Final Exam Solutions
1
. Find the derivative of
f
(
x
) =
x
tan
x
.
Using
a
b
=
e
b
ln
a
, this function is
f
(
x
) =
e
(tan
x
) ln
x
. So
f
(
x
) =
e
(tan
x
) ln
x
(sec
2
x
) ln
x
+ (tan
x
)
1
x
=
x
tan
x
(sec
2
x
) ln
x
+
tan
x
x
.
(Here, we used the chain rule:
d
dx
e
u
=
e
u
du
dx
, where
u
= (tan
x
) ln
x
has derivative
du
dx
= (sec
2
x
) ln
x
+ (tan
x
)
1
x
by the product rule.)
2
. Suppose we initially had 10 milligrams of Zirconium95.
How much will we have left in 100
days, assuming that it decays at a rate proportional to the amount present, with a halflife of
64 days?
If
y
is the amount of the isotope, we have
dy
dx
=
ky
where
k
is the unknown decay
constant.
This solves to give
y
=
y
0
e
kt
.
The initial amount is 10 milligrams, so
y
(0) = 10 =
y
0
e
k
·
0
=
y
0
.
So
y
= 10
e
kt
.
The halflife is 64 days, so
y
(64) = 5 or
10
e
64
k
= 5. We solve
e
64
k
= 0
.
5 to obtain 64
k
= ln 0
.
5 or
k
= (ln 0
.
5)
/
64 =

0
.
0108.
Therefore
y
= 10
e

0
.
0108
t
and
y
(100) = 10
e

1
.
08
= 3
.
38 milligrams.
3
. Find the limit: lim
t
→
0
1

cos
t
t
2
.
Use L’Hospital’s rule twice: lim
t
→
0
1

cos
t
t
2
= lim
t
→
0
sin
t
2
t
= lim
t
→
0
cos
t
2
=
cos 0
2
=
1
2
.
4
. Find
(2
x
+ 3) sin
x dx
.
Use integration by parts:
u dv
=
uv

v du
, with
u
= 2
x
+ 3,
du
= 2
dx
,
dv
= sin
x dx
and
v
=

cos
x
. We have
(2
x
+3) sin
x dx
= (2
x
+3)(

cos
x
)

2(

cos
x
)
dx
=

(2
x
+3) cos
x
+2 sin
x
+
C.
5
. Find
sec
x
tan
3
x dx
.
This is
tan
2
x
sec
x
tan
x dx
=
(sec
2
x

1) sec
x
tan
x dx
(using the identity
1 + tan
2
x
= sec
2
x
).
Let
u
= sec
x
so
du
= sec
x
tan
x dx
.
The integral becomes
(sec
2
x

1) sec
x
tan
x dx
=
(
u
2

1)
du
=
1
3
u
3

u
+
C
=
1
3
sec
3
x

sec
x
+
C
.
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6
. Find
dx
x
√
4

x
2
.
Use the trig substitution
x
= 2 sin
θ
, so
dx
= 2 cos
θ dθ
.
The integral becomes
dx
x
√
4

x
2
=
2 cos
θ dθ
2 sin
θ
4

4 sin
2
θ
=
2 cos
θ dθ
2 sin
θ
(2 cos
θ
)
=
1
2
csc
θ dθ
=
1
2
ln

csc
θ

cot
θ

+
C.
We can use a right triangle to find csc
θ
and cot
θ
.
Now
sin
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 Fall '08
 VALDIMARSSON
 Derivative, Cos, dx

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