Jolley, Garrett – Homework 6 – Due: Oct 2 2007, 3:00 am – Inst: R Heitmann
1
This
printout
should
have
20
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
Determine the derivative of
f
when
f
(
x
) =
4
5
¶
2
/
3
.
1.
f
0
(
x
) =
4
5
¶
x

1
/
3
2.
f
0
(
x
) =
2
3
4
5
¶

1
/
3
3.
f
0
(
x
) = 0
correct
4.
f
0
(
x
) =
8
15
x

1
/
3
5.
f
0
(
x
) does not exist
Explanation:
The derivative of any constant function is
zero. Consequently,
f
0
(
x
) = 0
.
keywords: derivative, constant function
002
(part 1 of 1) 10 points
Find the
x
coordinates of all the points on
the graph of
f
at which the tangent line is
horizontal when
f
(
x
) =
x
3
+
x
2

x
+ 2
.
1.
x
coord =

1
2.
x
coord = 1
3.
x
coords =

1
3
,
1
4.
x
coord =
1
3
5.
x
coord =

1
3
6.
x
coords =
1
3
,

1
correct
Explanation:
The
tangent
line
will
be
horizontal
at
P
(
x
0
, f
(
x
0
)) when
f
0
(
x
0
) = 0
.
Now
f
0
(
x
) = 3
x
2
+ 2
x

1
= (3
x

1)(
x
+ 1)
.
Consequently,
x
0
=
1
3
,

1
.
keywords:
horizontal tangent line, extrema,
polynomials, derivative
003
(part 1 of 1) 10 points
Differentiate the function
f
(
x
) =
√
5
x
8
1.
df
dx
=
8
√
5
x
9
2.
df
dx
=
√
5
8
x
7
3.
df
dx
=

9
√
5
x
9
4.
df
dx
=

8
√
5
x
9
correct
5.
df
dx
=

8
√
5
x
7
Explanation:
f
(
x
) =
√
5
x
8
=
√
5
x

8
f
0
(
x
) =

8
√
5
x

9
=

8
√
5
x
9
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Jolley, Garrett – Homework 6 – Due: Oct 2 2007, 3:00 am – Inst: R Heitmann
2
keywords: derivative, rational function
004
(part 1 of 1) 10 points
Find the derivative of
f
(
x
) = (
x
5
+ 1)(1

3
x
3
)
.
1.
f
0
(
x
) = 5
x
5
+ 9
x
3

24
x
7
2.
f
0
(
x
) = 5
x
4

9
x
2

24
x
7
correct
3.
f
0
(
x
) = 5
x
5

9
x
3

24
x
7
4.
f
0
(
x
) = 5
x
4

9
x
2

24
x
6
5.
f
0
(
x
) = 5
x
4
+ 9
x
2

24
x
6
Explanation:
By the Product rule
f
0
(
x
) = 5
x
4
(1

3
x
3
)

9
x
2
(
x
5
+ 1)
.
Thus
f
0
(
x
) = 5
x
4

9
x
2

24
x
7
.
keywords: derivatives, product rule
005
(part 1 of 1) 10 points
Find the derivative of
f
when
f
(
x
) =
√
x
(2
x
+ 5)
.
1.
f
0
(
x
) =
4
x

5
x
√
x
2.
f
0
(
x
) =
4
x
+ 5
2
√
x
3.
f
0
(
x
) =
6
x

5
x
√
x
4.
f
0
(
x
) =
6
x
+ 5
2
√
x
correct
5.
f
0
(
x
) =
6
x

5
2
√
x
6.
f
0
(
x
) =
4
x
+ 5
x
√
x
Explanation:
By the Product Rule
f
0
(
x
) =
2
x
+ 5
2
√
x
+ 2
√
x .
After simplification this becomes
f
0
(
x
) =
2
x
+ 5 + 4
x
2
√
x
=
6
x
+ 5
2
√
x
.
keywords: derivatives, product rule
006
(part 1 of 1) 10 points
Find
f
0
(
x
) when
f
(
x
) =
3
x

1
4
x

1
.
1.
f
0
(
x
) =

1
(4
x

1)
2
2.
f
0
(
x
) =
1
4
x

1
3.
f
0
(
x
) =
1
(4
x

1)
2
correct
4.
f
0
(
x
) =
12
x

3
(4
x

1)
2
5.
f
0
(
x
) =
4

3
x
(4
x

1)
2
Explanation:
Using the Quotient Rule for differentiation
we see that
f
0
(
x
) =
3(4
x

1)

4(3
x

1)
(4
x

1)
2
.
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 Spring '08
 schultz
 Derivative, Jolley

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