April 9, 2008 MATH 220 – TEST 2(A) (3.1 – 5.4) [Pilachowski] ANSWER KEY
1. a. (12 points) Find the first derivative of .
()
2ln
2
3+
=x
exg
The chain rule is needed. Note that ln(2) is a number, not a function, and therefore its derivative = 0.
() ( )
2ln2ln 22 623 ++ =∗=
′xx xexexg
1. b. (12 points) Find the first derivative of
()
x
e
x
xf 3
43+
=.
The quotient rule is needed. xx evevxuxu 3334 3,,4,3 =
′
==
′
+=
()
(
)
(
)
(
)
(
)
[
]
(
)
x
x
x
x
xx
e
xx
e
xxe
e
exxe
xf 3
34
2
3
433
2
3
3433 943934
334 −+−
=
−−
=
+−
=
′
Alternately, you could have written
()
(
)
x
exxf 34 3−
∗+= and used the product rule.
()
()( )()()
x
xx
e
xx
exexxf 3
34
3334 943
433 −+−
=+−+=
′−−
1. c. (12 points) Find the first derivative of
()
⎟
⎠
⎞
⎜
⎝
⎛
=+3
15 xexm x.
The product rule is needed.
3
2
3
1
3
1
,,5, 1515 −
++ =
′
==
′
=xgxgefef xx
()
3
15
32
15
1515 5
3
5
3
13
1
3
2xe
x
e
orxexexm x
x
xx +
+
+
−
+++=
′ --either version is acceptable--
2. a. (14 points) Use logarithmic differentiation to differentiate
() ( )( )
⎟
⎠
⎞
⎜
⎝
⎛
−+= 3
47 121 xxxxm .
Take ln of both sides and expand:
()
[]
()( ) () ( )
xxxxxxxm ln
3
1
12ln41ln7121lnln 3
1
47 +−++=
⎥
⎦
⎤
⎢
⎣
⎡−+=
Take derivative of both sides:
()
()
xxxxm
xm
3
1
12
8
1
7+
−
+
+
=
′
Solve algebraically for :
()
xm′
() ()( )
⎟
⎠
⎞
⎜
⎝
⎛
−+∗
⎥
⎦
⎤
⎢
⎣
⎡+
−
+
+
=
′3
47 121
3
1
12
8
1
7xxx
xxx
xm
2. b. The population of a city as a function of time is given by the equation t
e
P2.0
4
20
−
+
=.
i) (6 points) Find .
()
tP′
()
(
)
(
)
(
)
2
2.02.02.0
2
2.0 442.04120
−
−−−
−
−+=−+−=
′tttt eeeeP
ii) (4 bonus points) Is the population always increasing, always decreasing, or sometimes
increasing/sometimes decreasing? Explain how you know.
answer: always increasing; Each factor of
P
′
is always positive.
Math 220, Sections 01** & 02**, T Pilachowski, Spring 2008
