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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) + arctan 2
2
x
2
2 x
y = x + ln 2 + ln (2x) + arctan 2
2
x
2
2 x
- ln 2 + ln (2x) + arctan 2
dy
dx
= 2x + (ln 2)
x
2 x
2 x
· 2 Note that e
2
, ln 2, and arctan 2 are constants.
(b) y =
y = x · arctan (5x)
y = x · arctan (5x)
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(
cos(x
2
)
y = ln(tan(
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
- 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/dx
dy/dx dy/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
- Find the following.
(a) an antiderivative of y =
1 − 9 x
2
3
3
y =
1 − 9 x
2
3
3
y =
1 − 9 x
2
3
3
5 arcsin 3x
x
4
sin 2x
3
x + C
(b) tan(arccostan(arccostan(arccos xxx))) (rewritten as an algebraic expression - no trigonometric functions)
Let θ = arccos x. That is, θ is the angle whose cosine is x.
x
y
θ
x
2
2
2
⇒ y =
1 − x
2
tan(arccos x) = tan θ =
opposite
adjacent
y
x
1 − x
2
x
- Consider the function f(x) = x
4
e
x
f(x) = x
4
e
x
f(x) = x
4
e
x
with domain all real numbers.
(a) Find the x
x x-value(s) of all roots (x
x x-intercepts) of f
f f.
The equation x
4
e
x
= 0 means x
4
= 0 (that is, x = 0) or e
x
= 0 (no solution), so the only root is
at x = 0.
(b) Find the x
x x- and y
y y-value(s) of all critical points and identify each as a local max, local
min, or neither.
f
′
(x) = 4x
3
e
x
4
e
x
0 = x
3
e
x
(4 + x)
⇒ x = 0, − 4 Note that e
x
is never 0.
x < − 4 − 4 < x < 0 4 < x
f
′
positive negative positive
f ↗ ↘ ↗
y-values: f(−4) = 256e
− 4
≈ 4 .689, f(0) = 0
So, f has a local maximum at (− 4 , 256 e
− 4
) and a local minimum at (0, 0).
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
