Part II: The Answers
262
Answers
301–400
389.
x
x
x
x
x
x
x
x
−
(
)
+
+
(
)
−
+
−
+
(
)
−
+
(
)






1
2
4
1
1
1
2
2
3
2
4
3 2
sin
cot
Begin by rewriting the function and taking the natural logarithm of each side:
y
x
x
x
x
y
x
x
x
x
=
−
+
+
(
)
=
−
+
+






(
)sin
ln( ) ln
(
)sin
(
)
1
2
4
1
2
4
3 2
3 2
Use properties of logarithms to expand:
ln
ln(
) ln(sin )
ln(
)
ln(
)
y
x
x
x
x
=
− +
−
+ −
+
1
1
2
2 3
2
4
Then take the derivative of each side with respect to x:
1
1
1
1
1
2
2
3
2
4
y
dy
dx x
x
x
x
x
= −
+
−
+
−
+
sin
(cos ) (
)
(
)
Multiplying both sides by y produces
dy
dx
y x
x
x
x
x
=
−
+
(
)−
+
(
)
−
+
(
)


 


 
1
1
1
1
2
2
3
2
4
sin
cos
Replacing with y with (
)sin
(
)
x
x
x
x
−
+
+
1
2
4
3 2
gives you the answer:
dy
dx
x
x
x
x
x
x
x
x
=
−
+
+
−
+
−
+
−
+

 

 
(
)sin
(
)
cot
(
)
(
)
1
2
4
1
1
1
2
2
3
2
4
3 2
Note that you can find the derivative without logarithmic differentiation, but using this
technique makes the calculations easier.
390.
5e
5x
Apply the chain rule to find the derivative of f
 
(x) = e
5x
:
= ( )
=
e
e
x
x
( )
5
5
5
5
( )
x
f '
391.
e
x
x
x x
sin
cos
+
+
(
)
4
4
3
Applying the chain rule to f x
e
x x
( ) =
+
sin
4 gives you
=
+
(
)
+
e
x
x
x x
cos
sin
4
4
3
( )
x
f '
392.
2 3
2
1
2
3
x
x
x
+ ( ) +
(
)




ln
Applying the product rule to f x
x
x
( ) =
+
(
)
3
1 2 gives you the derivative as follows:
= ( ) + +
(
)(
)
=
+
+
(
)




x
x
x
x
x
x
x
ln
(ln )
3
2
1 2
2
2 3
2
1
2
3
2
3
( )
x
f '