Math 112 (Calculus I)
Midterm Exam 2 KEY
Multiple Choice. Fill in the answer to each problem on your computer–scored answer
sheet. Make sure your name, section and instructor are on that sheet.
1. Find the derivative of y =
a) y 0 =
3
+ x−4 .
x3
9
− 4x−5
x2
d) y 0 = −9x−2 − 4x−3
g) y 0 =
3x3 + 9x2 − 4x−5
x5
2. Find f 0 (2) if f (x) =
b) y 0 = 9x3 − 4x−5
c) y 0 = −9x−4 − 4x−5
e) y 0 = 9x2 + 4x3
f) y 0 =
h) y 0 = 9x2 − 4x3
i)
None of the above
3
+ x−5
x4
2x
.
x2 + 1
a)
0
b)
1
2
c)
−1
2
d)
3
17
e)
−2
9
f)
−6
25
g)
1
h)
None of the above
3. How many years would it take an amount of money to triple if it was invested at a rate of 6
percent compounded continuously?
a) .06 ln(3)
b)
ln(.06)/3
c)
(ln(3)).06
d)
ln(3)/.06
e)
3 ln(.06)
f)
6 ln(3)
g)
None of the above
4. The derivative of f (x) = tan(x) sec(x) is
a)
tan3 (x) + tan2 (x) sec(x)
b)
tan3 (x) + tan(x) sec2 (x)
c)
cos(x)
d)
tan(x) sec3 (x)
e)
sec3 (x) + tan2 (x) sec(x)
f)
− sec3 (x) + tan2 (x) sec(x)
h)
None of the above
g) − tan3 (x) + tan2 (x) sec(x)
5. Find the slope of the tangent line to the curve f (x) = sin(sin(x)) at the point (π, 0).
a)
d)
g)
0
√
b)
2/2
Undefined
−1
c) 1
√
e) − 3/2
h)
f)
1/2
None of the above
6. Find the linear approximation of the function f (x) = ln(x) at a = 1, and use it to approximate
ln(1.05).
7.
a)
ln(1.05) ≈ 0.05
b)
ln(1.05) ≈ 1.05
c)
ln(1.05) ≈ 0.95
d)
ln(1.05) ≈ −0.05
e)
ln(1.05) ≈ −0.95
f)
ln(1.05) ≈ 0.025
g)
ln(1.05) ≈ 0.15
h)
None of the above
b)
2
1 + x2
c)
1
1 + 4x2
f)
1
√
1 − 4x2
d
(arccos(2x)) =
dx
a)
2
1 + 4x2
d)
√
g)
None of the above
2
1 − 4x2
e) − √
2
1 − 4x2
8. Each side of a square is increasing at a rate of 6 cm/s. At what rate is the area of the square
increasing when the area of the square is 25 cm2 ?
a)
60 cm2 /s
b)
12 cm2 /s
c)
5 cm2 /s
d)
6 cm2 /s
e)
25 cm2 /s
f)
30 cm2 /s
g)
Problem cannot be solved
without additional information
9. If f (x) = ln(1 + e2x ), find f 0 (0).
a)
1+e
b) e
d)
2
1+e
e)
2
1 + e2
g)
1
h)
None of the above
c) e2
f)
0
10. Evaluate the numerical value of sinh(ln(1/3)).
b)
5/3
c)
−4/3
d) −5/3
e)
3/5
f)
3/4
g) −3/5
h)
−3/4
i)
None of the above
a)
4/3
Solution:
1. c)
3. d)
5. b)
7. e)
9. g)
2. f)
4. e)
6. a)
8. a)
10. c)
Free response: Write your answer in the space provided. Answers not placed in
this space will be ignored.
11. (8 points) Find the derivative of the following functions.
(a) y = x3/2 (x3 + 2x)
3
3 1
y 0 = x 2 (x3 + 2x) + x 2 (3x2 + 2)
2
3
7
3
3 7
= x 2 + 3x 2 + 3x 2 + 2x 2
2
3
9 7
= x 2 + 5x 2 .
2
(b) h(t) =
t
(t − 1)2
(t − 1)2 (1) − t · 2(t − 1)(1)
(t − 1)2 − 2t(t − 1)
=
(t − 1)4
(t − 1)4
2
2
2
−t + 1
t − 2t + 1 − 2t + 2t
=
=
(t − 1)4
(t − 1)4
−(t + 1)
.
=
(t − 1)3
h0 (t) =
12. (10 points) Find the derivative of the following functions.
2)
(a) f (x) = 2sec(x
0
sec(x2 )
f (x) = 2
ln 2 · (tan(x2 ) sec(x2 )) · (2x)
2
= (2 ln 2) · x · 2sec(x ) tan(x2 ) sec(x2 ).
√
(b) g(x) = log5 (1 + cos( x))
√
1
1
√
· (− sin( x)) · √
(1 + cos( x)) ln 5
2 x
√
− sin( x)
√
√ .
=
2 ln 5 x(1 + cos( x))
g 0 (x) =
13. (6 points) Find the derivative of the function h(x) = xsin(x) .
ln(h(x)) = ln xsin(x) = sin(x) ln(x)
1 0
1
h (x) = cos(x) ln(x) + sin(x) ·
h(x)
x
sin(x)
0
sin(x)
h (x) = x
cos(x) ln(x) +
.
x
14. (5 points) Let f (x) = x + sin(x) for all x. Find the points x for which the graph of f at
(x, f (x)) has slope zero.
Solution: Slope 0 at x iff f 0 (x) = 0. We have:
f 0 (x) = 1 + cos(x) = 0
cos(x) = −1
x = ±(2k + 1)π, k ∈ Z.
15. (6 points) Find the limit. If necessary, state whether the limit is ∞, −∞ or does not exist.
tan(2x)
lim
x→0 sin(x)
Solution:
tan(2x)
sin(2x)
1
lim
= lim
·
x→0 sin(x)
x→0 cos(2x) sin(x)
2x
sin(2x)
·
= lim
x→0
2x
cos(2x) sin(x)
sin(2x)
x
2
= lim
·
x→0
2x
sin(x) cos(2x)
2
= 1 · 1 · = 2.
1
√
x x3 − 2
.
16. (7 points) Use logarithmic differentiation to find the derivative of y =
(5x + 4)6
Solution:
√ 3
x x −1
1
ln(y) = ln
= ln(x) + ln(x3 − 2) − 6 ln(5x + 4)
6
(5x + 4)
2
1 1
1
1
1 dy
= + · 3
· 3x2 − 6 ·
·5
y dx
x 2 x −2
5x + 4
√
dy
x x3 − 1
3x2
30
1
=
+
−
.
dx
(5x + 4)6
x 2(x3 − 2) 5x + 4
17. (8 points) A space traveler is moving from left to right along the curve y = x2 . When she
shuts off the engines, the ship begins instead to travel to the right along the tangent line to
the curve, at the point where the engines were shut off. At what point (a, a2 ) should she shut
off the engines in order to reach the point (4, 15)?
Solution: We want to find the point (a, a2 ) on the curve y = x2 such that the tangent line
to the curve at (a, a2 ) passes through (4, 15). The tangent line at (a, a2 ) has slope y 0 (a) = 2a.
Hence, its equation is
y = a2 + 2a(x − a) = 2ax − a2 .
We want this line to pass through (4, 15), so,
15 = 2a · 4 − a2
a2 − 8a + 15 = 0
(a − 3)(a − 5) = 0
a = 3 or a = 5.
Since the ship travels to the right along the tangent line, a = 3. So, the point we want to find
is (a, a2 ) = (3, 9).
18. (5 points) Find the linearization L(x) of the function f (x) = x5 + 4x2 at the point x = 1.
Solution: We have:
L(x) = f (1) + f 0 (1)(x − 1),
where f (1) = 1 + 4 = 5, f 0 (x) = 5x4 + 8x, so f 0 (1) = 5 + 8 = 13. Hence,
L(x) = 5 + 13(x − 1).
19. (7 points) Use implicit differentiation to show that
you may use the fact that 1 + tan2 (x) = sec2 (x).)
1
d
(arctan(x)) =
. (Hint: If desired,
dx
1 + x2
Solution:
y = arctan(x)
tan(y) = x
d
d
(tan(y)) =
(x) = 1
dx
dx
dy
sec2 (y)
=1
dx
1
dy
=
dx
sec2 (y)
1
dy
=
dx
1 + tan2 (y)
dy
1
=
.
dx
1 + x2
Alternatively, using triangle, tan(y) = x implies that sec(y) =
√
1 + x2 . So,
1
=
sec2 (y)
1
1
√
=
.
2
2
1 + x2
( 1+x )
20. (8 points) At the creamery, soft serve ice cream is being poured into a cone with height 5
inches, and radius 1 inch at its widest part. If the cone is being filled at a rate of 1 in3 /s, find
1
the rate at which the ice cream level is rising when the ice cream is 2 inches deep. (V = πr2 h)
3
r = 15
dV
1
= 1 in3 /s, h = 2 in. V = πr2 h. Must substitute for r in terms of h.
dt
3
h
r
h
Using similar triangles: = , so r = .
5
1
5
2
1
h
1 h3
V = π
h= π .
3
5
3 25
Solution:
So,
πh2 dh
dV
=
.
dt
25 dt
Plug the numbers in:
1=
π(2)2 dh
.
25 dt
So,
dh
25
=
in/s
dt
4π
END OF EXAM