Part II: The Answers
258
Answers
301–400
Now apply the chain rule to each term to get the derivative:
=
− +
−
+
=
x
x
x
x
x
(ln )(
sin )
( cos ) (ln )(
sin )
( cos )
cos
1
6 1 2
2
1
6 4 3
3
2
( (ln )(
sin )
cos
(ln )(
sin )
6 1 2
3
6 4 3
− +
−
+
x
x
x
( )
x
f '
382.
1
7
(ln ) ln
x x
Begin by rewriting the function using properties of logarithms:
f x
x
x
x
( ) log log
log
log
log
log log
=
(
)
=
(
)
=
+
(
)
7
8
5
7
8
7
7
8
5
5
The derivative becomes
= +
(
)






=
(
)
x
x
x
x
ln log
(ln )
(ln )(ln )( ) log
0
1
7
1
8
1
7
8
8
8
( )
x
f '
Note that log 7 5 is a constant, so its derivative is equal to zero.
You can further simplify by using the change of base formula to write log
ln
ln
8
8
x
x
=
:
f x
x
x
x
x
x
’( )
ln
ln
log
ln
ln
ln
ln
ln
=
( )( )( )(
)
= ( )( )( ) ( )
=
( )
1
7
8
1
7
8
8
1
7
8
l ln x
383.
sec x
Applying the chain rule to f x
x
x
( ) ln sec
tan
=
+
gives you
=
+
+
(
)
=
+
+
x
x
x
x
x
x
x
x
x
sec
tan
sec tan
sec
(sec )(tan
sec )
sec
tan
1
2
x x
x
= sec
( )
x
f '