Math 161 - Solutions To Sample Exam 2 Problems Worksheet

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Solutions To Sample Exam 2 Problems – Math 161
1
Math 161 – Solutions To Sample Exam 2 Problems
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1. For each of the following, find
. You may assume that a and b are constants. (REMINDER: No calculators
dx
would be allowed for an exam question resembling this one.)
ax cos x
x
(a) y = e
(c) x ln(xy) + e
= y
x
1/x
(b) y = sin(tan(bxe
))
(d) y = x
Solution:
d
ax cos x
ax cos x
(a) y = e
(ax cos x) = e
· ( ax sin x + a cos x).
·
dx
ax cos x
Therefore, our final answer is y = e
· ( ax sin x + a cos x).
(b)
d
x
x
y
= cos(tan(bxe
)) ·
(tan(bxe
))
dx
d
x
2
x
x
= cos(tan(bxe
)) · sec
(bxe
) ·
(bxe
)
dx
x
2
x
x
x
= cos(tan(bxe
)) · sec
(bxe
) · (bxe
+ be
)
x
2
x
x
x
Therefore, our final answer is y = cos(tan(bxe
)) · sec
(bxe
) · (bxe
+ be
).
(c) We have
1
d
1
x
x
x ·
(xy) + ln(xy) · 1 + e
= y
=⇒
· (xy + y) + ln(xy) + e
= y
·
xy
dx
y
x
=⇒
xy + y + y ln(xy) + ye
= yy
x
=⇒
y (y
x) = y + y ln(xy) + ye
x
y + y ln(xy) + ye
=⇒
y =
y
x
y+y ln(xy)+ye
x
Therefore, our final answer is y =
.
y x
(d) Since this function has variables in the base and in the exponent, we need to use logarithmic
differentiation. Therefore, we have
1/x
y = x
=⇒
ln y = (1/x) ln x
y
2
=⇒
= (1/x)(1/x) + ln x( x
) (using implicit differentiation and the product rule)
y
2
=⇒
y = y[(1/x)(1/x) + ln x( x
)]
1/x
2
1/x
=⇒
y = x
[(1/x)(1/x) + ln x( x
)] (since y = x
)
1/x
2
Therefore, our final answer is y = x
· [(1/x)(1/x) + ln x( x
)].
2. If f (x) and g(x) are the functions whose graphs are shown below, let
8
u(x) = f (g(x)), let v(x) = g(f (x)), and let w(x) = f (f (x)). Calculate
f(x)
7
each of the following, if it exists: u (6), v (1), and w (1). If one or more
of these derivatives does not exist, explain why.
6
5
4
g(x)
3
2
1
1
3
4
5
6
7
8
2

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