Exam 1
Question
What is the derivative of
f
(
x
)
=
x
8
e
3
.
4
x
, evaluated at
x
=
0
.
7
?
Solution
Using the product rule for
f
(
x
)
=
g
(
x
)
·
h
(
x
)
, where
g
(
x
)
:
=
x
8
and
h
(
x
)
:
=
e
3
.
4
x
, we obtain
f
'
(
x
)
=
[
g
(
x
)
·
h
(
x
)
]
'
=
g
'
(
x
)
·
h
(
x
)
+
g
(
x
)
·
h
'
(
x
)
=
8
x
8
-
1
·
e
3
.
4
x
+
x
8
·
e
3
.
4
x
·
3
.
4
=
e
3
.
4
x
·
(
8
x
7
+
3
.
4
x
8
)
=
e
3
.
4
x
·
x
7
·
(
8
+
3
.
4
x
)
.
Evaluated at
x
=
0
.
7
, the answer is
e
3
.
4
·
0
.
7
·
0
.
7
7
·
(
8
+
3
.
4
·
0
.
7
)
=
9
.
236438
.
Thus, rounded to two digits we have
f
'
(
0
.
7
)
=
9
.
24
.