Tran, Tony – Homework 6 – Due: Oct 2 2007, 3:00 am – Inst: Samuels
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
) =
1
2
¶
2
/
3
.
1.
f
0
(
x
) =
1
2
¶
x

1
/
3
2.
f
0
(
x
) =
1
3
x

1
/
3
3.
f
0
(
x
) =
2
3
1
2
¶

1
/
3
4.
f
0
(
x
) does not exist
5.
f
0
(
x
) = 0
correct
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
) = 2
x
3
+ 8
x
2

32
x
+ 3
.
1.
x
coords =

4
3
,
4
2.
x
coord =
4
3
3.
x
coord = 4
4.
x
coord =

4
3
5.
x
coords =
4
3
,

4
correct
6.
x
coord =

4
Explanation:
The
tangent
line
will
be
horizontal
at
P
(
x
0
, f
(
x
0
)) when
f
0
(
x
0
) = 0
.
Now
f
0
(
x
) = 6
x
2
+ 16
x

32
= 2(3
x

4)(
x
+ 4)
.
Consequently,
x
0
=
4
3
,

4
.
keywords:
horizontal tangent line, extrema,
polynomials, derivative
003
(part 1 of 1) 10 points
Differentiate the function
f
(
x
) =
√
5
x
3
1.
df
dx
=

3
√
5
x
4
correct
2.
df
dx
=

√
5
3
x
2
3.
df
dx
=
4
√
5
x
4
4.
df
dx
=
3
√
5
x
2
5.
df
dx
=
3
√
5
x
4
Explanation:
f
(
x
) =
√
5
x
3
=
√
5
x

3
f
0
(
x
) =

3
√
5
x

4
=

3
√
5
x
4
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Tran, Tony – Homework 6 – Due: Oct 2 2007, 3:00 am – Inst: Samuels
2
keywords: derivative, rational function
004
(part 1 of 1) 10 points
Find the derivative of
f
(
x
) = (
x
5
+ 2)(1

2
x
3
)
.
1.
f
0
(
x
) = 5
x
5

12
x
3

16
x
7
2.
f
0
(
x
) = 5
x
5
+ 12
x
3

16
x
7
3.
f
0
(
x
) = 5
x
4

12
x
2

16
x
6
4.
f
0
(
x
) = 5
x
4

12
x
2

16
x
7
correct
5.
f
0
(
x
) = 5
x
4
+ 12
x
2

16
x
6
Explanation:
By the Product rule
f
0
(
x
) = 5
x
4
(1

2
x
3
)

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

12
x
2

16
x
7
.
keywords: derivatives, product rule
005
(part 1 of 1) 10 points
Find the derivative of
f
when
f
(
x
) =
√
x
(
x

2)
.
1.
f
0
(
x
) =
3
x
+ 2
x
√
x
2.
f
0
(
x
) =
2
x

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

2
2
√
x
6.
f
0
(
x
) =
3
x

2
2
√
x
correct
Explanation:
By the Product Rule
f
0
(
x
) =
x

2
2
√
x
+
√
x .
After simplification this becomes
f
0
(
x
) =
x

2 + 2
x
2
√
x
=
3
x

2
2
√
x
.
keywords: derivatives, product rule
006
(part 1 of 1) 10 points
Find
f
0
(
x
) when
f
(
x
) =
4
x

1
5
x

1
.
1.
f
0
(
x
) =

1
(5
x

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

4
x
(5
x

1)
2
3.
f
0
(
x
) =
1
5
x

1
4.
f
0
(
x
) =
20
x

4
(5
x

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

1)
2
correct
Explanation:
Using the Quotient Rule for differentiation
we see that
f
0
(
x
) =
4(5
x

1)

5(4
x

1)
(5
x

1)
2
.
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 Fall '08
 schultz
 Derivative, Sin, Tran

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