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Math 105: Review for Exam II - Solutions
- Find dy/dx
dy/dx dy/dx for each of the following.
(a) y = x
2
x
2
2 x
y = x + ln 2 + ln(2x) + (ln 2)x + arctan 2
2
x
2
2 x
y = x + ln 2 + ln(2x) + (ln 2)x + arctan 2
2
x
2
2 x
- ln 2 + ln(2x) + (ln 2)x + arctan 2
dy
dx
= 2x + (ln 2)
x
2 x
2 x
· 2 + ln 2 Note that e
2
, ln 2, and arctan 2 are constants.
(b) y =
x · arctan(5x)
y =
x · arctan(5x) y =
x · arctan(5x)
dy
dx
x
− 1 / 2
arctan(5x) +
x ·
1 + (5x)
2
arctan(5x)
2 x
1 / 2
x
1 + 25x
2
(c) y = ln(tan(
cos(x
2
)
y = ln(tan(2 ))
cos(x
2
)
y = ln(tan(2 ))
cos(x
2
)
dy
dx
tan(
cos(x
2 )
· sec
2
cos(x
2
)
) · ln 2(
cos(x
2
)
) · (− sin(x
2
)) · 2 x
(d) y =
x + e
π
cos 4 + sin
5
(6x)
y =
x + e
π
cos 4 + sin
5
(6x)
y =
x + e
π
cos 4 + sin
5
(6x)
Note that e
π
and cos 4 are constants.
dy
dx
(1)(cos 4 + sin
5
(6x)) − (x + e
π
)(5 sin
4
(6x) · cos(6x) · 6)
(cos 4 + sin
5
(6x))
2
Recall that sin
5
(6x) = (sin(6x))
5
(e) y = (x
2
sin x
y = (x
2
sin x
y = (x
2
sin x
We need logarithmic differentiation here.
ln y = sin x · ln(x
2
- Take natural log of each side.
y
dy
dx
= cos x · ln(x
2
x
2
· 2 x Differentiate.
dy
dx
[
cos x · ln(x
2
2 x sin x
x
2
]
y Solve for
dy
dx
dy
dx
[
cos x · ln(x
2
2 x sin x
x
2
]
· (x
2
sin x
Replace y.
- Consider the curve defined by x
3
3
xy
x
3
3
xy x
3
3
xy (known as the Folium of Descartes).
(a) Find dy/dxdy/dxdy/dx. Use implicit differentiation.
3 x
2
2
dy
dx
y +
x
dy
dx
3 y
2
dy
dx
x
dy
dx
y − 3 x
2
dy
dx
3 y
2
x
y − 3 x
2
dy
dx
9
2
y − 3 x
2
3 y
2
9
2
x
(b) Verify that the point (1,2) is on the curve above.
We must check to see if the values x = 1 and y = 2 satisfy the equation above.
x
3
3
?
xy
3
3
?
?
Thus, the point (1,2) is on the curve.
(c) Find the equation of the tangent line at the point (1,2).
We want y = mx + b.
m =
9
2
2
2
9
2
, so y =
x + b.
Now plug in x = 1 and y = 2 to find b.
· 1 + b ⇒
= b
Therefore, we have y =
x +
- Evaluate the following limits.
Throughout this solution, the symbol F will stand for whatever notation your instructor prefers for
using L’Hopital’s Rule on the indeterminate form 0/0; this may be
“0/0”
= or
L
′
H
= or
H
= or = “0/0” or
“has the form ‘
’ and so, by L’Hopital’s Rule, is equal to” or something else. The symbol ♥ will serve
the same purpose for the indeterminate forms ∞/∞ and −∞/∞.
(a) lim
x→ 1
x
3
7 − 7 x
lim
x→ 1
x
3
7 − 7 x
lim
x→ 1
x
3
7 − 7 x
F lim
x→ 1
3 x
2
(b) lim
x→ 0
1 − cos(2x)
x
lim
x→ 0
1 − cos(2x)
x
lim
x→ 0
1 − cos(2x)
x
= 0 Can’t use (and don’t need) L’Hopital’s Rule!
(c) lim
x→ 0
1 − cos(4x)
5 x
2
lim
x→ 0
1 − cos(4x)
5 x
2
lim
x→ 0
1 − cos(4x)
5 x
2
F lim
x→ 0
4 sin(4x)
10 x
F lim
x→ 0
16 cos(4x)
(d) lim
x→∞
x
2
x
lim
x→∞
x
2
x
lim
x→∞
x
2
x
♥ lim
x→∞
2 x
ln 2 · 2
x
♥ lim
x→∞
ln 2 · ln 2 · 2
x
(e) lim
x→∞
x
503 x
lim
x→∞
x
503 x
lim
x→∞
x
503 x
[Students in the 8:00 and 9:30 sections may omit this problem.]
Let L = lim
x→∞
x
503 x
This is of the form 1
∞
, so we take the ln of each side to turn it into a L’Hopital’s Rule problem.
ln L = lim
x→∞
ln
x
503 x
ln L = lim
x→∞
(503x) ln
x
Apply the log rule ln(a
b
) = b ln(a).
ln L = lim
x→∞
ln
4
x
1
503 x
Rewrite so it takes the form
ln L = lim
x→∞
1
1+
4
x
− 4
x
2
− 1
503 x
2
Apply L’Hopital’s Rule.
ln L = lim
x→∞
1
4
x
1
503
Cancel terms.
ln L =
1
1+
1
503
Take the limit.
ln L = 4 · 503
Now, since ln L = 4 · 503 = 2012, we have L = e
2012
0
-12 -6 -4 -2 0
4
10
8
6
2
- How would your answers to the previous question change if the domain of f
f f were [− 10 , 10]
[− 10 , 10]
[− 10 , 10]?
There would be a global maximum at (10, 10
4
e
10
). (And the graph would be restricted to − 10 ≤ x ≤
- You are planning to build a box-shaped aquarium with no top and with two square ends.
Your budget is $288. If the glass for the sides costs $12 per square foot and the opaque
material for the bottom costs $3 per square foot, what dimensions will maximize the
volume? Be sure to show how you know you have found the maximum.
x
x
y
Goal : Maximize volume
Objective function: volume = V = x · x · y = x
2
y
We need to get this down to a function of just one variable, so we use the constraint equation:
total cost =(cost of base) + (cost of two square ends) + (cost of two other sides)
288 = 3 xy + 12 · 2 x
2
288 = 27xy + 24x
2
288 − 24 x
2
= 27xy
288 − 24 x
2
27 x
= y
Substituting this back into the objective function gives
V = x
2
y = x
2
288 − 24 x
2
27 x
= x ·
288 − 24 x
2
(288x − 24 x
3
Now that we have V as a function of just one variable, we find its maximum.
V
′
(x) =
(288 − 72 x
2
(288 − 72 x
2
0 = (288 − 72 x
2
72 x
2
x
2
x = 2 We discard x = −2 because lengths must be nonnegative.
Since V
′
is positive for x < 2 and negative for 2 < x, we know that the maximum occurs at x = 2.
And y =
288 − 24 x
2
27 x
2
, so the dimensions are 2 by 2 by
