Solution:
Using the quotient and product rules:
d
dx
x
2
+ 1
sin
x
cos
x
+ 1
=
(sin
x
cos
x
+ 1)
d
dx
(
x
2
+ 1)

(
x
2
+ 1)
d
dx
(sin
x
cos
x
+ 1)
(sin
x
cos
x
+ 1)
2
=
(sin
x
cos
x
+ 1)(2
x
)

(
x
2
+ 1)(cos
2
x

sin
2
x
)
(sin
x
cos
x
+ 1)
2
.
Here we use that
d
dx
(sin
x
cos
x
+ 1) = (sin
x
)
0
cos
x
+ sin
x
(cos
x
)
0
= cos
2
x

sin
2
x
.
(b) [6 points] Find
d
dx
sec(sin(2
x
+ 1)).
Solution:
Using the chain rule:
d
dx
sec(sin(2
x
+ 1)) = sec(sin(2
x
+ 1)) tan(sin(2
x
+ 1))
±
d
dx
sin(2
x
+ 1)
²
= sec(sin(2
x
+ 1)) tan(sin(2
x
+ 1)) cos(2
x
+ 1)
±
d
dx
2
x
+ 1
²
= sec(sin(2
x
+ 1)) tan(sin(2
x
+ 1)) cos(2
x
+ 1)
·
2
.
5.
(a) [6 points] Find the linearization of
f
(
x
) =
x
1
/
2
at
x
= 9.
Solution:
f
0
(
x
) = 1
/
2
x

1
/
2
, so
f
0
(9) = 1
/
6. Thus
L
(
x
) =
f
(9)+
f
0
(9)(
x

9) =
3 + 1
/
6(
x

9).
Page 2
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View Full DocumentMath 171: Final Exam
(b) [3 points] Use your answer to part (a) to approximate
√
10.
Simplify your answer.
Solution:
We use the fact that 10
1
/
2
=
f
(10)
≈
L
(10). Evaluating,
L
(10) =
3 + 1
/
6(10

9) = 3 + 1
/
6 = 19
/
6.
6. You are ﬂying a kite. Assume that the string is a straight line.
(a) [3 points] If the angle of the string to level ground is
π/
3 radians and the kite string
has been let out to 300ft, how high is the kite?
Solution:
In both parts of the problem we have the following picture:
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
θ
h
300
Here
θ
is in radians and
h
is in feet. The relationship between
θ
and
h
is
sin(
θ
) =
h/
300 (because sin is opposite/hypotenuse) or
h
= 300 sin(
θ
).
When
θ
=
π/
3, we have
h
= 300 sin(
π/
3) = 300
√
3
2
= 150
√
3ft.
(b) [8 points] If the length of the string and the angle of the string to level ground are
as above and the angle is changing at

0
.
1 radians a minute, what is the rate of
change of the kite’s height?
Solution:
Diﬀerentiating both sides of
h
= 300 sin(
θ
) with respect to
t
(in
minutes) we have
dh
dt
= 300 cos(
θ
)
dθ
dt
. At the instant we are interested in,
θ
=
π/
3 and
dθ
dt
=

0
.
1, so the height is changing at
dh
dt
= 300 cos(
π/
3)(

0
.
1) = 300
1
2
(

0
.
1) =

15ft/min.
7. [8 points] Find all solutions to
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 Fall '07
 GOMEZ,JONES
 Math, Calculus, Algebra, Trigonometry, Sin, Quadratic equation, Mathematics in medieval Islam

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