Calculus I (MAC2311-17)
Test 2 (2015/02/19)
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Page
Points Score
2
15
3
16
4
16
5
12
6
15
7
16
8
10
Total:
100
Page 1 of 8
1. In each part, find the derivative of the given function.
(a) (5 points) g(x) = x−2 ex
Solution:
g 0 (x) = −2x−3 ex + x−2 ex
√
(b) (5 points) f (t) =
t+t−1
t2
Solution:
√
t+t−1
= t1/2−2 + t1−2 − t−2
2
t
= t−3/2 + t−1 − t−2
−3 −5/2
t
− t−2 + 2t−3
f 0 (t) =
2
f (t) =
(c) (5 points) h(θ) =
1 − sin(θ)
1 + cos(θ)
Solution:
(− cos(θ))(1 + cos(θ)) − (1 − sin(θ))(− sin θ)
(1 + cos(θ))2
sin(θ) − cos(θ) − sin2 (θ) − cos2 (θ)
=
(1 + cos(θ))2
sin(θ) − cos(θ) − 1
=
(1 + cos(θ))2
h0 (θ) =
Calculus I (MAC2311-17)
Page 2 of 8
(d) (7 points) h(x) =
x−1/3 − 2x2
√
x+1
Solution:
√
( −1
x−4/3 − 4x)( x + 1) − (x−1/3 − 2x2 )( 2√1 x )
3
√
h (x) =
( x + 1)2
0
√
(e) (9 points) g(x) = arcsin( 1 − 2x)
1
(Hint: Use arcsin0 (x) = √1−x
2 .)
Solution:
1
1
√
g 0 (x) = q
(−2)
√
1 − ( 1 − 2x)2 2 1 − 2x
1
1
√
=p
1 − (1 − 2x) 1 − 2x
−1
=√ √
2x 1 − 2x
√
Note that g(x) is only defined when −1 ≤ 1 − 2x ≤ 1 (the domain of
arcsin in real numbers); that is,
√
1 − 2x ≤ 1
1 − 2x ≤ 1
x ≥ 0,
and for this particular function this is also where g 0 (x) is defined.
Calculus I (MAC2311-17)
Page 3 of 8
(f) (7 points) f (x) = etan(x
2)
Solution:
2
f 0 (x) = etan(x ) sec2 (x2 )(2x)
2
= 2xetan(x ) sec2 (x2 )
(g) (9 points) y = xx
(Hint: Use “logarithmic differentiation.”)
Solution:
ln(y) = ln(xx )
= x ln(x)
1 0
1
y = (1) ln(x) + x
y
x
y0
= ln(x) + 1
y
y 0 = y(ln(x) + 1)
= xx (ln(x) + 1)
Calculus I (MAC2311-17)
Page 4 of 8
2. (12 points) Assuming that tan(x − y 2 ) = x2 y defines y as a function of x around
dy
some points, use implicit differentiation to find
.
dx
Solution:
d
d 2
tan(x − y 2 ) =
(x y)
dx
dx
d
d 2
d
sec2 (x − y 2 ) (x − y 2 ) =
(x )y + x2 (y)
dx
dx
dx
dy
dy
sec2 (x − y 2 ) 1 − 2y
= 2xy + x2
dx
dx
dy
= sec2 (x − y 2 ) − 2xy
x2 + 2y sec2 (x − y 2 )
dx
dy
sec2 (x − y 2 ) − 2xy
= 2
dx
x + 2y sec2 (x − y 2 )
Calculus I (MAC2311-17)
Page 5 of 8
√ 2
x −4
3. Let f (x) = ln
.
2x4 + 1
(a) (5 points) Find the domain of f .
Solution: The argument of ln has to be positive:
√
x2 − 4
> 0.
2x4 + 1
The denominator of the left-hand side is always > 0 and the numerator
√
is always ≥ 0, so we only need to make sure the argument of
is
positive:
x2 − 4 > 0,
(x − 2)(x + 2) > 0,
which happens if x − 2 > 0 and x + 2 > 0 or if x − 2 < 0 and x + 2 < 0;
that is, if x > 2 and x > −2 or if x < 2 and x < −2, which, dropping
the redundant conditions, is x > 2 or x < −2. So,
Domf = (−∞, −2) ∪ (2, ∞).
(b) (10 points) Using the properties of logarithms formally, find f 0 (x).
Solution:
1
f (x) = ln((x2 − 4) 2 ) − ln(2x4 + 1)
1
= ln(x2 − 4) − ln(2x4 + 1)
2
1
1
1
f 0 (x) =
(2x) − 4
(8x3 )
2
2x −4
2x + 1
x
8x3
= 2
− 4
.
x − 4 2x + 1
We know it is always the case that Domf 0 ⊆ Domf . For this particular
function, in fact we have Domf 0 = Domf .
Calculus I (MAC2311-17)
Page 6 of 8
4. (6 points) Let f (θ) = sin(5θ). Find f (3) (θ) =
d3 f
.
dθ3
Solution:
d(sin(5θ))
df
=
= 5 cos(5θ)
dθ
dθ
d2 f
d(5 cos(5θ))
d df
=
= −25 sin(5θ)
=
2
dθ
dθ dθ
dθ
d3 f
d d2 f
d(−25 sin(5θ))
= −125 cos(5θ)
=
=
3
2
dθ
dθ dθ
dθ
sin(5x)
sin(θ)
, using what we know about limθ→0
.
x→0 tan(3x)
θ
5. (10 points) Evaluate lim
Solution:
sin(5x)
sin(5x)
5x
5
sin(5x)
sin(5x)
5x
= lim
= lim 5x
lim
= lim
x→0
x→0 tan(3x)
x→0 sin(3x)
sin(3x)
3 x→0 sin(3x)
3x
cos(3x)
3x
3x
cos(3x)
cos(3x)
sin(5x)
5 limx→0 5x
5
=
= ,
sin(3x)
3
3
limx→0
3x
limx→0 cos(3x)
since limθ→0
sin(θ)
= 1.
θ
Calculus I (MAC2311-17)
Page 7 of 8
6. A particle moves with the position function s(t) = t3 − 6t2 + 9t (m), t (s) ≥ 0.
(Don’t forget units.)
(a) (5 points) What is its velocity at t = 1 (s)?
Solution:
v(t) = s0 (t)
= 3t2 − 12t + 9
v(1) = 3(1)2 − 12(1) + 9 = 0 (m/s)
(b) (5 points) What is its acceleration at t = 1 (s)?
Solution:
a(t) = v 0 (t)
= 6t − 12
a(1) = 6(1) − 12 = −6 (m/s2 )
Calculus I (MAC2311-17)
Page 8 of 8