Part II: The Answers
240
Answers
301–400
323.
x g x xg x
2 ( )+ ′ ( )
 
 
Using the product rule on f (x) = x
2
g
 
(x) and then factoring gives you the derivative
as follows:
f x
xg x x g x
x g x xg x
'( )
( )
' ( )
( )
'( )
=
+ [
]
=
+
[
]
2
2
2
324.
g x xg x
x
( )+ ( )−
'
1
2
Begin by breaking up the fraction and simplifying:
f x
x g x
x
x
x g x
x
x
xg x
( )
( )
( )
( )
=
+
= +
=
+
−
1
1
2
2
1
Next, apply the power rule to the first term and the product rule to the second term:
= −
+
+
 
 
=
+
−
−
x
g x xg x
g x xg x
x
( )
( )
( )
( )
( )
1
1
1
2
2
x
f '
'
'
325.
–46
The product rule tells you that
( ) ( )
( ) ( ) ( ) ( )
fg x
f x g x f x g x
=
+
'
'
'
To find (fg)'(3), enter the numbers and solve:
( ) ( )
( ) ( ) ( ) ( )
( )( ) ( )( )
fg
f
g
f
g
=
+
=
− + −
= −
3
3 3
3
3
4
8
2 7
46
'
'
'
326.
2x cos x sin x – x
2 
sin
2
x + x
2 
cos
2 
x
Recall that the product rule states
d
dx
f x g x
f x g x f x g x
( ) ( )
 
  = ( ) ( )+ ( ) ( )
'
'
You can group the factors however you want and then apply the product rule within
the product rule. If you group together the trigonometric functions and apply the
product rule, you have f
 
(x) = x
2
(cos x sin x) so that
=
+
−
+
)
(
=
x
x
x x
x
x
x
x
x
x
( ) ( )(cos sin )
( sin )(sin ) (cos )(cos )
cos
2
2
2
s sin
s in
cos
x x
x x
x
−
+
2
2
2
2
x
f '