final3 - Granillo Yvette Final 1 Due 10:00 pm Inst E...

Granillo, Yvette – Final 1 – Due: May 10 2006, 10:00 pm – Inst: E Schultz 1
This print-out should have 25 questions.
Multiple-choice 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
Find the derivative of
f ( t ) =
ln t
4 + ln t
.
1. f
0
( t ) =
4 ln t
4 + ln t
2. f
0
( t ) =
4
t (4 + ln t ) 2
correct
3. f
0
( t ) =
ln t
4 + ln t
4. f
0
( t ) =
t (4 + ln t )
2
5. f
0
( t ) =
t (4 + ln t )
2
6. f
0
( t ) =
4 ln t
(4 + ln t )
2
Explanation:
By the Quotient Rule,
f
0
( t ) =
(4 + ln t )(1 /t ) - (ln t )(1 /t )
(4 + ln t ) 2
.
Consequently,
f
0
( t ) =
4
t (4 + ln t ) 2
.
keywords: Stewart5e, derivative, Quotient
Rule, log function,
002 (part 1 of 1) 10 points
Find the derivative of f when
f ( x ) = 3 x cos 2 x - 4 sin 2 x .
1. f
0
( x ) = - 8 cos 2 x + 6 x sin 2 x
2. f
0
( x ) = - 6 x sin 2 x - 8 cos 2 x
3. f
0
( x ) = - 6 x sin 2 x - 5 cos 2 x correct
4. f
0
( x ) = 8 cos 2 x - 5 x sin 2 x
5. f
0
( x ) = 6 x sin 2 x - 5 cos 2 x
Explanation:
Using formulas for the derivatives of sine
and cosine together with the Product and
Chain Rules, we see that
f
0
( x ) = 3 cos 2 x - 6 x sin 2 x - 8 cos 2 x
= - 6 x sin 2 x - 5 cos 2 x .
keywords: Stewart5e,
003 (part 1 of 1) 10 points
Determine f
0
( x ) when
f ( x ) = cos( e
- x
) .
1. f
0
( x ) = - sin( e
- x
)
2. f
0
( x ) = - e
- x
sin( e
- x
)
3. f
0
( x ) = e
- x
sin( e
- x
) correct
4. f
0
( x ) = e
- x
cos( e
- x
)
5. f
0
( x ) = cos( e
- x
) + e
- x
6. f
0
( x ) = - cos( e
- x
) - e
- x
Explanation:
By the Chain Rule,
f
0
( x ) = sin( e
- x
)
d
dx
e
- x
.
Consequently,
f
0
( x ) = e
- x
sin( e
- x
) .

Granillo, Yvette – Final 1 – Due: May 10 2006, 10:00 pm – Inst: E Schultz 2
keywords: Stewart5e,
004 (part 1 of 1) 10 points
Find the derivative of f when
f ( x ) =
1
2
ln
s
2 + x 2
2 - x
2
.
1. f
0
( x ) =
x
2 - x 4
2. f
0
( x ) =
x
4 - x 4
3. f
0
( x ) = -
x
2 - x
4
4. f
0
( x ) =
2 x
4 - x
4
correct
5. f
0
( x ) =
2 x
2 - x 4
6. f
0
( x ) = -
2 x
2 - x 4
7. f
0
( x ) = -
2 x
4 - x
4
8. f
0
( x ) = -
x
4 - x
4
Explanation:
Properties of logs ensure that
1
2
ln
s
2 + x 2
2 - x 2
=
1
4
n
ln(2 + x
2
) - ln(2 - x
2
)
o
.
By the Chain rule, therefore,
f
0
( x ) =
1
4
n
2 x
2 + x 2
+
2 x
2 - x 2
o
.
Consequently,
f
0
( x ) =
2 x
4 - x
4
.
keywords: Stewart5e,
005 (part 1 of 1) 10 points
Find the derivative of f when
f ( x ) = 4 tan
- 1
( e
3 x
) .
1. f
0
( x ) =
4
9 + e 6 x
2. f
0
( x ) =
3 e
3 x
9 + e 6 x
3. f
0
( x ) =
3 e
3 x
1 + e 3 x
4. f
0
( x ) =
12 e
3 x
9 + e 3 x
5. f
0
( x ) =
4
1 + e
3 x
6. f
0
( x ) =
12 e
3 x
1 + e
6 x
correct
Explanation:
By the Chain Rule,
f
0
( x ) =
3 · 4 e
3 x
1 + e
6 x
since
d
dx
tan
- 1
x =
1
1 + x
2
, ( e
3 x
)
2
= e
6 x
.
Consequently,
f
0
( x ) =
12 e
3 x
1 + e
6 x
.
keywords: Stewart5e,
006 (part 1 of 1) 10 points
Determine dy/dx when
3 cos x sin y = 5 .
1.
dy
dx
= tan x

Granillo, Yvette – Final 1 – Due: May 10 2006, 10:00 pm – Inst: E Schultz 3
2.
dy
dx
= tan xy
3.
dy
dx
= cot x cot y
4.
dy
dx
= tan x tan y correct
5.
dy
dx
= cot x tan y
Explanation:
Differentiating implicitly with respect to x
we see that
3
n
cos x cos y
dy
dx
- sin y sin x
o
= 0 .
Thus
dy
dx
cos x cos y = sin x sin y .
Consequently,
dy
dx
=
sin x sin y
cos x cos y
= tan x tan y .
keywords: Stewart5e, implicit differentiation,
trig functions, product rule
007 (part 1 of 1) 10 points
Determine if the limit
lim
x → -∞
3 + x + 2 x
3
2 - 5 x 3
exists, and if it does, find its value.