v ' ( x ) u ( x ) + u ' ( x ) v ( x ).
)234(ln8)2)(8ln( 8234
)38(' 22
2
2
xxx
xx
xy xx
In mathematics f ( x ), g ( x ), etc. are just names. If it becomes to confusing using the same name in a problem you may want to change one of the names to h ( x ) or some other functional name. I will do that in the next example.
)132(log 2543
2 xxey x
Solution: We first determine that the main rule to follow is the product rule.
Therefore
54 2)( xexu and )132(log)( 2
3 xxxv
Use the derivative of e g (x):
D x [ e g (x) ] = g ' ( x ) · e g (x)and
Use the derivative of
log a | h ( x ) | which is
)(
)('
)ln(
1])(log[
xh
xh
axhD ax g ( x ) = 4 x 2 – 5, then
g ' ( x ) = 8 x
54 28)(' xexxu
a = 3
h ( x ) = 2x 2 – 3 x + 1
h ' ( x ) = 4 x – 3
132
34
)3ln(
1)('
2
xx
xxv
Now substitute these answers into the product rule
f ' ( x ) = v ' ( x ) u ( x ) + u ' ( x ) v ( x ).
)132(log8132
34
)3ln(
1' 25454
2 3
22
xxexe
xx
xy xx
34
)756ln( 2 xxy
Use the derivative of ln | g ( x ) | which is )(
)('])(ln[
xg
xgxgDx
34
)756()(Let 2 xxxg
To find g ' ( x ) use the chain rule alternative form but to reduce some of the confusion with all of the g ( x )’s and g ' ( x )’s let us change the notation to
d yd x
= h ' ( x ) · f ' [ h ( x ) ].
31
34
34)('
)(
Let
xxf
xxf
Let
h ( x ) = 6x 2 – 5 x + 7
h ' ( x ) = 12 x – 5
34
31
)756(
)756()512('
2
234
xx
xxxy
)756(3
)512(4
2
xx
x
( 6x 2 – 5 x + 7 )
)(')756()512(3
4 312 xgxxx
xd
ydTherefore
Substituting answers into the derivative:
Rules of algebra when canceling like terms: take the smallest exponent from the largest and write the answer where the largest exponent existed.The exponent of one is not written because it is understood.
133
31
34
Compare this new problem with the previous problem. They may look similar but there are subtle differences. Do not try to generalize or take any false short cuts. Make certain you understand the algebra rule of exponents and arithmetic.
52
)437(ln 2 xxy
Determine to use the derivative of ln | g ( x ) | is )(
)('])(ln[
xg
xgxgDx
52
)437()(Let 2 xxxg
To find g ' ( x ) use the chain rule alternative form.d yd x
= h ' ( x ) · f ' [ h ( x ) ].
53
52
52)('
)(
Let
xxf
xxf
Let
h ( x ) = 7x 2 + 3 x – 4
h ' ( x ) = 14 x + 3
)(')437()314(5
2 532 xgxxx
xd
yd
Therefore
52
53
)437(
)437()314('
2
252
xx
xxxy
53
52
)437()437(5
)314(2'
22
xxxx
xy
)437(5
)314(2'
2
xx
xy
I suggest that you try to solve these problems in small steps until you become very familiar with each type. Try to refrain from trying to take short cuts or immediately trying to write the final answers. It may take a little longer but probably you will be more successful. Speed only comes from many hours of practice.
22 )372)(29( xxxy
u ( x ) = 9 x + 2 v ( x ) = ( 2 x 2 – 7 x + 3 ) 2
u ' ( x ) = 9 v ' ( x ) = 2 ( 4 x – 7 )( 2 x 2 – 7 x + 3 )
Now substitute these answers into the product rule
f ' ( x ) = v ' ( x ) u ( x ) + u ' ( x ) v ( x ).
y ' = 2 ( 4 x – 7 )( 2 x 2 – 7 x + 3 )( 9 x + 2 ) + 9 ( 2 x 2 – 7 x + 3 ) 2
y ' = ( 2 x 2 – 7 x + 3 ) [ 2 ( 4 x – 7 )( 9 x + 2 ) + 9 ( 2 x 2 – 7 x + 3 ) ]
y ' = ( 2 x 2 – 7 x + 3 ) [ ( 8 x – 14 )( 9 x + 2 ) + 9 ( 2 x 2 – 7 x + 3 ) ]
y ' = ( 2 x 2 – 7 x + 3 ) [ 72 x 2 + 16 x – 126 x – 28 + 18 x 2 – 63 x + 27 ]
y ' = ( 2 x 2 – 7 x + 3 ) ( 90 x 2 – 173 x – 1 )