ÇANKAYA UNIVERSITY
Department of Mathematics and Computer Science
MATH 155 Calculus for Engineering I
Summer 2008
Problems and Solutions for Recitation 5
July 25, 2008
9:40
––––––––––––––––––––––––––––––––––––––––––
(p.474)
1. D I F
31. Let f (x) = x3 − 3x2 , x ≥ 2. Find the value of df −1 /dx at the point x = −1 = f (3).
Solution:
df −1
1
1
df
= 3x2 − 6x =⇒
|x=f (3) = df |x=3 =
dx
dx
9
dx
––––––––––––––––––––––––––––––––––––––––––
32. Let f (x) = x2 − 4x, x > 2. Find the value of df −1 /dx at the point x = 0 = f (5).
Solution:
1
1
df
df −1
= 2x − 4 =⇒
|x=f (5) = df |x=5 =
dx
dx
6
dx
––––––––––––––––––––––––––––––––––––––––––
33. Suppose that the differentiable function y = f (x) has an inverse and that the graph of f
passes through the point (2, 4) and has a slope of 1/3 there. Find the value of df −1 /dx at the
point x = 4.
Solution:
df −1
df −1
1
1
|x=4 =
|x=f (2) = df |x=2 = 1 = 3.
dx
dx
3
dx
––––––––––––––––––––––––––––––––––––––––––
34. Suppose that the differentiable function y = g (x) has an inverse and that the graph of g
passes through the origin with slope 2. Find the slope of the graph of g −1 at the origin.
Solution:
dg −1
dg −1
1
1
|x=0 =
|x=f (0) = dg |x=0 = .
dx
dx
2
dx
––––––––––––––––––––––––––––––––––––––––––
2. U P L
(p.484)
1. Express the following logarithms in terms of ln 2 and ln 3.
a. ln 0.75 b. ln (4/9) c. ln (1/2)
√
√
√
3
d. ln 9 e. ln 3 2 f. ln 13.5
Solution:
3
(a) ln 0.75 = ln = ln 3 − ln 4 = ln 3 − ln 22 = ln 3 − 2 ln 2
4
(b) ln (4/9) = ln 4 − ln 9 = ln 22 − ln 32 = 2 ln 2 − 2 ln 3
(c) ln (1/2) = ln 1 − ln 2 = − ln 2
√
1
2
1
3
(d) ln 9 = ln 9 = ln 32 = ln 3
3
3
3
√
1
1/2
(e) ln 3 2 = ln 3 + ln 2 = ln 3 + ln 2
2
√
1
1
1 27
1 3
(f) ln 13.5 = ln 13.5 = ln
=
ln 3 − ln 2 = (3 ln 3 − ln 2).
2
2
2
2
2
––––––––––––––––––––––––––––––––––––––––––
2. Express the following logarithms
in terms of ln 5 and ln 7.
√
a. ln (1/125) b. ln 9.8 c. ln 7 7
d. ln 1225 e. ln 0.056 f. (ln 35 + ln (1/7)) / ln 25
Solution:
(a) ln (1/125) = ln 1 − 3 ln 5 = −3 ln 5
49
(b) ln 9.8 = ln
= ln 72 − ln 5 = 2 ln 7 − ln 5
5
√
3
(c) ln 7 7 = ln 73/2 = ln 7
2
(d) ln 1225 = ln 352 = 2 ln 35 = 2 ln 5 + 2 ln 7
7
(e) ln 0.056 = ln
= ln 7 − ln 53 = ln 7 − 3 ln 5
125
ln 5 + ln 7 − ln 7
1
(f) (ln 35 + ln (1/7)) / ln 25 =
= .
2 ln 5
2
––––––––––––––––––––––––––––––––––––––––––
Use properties of logarithms
to simplify the expressions
in Exercises 3 and 4.
2
sin θ
1
3. a. ln sin θ − ln
b. ln 3x − 9x + ln
5
3x
1 4
c. ln 4t − ln 2
2
Solution:
sin θ
sin θ
= ln sin θ = ln 5
(a) ln sin θ − ln
5
5
2
3x2 − 9x
1
= ln
= ln (x − 3)
(b) ln 3x − 9x + ln
3x
3x
2
√
2t
1 4
2
(c) ln 4t − ln 2 = ln 4t4 − ln 2 = ln 2t − ln 2 = ln
= ln t2
2
2
––––––––––––––––––––––––––––––––––––––––––
4. a. ln sec
θ + ln cos θ b. ln (8x + 4) − 2 ln 2
√
3 2
c. 3 ln t − 1 − ln (t + 1)
Solution:
(a) ln sec θ + ln cos θ = ln [(sec θ) (cos θ)] = ln 1 =
0
8x + 4
(b) ln (8x + 4) − 2 ln 2 = ln (8x + 4) − ln 4 = ln
= ln (2x + 1)
4
√
2
1/3
1
3 2
ln t2 − 1 − ln (t + 1) =
(c) 3 ln t − 1 − ln (t + 1) = 3 ln t − 1
− ln (t + 1) = 3
3
(t + 1) (t − 1)
= ln (t − 1).
ln
(t + 1)
––––––––––––––––––––––––––––––––––––––––––
3. D L
In Exercises 5-36, find the derivative of y with respect to x, t, or θ, as appropriate.
5. y = ln 3x
Solution:
1
1
y = ln 3x =⇒ y =
(3) = .
3x
x
–––––––––––––––––––––––––––––––––––––––––
6. y = ln kx, k constant
Solution:
1
1
y = ln kx =⇒ y =
(k) = .
kx
x
–––––––––––––––––––––––––––––––––––––––
7. y = ln t2
Solution:
2
1
2
y = ln t =⇒ y = 2 (2t) = .
t
t
–––––––––––––––––––––––––––––––––––––––
8. y = ln t3/2
Solution:
3/2 dy
1
3 1/2
3
=⇒
= 3/2
t
= .
y = ln t
dt
t
2
2t
––––––––––––––––––––––––––––––––––––––––
3
9. y = ln
x
Solution:
1
1
dy
3
−1
−2
=− .
=
−3x
y = ln = ln 3x =⇒
−1
x
dx
3x
x
–––––––––––––––––––––––––––––––––––––––––
3
10. y = ln
x
Solution:
1
3
y = ln = ln 3 − ln x =⇒ y = −
x
x
–––––––––––––––––––––––––––––––––––––––––
24. y = ln (ln (ln x))
Solution:
1
1
d
1 d
1
(ln (ln x)) =
(ln x) =
y = ln (ln (ln x)) =⇒ y =
ln (ln x) dx
ln (ln x) ln x dx
x ln x ln (ln x)
––––––––––––––––––––––––––––––––––––––––––
25. y = θ (sin (ln θ) + cos (ln θ))
Solution:
dy
1
1
y = θ (sin (ln θ) + cos (ln θ)) =⇒
= [sin (ln θ) + cos (ln θ)] + θ cos (ln θ) · − sin (ln θ) ·
dθ
θ
θ
= sin (ln θ) + cos (ln θ) + cos (ln θ) − sin (ln θ) = 2 cos (ln θ)
––––––––––––––––––––––––––––––––––––––––––
26. y = ln (sec θ + tan θ)
Answer:
y = sec θ
––––––––––––––––––––––––––––––––––––––––––
1
27. y = ln √
x x+1
Answer:
1
3x + 2
.
y = − ln x − ln (x + 1) =⇒ y = −
2
2x (x + 1)
––––––––––––––––––––––––––––––––––––––––––
1 1+x
ln
2 1−x
Answer:
1
.
y =
1 − x2
––––––––––––––––––––––––––––––––––––––––––
28. y =
1 + ln t
1 − ln t
Answer:
2
y =
.
t (1 − ln t)2
––––––––––––––––––––––––––––––––––––––––––
√
30. y = ln t
Solution:
√
1/2
1 1/2 −1/2 d 1/2 1 1/2 −1/2 1
dy
d 1/2 y = ln t = ln t1/2
=
=
ln t
ln t
ln t
t
=⇒
·
· 1/2 ·
dt
2
dt
2
t
dt
1
y = √ .
4t ln t
––––––––––––––––––––––––––––––––––––––––––
29. y =
31. y = ln (sec (ln θ))
Solution:
1
d
tan (ln θ)
dy
=
·
(sec (ln θ)) =
.
y = ln (sec (ln θ)) =⇒
dθ
sec (ln θ) dθ
θ
––––––––––––––––––––––––––––––––––––––––––
√
sin θ cos θ
32. y = ln
1 + 2 ln θ
Solution: √
sin θ cos θ
1
dy
1 cos θ sin θ
y = y = ln
= (ln sin θ + ln cos θ) − ln (1 + 2 ln θ) =⇒
=
−
−
1 + 2 ln θ
2
dθ
2 sin θ cos θ
2
θ
1 + 2 ln θ 4
1
=⇒ y =
cot θ − tan θ −
.
2
θ (1 + 2 ln θ)
––––––––––––––––––––––––––––––––––––––––––
5
(x2 + 1)
33. y = ln √
1−x
Solution:
5
2
1
1
1
5 · 2x
(x2 + 1)
−
(−1) =
y = ln √
= 5 ln x + 1 − ln (1 − x) =⇒ y = 2
2
x +1
2 1−x
1−x
10x
1
+
.
2
x + 1 2 (1 − x)
––––––––––––––––––––––––––––––––––––––––––
(x + 1)5
34. y = ln
(x + 2)20
Solution:
20
5
1
1
(x + 1)5
−
.
= [5 ln (x + 1) − 20 ln (x + 2)] =⇒ y =
y = ln
2
2 x+1 x+2
(x + 2)20
––––––––––––––––––––––––––––––––––––––––––
35. y =
x2
√
3x
√
ln t dt
x2 /2
Solution:
x2
√
x2
d x2
|x|
dy √ 2 d 2 y=
·
= ln x ·
x − ln
= 2x ln |x| − x ln √
ln t dt =⇒
dx
dx
2
dx 2
2
x2 /2
––––––––––––––––––––––––––––––––––––––––––
36. y =
√
x
ln t dt
Solution:
√
3x
√
d √
√ d √ √
1 −2/3
dy
3
3
3
y = √ ln t dt =⇒
= ln x ·
x
x − ln x ·
x = ln x
−
dx
dx
dx
3
x
√ 1 −1/2
x
ln x
2
√
√
ln 3 x ln x
y = √
− √ .
3
2 x
3 x2
––––––––––––––––––––––––––––––––––––––––––
4. I
Evaluate
−2 the integrals in Exercises 37-54.
dx
37.
−3 x
Solution:
−2
2
dx
= [ln |x|]−2
−3 = ln .
3
−3 x
––––––––––––––––––––––––––––––––––––––––––
0
3 dx
−1 3x − 2
Solution:
0
3 dx
2
= [ln |3x − 2|]0−1 = ln .
5
−1 3x − 2
––––––––––––––––––––––––––––––––––––––––––
38.
2y dy
y 2 − 25
Solution:
2y dy
= ln y 2 − 25 + C.
2
y − 25
––––––––––––––––––––––––––––––––––––––––––
8y dy
40.
4y 2 − 5
Solution:
2
8y dy
4y − 5 + C.
=
ln
4y 2 − 5
––––––––––––––––––––––––––––––––––––––––––
π
sin t dt
41.
0 2 − cos t
Solution:
π
sin t dt
= ln [2 − cos t]π0 = ln 3.
0 2 − cos t
––––––––––––––––––––––––––––––––––––––––––
dx
√
53.
2 x + 2x
Solution:
√
dx
dx
1
√
√
√ ; let u = 1 + x =⇒ du = √ dx;
=
2 x (1 + x)
2 x
2 x + 2x
√ √
dx
du
√
√ =
= ln |u| + C = ln 1 + x + C = ln 1 + x + C
u
2 x (1 + x)
––––––––––––––––––––––––––––––––––––––––––
39.
5. L D
In Exercises 55-68, use logarithmic differentiation to find the derivative of y with respect to the
given independent
variable.
55. y = x (x + 1)
Solution:
1
y = x (x + 1) = (x (x + 1))1/2 =⇒ ln y = ln (x (x + 1)) =⇒ 2 ln y = ln x + ln (x + 1)
2 1
1
2y 1
1
1
2x + 1
x (x + 1)
=⇒
= +
=⇒ y =
+
= y
x x+1
2
x x+1
2 x (x + 1)
––––––––––––––––––––––––––––––––––––––––––
56. y = (x2 + 1) (x − 1)2
Solution:
1 2
y = (x2 + 1) (x − 1)2 =⇒ ln y =
ln x + 1 + 2 ln (x − 1)
2
1
2
1
2x
x
y
2
2
=
+
+
=⇒ y = (x + 1) (x − 1)
=⇒
y
2 x2 + 1 x − 1
x2 + 1 x − 1
––––––––––––––––––––––––––––––––––––––––––
t
57. y =
t+1
Solution:
1/2
t
1
t
y=
=⇒ ln y = [ln t − ln (t + 1)]
=
t+1
t+1
2
1 1
1
1
1
dy
t
1 dy
=
−
=
=⇒
=⇒
y dt
2 t t+1
dt
2 t + 1 t (t + 1)
––––––––––––––––––––––––––––––––––––––––––
1
58. y =
t (t + 1)
Answer:
1
y=
t (t + 1)
dy
2t + 1
=⇒
=−
.
dt
2 (t2 + 1)3/2
––––––––––––––––––––––––––––––––––––––––––
√
59. y = θ + 3 (sin θ)
Answer:
√
y = θ + 3 (sin θ)
1
dy √
=⇒
= θ + 3 (sin θ)
+ cot θ .
dθ
2 (θ + 3)
––––––––––––––––––––––––––––––––––––––––––
√
60. y = (tan θ) 2θ + 1
Answer: √
y = (tan θ) 2θ + 1
tan θ
dy 2 √
.
= sec θ 2θ + 1 + √
=⇒
dθ
2θ + 1
––––––––––––––––––––––––––––––––––––––––––
61. y = t (t + 1) (t + 2)
Answer:
y = t (t + 1) (t + 2)
dy
=⇒
= 3t2 + 6t + 2.
dt
––––––––––––––––––––––––––––––––––––––––––
1
t (t + 1) (t + 2)
Solution:
1
1 dy
1
1
1
y=
=⇒ ln y = ln 1 − ln t − ln (t + 1) − ln (t + 2) =⇒
=− −
−
t (t + 1) (t + 2)
y dt
t t+1 t+2
1
1
1
1
3t2 + 6t + 2
dy
=
−
− −
=−
.
=⇒
dt
t (t + 1) (t + 2)
t t+1 t+2
(t3 + 3t2 + 2t)2
––––––––––––––––––––––––––––––––––––––––––
62. y =
θ+5
63. y =
θ cos θ
Answer:
θ+5
y=
θ cos θ
θ+5
1
1
dy
=
− + tan θ .
=⇒
dθ
θ cos θ
θ+5 θ
––––––––––––––––––––––––––––––––––––––––––
θ sin θ
64. y = √
sec θ
Answer:
θ sin θ
y=√
sec θ
dy
θ sin θ 1
1
=⇒
=√
+ cot θ − tan θ .
dθ
2
sec θ θ
––––––––––––––––––––––––––––––––––––––––––
√
x x2 + 1
65. y =
(x + 1)2/3
Answer:
√
x x2 + 1
y=
(x + 1)2/3
√
x x2 + 1 1
x
2
dy
=
+
−
.
=⇒
dx
(x + 1)2/3 x x2 + 1 3 (x + 1)
––––––––––––––––––––––––––––––––––––––––––
(x + 1)10
66. y =
(2x + 1)5
Answer:
(x + 1)10
y=
(2x + 1)5
5
dy
(x + 1)10
5
=⇒
=
−
.
dx
(2x + 1)5 x + 1 2x + 1
––––––––––––––––––––––––––––––––––––––––––
3 x (x − 2)
67. y =
x2 + 1
Answer:
3 x (x − 2)
y=
x2 +1
1 3 x (x − 2) 1
1
2x
dy
=
+
−
.
=⇒
dx
3
x2 + 1 x x − 2 x2 + 1
––––––––––––––––––––––––––––––––––––––––––
x (x + 1) (x − 2)
68. y = 3 2
(x + 1) (2x + 3)
Solution:
1
x (x + 1) (x − 2)
=⇒ ln y =
ln x + ln (x + 1) + ln (x − 2) − ln x2 + 1 − ln (2x + 3)
y= 3 2
(x + 1) (2x + 3)
3
1 3 x (x + 1) (x − 2) 1
1
1
2x
2
=⇒ y =
+
+
−
−
.
3 (x2 + 1) (2x + 3) x x + 1 x − 2 x2 + 1 2x + 3
––––––––––––––––––––––––––––––––––––––––––
(p.493)
6. D
In Exercises 17-36, find the derivative of y with respect to x, t, θ as appropriate.
17. y = e−5x
Solution:
y = e−5x =⇒ y = −5e−5x .
––––––––––––––––––––––––––––––––––––––––––
18. y = e2x/3
Solution:
2
y = e2x/3 =⇒ y = e2x/3 .
3
––––––––––––––––––––––––––––––––––––––––––
36. y =
e2x
√
e4
ln t dt
x
Solution:
e2x
2x d 2x 4√x d 4√x 2x √ 4√x ·
= (2x) e − 4 x e
·
e
e − ln e
y = √ ln t dt =⇒ y = ln e ·
dx
dx
e4 x
d √ 4 x
dx
√
√ 4√x 2
2x
√
=⇒ y = 4xe2x − 8e4 x .
=⇒ y = 4xe − 4 xe
x
––––––––––––––––––––––––––––––––––––––––––
7. E L
(p.549)
Evaluate the
limits in Exercises 85-96.
10x − 1
85. lim
x→0
x
Solution:
10x − 1
(ln 10) 10x
0
= lim
= ln 10.
The limit leads to the indeterminate form : lim
x→0
0 x→0
x
1
––––––––––––––––––––––––––––––––––––––––––
3θ − 1
θ→0
θ
Solution:
86. lim
0
3θ − 1
(ln 3) 3θ
The limit leads to the indeterminate form : lim
= lim
= ln 3.
θ→0
0 θ→0 θ
1
––––––––––––––––––––––––––––––––––––––––––
2sin x − 1
x→0 ex − 1
Solution:
87. lim
0
2sin x − 1
2sin x (ln 2) (cos x)
The limit leads to the indeterminate form : lim x
= ln 2.
= lim
x→0
0 x→0 e − 1
ex
––––––––––––––––––––––––––––––––––––––––––
2− sin x − 1
x→0
ex − 1
Solution:
88. lim
The limit leads to the indeterminate form
0
2− sin x − 1
2− sin x (ln 2) (− cos x)
: lim
=
lim
=
x→0
0 x→0 ex − 1
ex
− ln 2.
––––––––––––––––––––––––––––––––––––––––––
5 − 5 cos x
x→0 ex − x − 1
Solution:
0
5 − 5 cos x
5 sin x
5 cos x
The limit leads to the indeterminate form : lim x
= lim x
= lim
= 5.
0 x→0 e − x − 1 x→0 e − 1 x→0 ex
–––––––––––––––––––––––––––––––––––––––––
4 − 4ex
90. lim
x→0
xex
Solution:
0
4 − 4ex
−4ex
= lim x
= −4.
The limit leads to the indeterminate form : lim
x→0 e + xex
0 x→0 xex
–––––––––––––––––––––––––––––––––––––––––
t − ln (1 + 2t)
91. lim+
t→0
t2
Solution:
2
1 − 1+2t
0
t − ln (1 + 2t)
The limit leads to the indeterminate form : lim+
= −∞.
= lim+
t→0
0 t→0
t2
2t
–––––––––––––––––––––––––––––––––––––––––
sin2 (πx)
92. lim x−4
x→4 e
+3−x
Solution:
sin2 (πx)
2π sin (πx) cos (πx)
0
= lim
The limit leads to the indeterminate form : lim x−4
x→4
x→4
0
e
+3−x
ex−4 − 1
2
π sin (2πx)
2π cos (2πx)
= lim x−4
= lim
= 2π 2 .
x−4
x→4 e
x→4
−1
e
–––––––––––––––––––––––––––––––––––––––––
t
1
e
−
93. lim+
t→0
t
t
Solution:
t
t
t
e
e −1
e
0
1
The limit leads to the indeterminate form : lim+
−
= lim+
= lim+
=
t→0
t→0
0 t→0
t
t
t
1
1.
–––––––––––––––––––––––––––––––––––––––––
94. lim+ e−1/y ln y
89. lim
y→0
Solution:
∞
ln y
y −1
The limit leads to the indeterminate form : lim+ e−1/y ln y = lim+ y−1 = lim+
=
−1
y→0 e
y→0 −ey
∞ y→0
(y −2 )
y
− lim+
−1 = 0.
y→0 −ey
–––––––––––––––––––––––––––––––––––––––
(p.550)
√ 1/x
3. lim+ cos x
x→0
Solution:
√ 1/x
1 √ =⇒ ln y = ln cos x and
y = cos x
x √
√
√
√
1 −1/2
x
sec2 x
−1
1
ln (cos x)
− sin ( x)
tan x
1
2
√ =
lim+
= lim+ √
lim+ √
= − lim4
=−
1 −1/2
x→0
x→0 2 x cos x
x
2 x→0
2 x→0
2
x
x
2
√ 1/x
1
=⇒ lim+ cos x
= e−1/2 = √ .
x→0
e
–––––––––––––––––––––––––––––––––––––––
4. lim (x + ex )2/x
x→∞
Solution:
2 ln (x + ex )
2 (1 + ex )
2ex
=⇒ lim ln y = lim
=
lim
=
y = (x + ex )2/x =⇒ ln y =
x→∞
x→∞ x + ex
x→∞ 1 + ex
x
x
2e
lim x = 2
x→∞ e
=⇒ lim (x + ex )2/x = lim ey = e2
x→∞
x→∞
–––––––––––––––––––––––––––––––––––––––
x
11. For what x > 0 does x(x ) = (xx )x ? Give reasons for your answer.
Solution:
x
ln x(x ) = xx ln x and ln (xx )x = x ln xx = x2 ln x; then, xx ln x = x2 ln x =⇒ xx − x2 ln x =
0 =⇒ xx = x2 or ln x = 0.
x
ln x = 0 =⇒ x = 1; xx = x2 =⇒ x ln x = 2 ln x =⇒ x = 2. Therefore, x(x ) = (xx )x when x = 2
or x = 1.
–––––––––––––––––––––––––––––––––––––––
x
t
g(x)
13. Find f (2) if f (x) = e
and g (x) =
dt.
4
2 1+t
Solution:
2
2
x
g(x)
g(x) 0
=
=⇒ f (x) = e g (x), where g (x) =
=⇒ f (2) = e
f (x) = e
4
1+x
1 + 16
17
–––––––––––––––––––––––––––––––––––––––
(p.547)
−1
24. Find the derivative of y = 1 + x2 etan x
Solution:
tan−1 x
e
−1
−1
−1
−1
y = 1 + x2 etan x =⇒ y = 2xetan x + 1 + x2
= 2xetan x + etan x .
2
1+x
–––––––––––––––––––––––––––––––––––––––
√
29. Find the derivative of y = (sin θ) θ .
Solution: √
√
ln
y
=
y = (sin θ) θ =⇒
θ ln (sin θ)
√
1
cos θ
1 dy
= θ
+ θ−1/2 ln (sin θ)
=⇒
y dθ
sin
2
θ
√
√
dy
ln (sin θ)
θ
√
=⇒
= (sin θ)
θ cot θ +
dθ
2 θ
–––––––––––––––––––––––––––––––––––––––
30. Find the derivative of y = (ln x)1/ ln x .
Solution:
1
1/ ln x
=⇒ ln y =
y = (ln x)
ln (ln x)
lnx
1
1
1
−1
1
y
=
+ ln (ln x)
=⇒
2
y
ln x
ln x
x
x
(ln x)
1/ ln x 1 − ln (ln x)
=⇒ y = (ln x)
.
x (ln x)2
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