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Jolley, Garrett – Homework 7 – Due: Oct 9 2007, 3:00 am – Inst: R Heitmann
1
This
printout
should
have
18
questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
The due time is Central
time.
001
(part 1 oF 1) 10 points
±ind the derivative oF
f
when
f
(
x
) = 3
x
cos 5
x.
1.
f
0
(
x
) = 3 cos 5
x
+ 15
x
sin 3
x
2.
f
0
(
x
) = 3 cos 5
x

15
x
sin 5
x
correct
3.
f
0
(
x
) = 15 cos 5
x
+ 5
x
sin 5
x
4.
f
0
(
x
) = 3 cos 3
x

3
x
sin 5
x
5.
f
0
(
x
) = 15 cos 5
x

3
x
sin 5
x
Explanation:
Using the Formulas For the derivatives oF
sine and cosine together with the Chain Rule
we see that
f
0
(
x
) = (3
x
)
0
cos 5
x
+ 3
x
(cos 5
x
)
0
=
3 cos 5
x

15
x
sin 5
x.
keywords: derivative, trig Function, chain rule
002
(part 1 oF 1) 10 points
±ind the value oF
f
0
(0) when
f
(
x
) = (1

3
x
)

5
.
Correct answer: 15 .
Explanation:
Using the chain rule and the Fact that
(
x
α
)
0
=
α x
α

1
,
we obtain
f
0
(
x
) =
15
(1

3
x
)
6
.
At
x
= 0, thereFore,
f
0
(0) = 15
.
keywords: derivative, chain rule
003
(part 1 oF 1) 10 points
±ind
f
0
(
x
) when
f
(
x
) =
p
x
2

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

2)
p
x
2

4
x
2.
f
0
(
x
) =
x

2
2
√
x
2

4
x
3.
f
0
(
x
) = 2(
x

2)
p
x
2

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

2)
p
x
2

4
x
5.
f
0
(
x
) =
x

2
√
x
2

4
x
correct
6.
f
0
(
x
) =
2(
x

2)
√
x
2

4
x
Explanation:
By the Chain Rule,
f
0
(
x
) =
1
2
√
x
2

4
x
(2
x

4)
.
Consequently,
f
0
(
x
) =
x

2
√
x
2

4
x
.
keywords: derivative, square root, chain rule
004
(part 1 oF 1) 10 points
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View Full Document Jolley, Garrett – Homework 7 – Due: Oct 9 2007, 3:00 am – Inst: R Heitmann
2
Find
f
0
(
x
) when
f
(
x
) =
1
√
x
2

8
x
.
1.
f
0
(
x
) =
x

4
(8
x

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

x
(
x
2

8
x
)
3
/
2
correct
3.
f
0
(
x
) =
4

x
(8
x

x
2
)
3
/
2
4.
f
0
(
x
) =
x

4
(8
x

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

4
(
x
2

8
x
)
3
/
2
6.
f
0
(
x
) =
4

x
(
x
2

8
x
)
1
/
2
Explanation:
By the Chain Rule,
f
0
(
x
) =

1
2(
x
2

8
x
)
3
/
2
(2
x

8)
.
Consequently,
f
0
(
x
) =
4

x
(
x
2

8
x
)
3
/
2
.
keywords: derivative, square root, chain rule
005
(part 1 of 1) 10 points
Determine
f
0
(
x
) when
f
(
x
) =
1

2
x
√
1

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

2
(1

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

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

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

2
(1

x
2
)
3
/
2
correct
5.
f
0
(
x
) =
2

x
(1

x
2
)
3
/
2
6.
f
0
(
x
) =
2

x
(1

x
2
)
1
/
2
Explanation:
By the Product and Chain Rules,
f
0
(
x
) =

2
(1

x
2
)
1
/
2
+
2
x
(1

2
x
)
2(1

x
2
)
3
/
2
=

2(1

x
2
) +
x
(1

2
x
)
(1

x
2
)
3
/
2
.
Consequently,
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This note was uploaded on 10/29/2008 for the course M 408k taught by Professor Schultz during the Spring '08 term at University of Texas at Austin.
 Spring '08
 schultz

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