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C H A P T E R 2Differentiation
Section 2.1 The Derivative and the Tangent Line Problem . . . 53
Section 2.2 Basic Differentiation Rules and Rates of Change . 60
Section 2.3 The Product and Quotient Rules andHigher-Order Derivatives . . . . . . . . . . . . . . 67
Section 2.4 The Chain Rule . . . . . . . . . . . . . . . . . . . 73
Section 2.5 Implicit Differentiation . . . . . . . . . . . . . . . 79
Section 2.6 Related Rates . . . . . . . . . . . . . . . . . . . . 85
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 92
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . 98
-
53
C H A P T E R 2Differentiation
Section 2.1 The Derivative and the Tangent Line ProblemSolutions to Odd-Numbered Exercises
1. (a)
(b) m 3
m 0 3. (a), (b) (c)
x 1
1x 1 2
33
x 1 2
6
5
4
3
2
654321
1
y
x
1)1f ) x
)1 3f ))f )4
11)f )
1)x
))f 44
y
4) , )5
2)
21) ,
f )
)
1
f ) )4 5
)
y f 4 f 1
4 1x 1 f 1)
5. is a line. Slope 2f x 3 2x 7. Slope at
limx0
2 2x 2
limx0
1 2x x2 1
x
limx0
1 x2 4 3
x
1, 3 limx0
g1 x g1
x
9. Slope at
limt0
3 t 3
limt0
3t t2 0
t
0, 0 limt0
f 0 t f 0
t11.
limx0
0 0
limx0
3 3
x
f x limx0
f x x f x
x
f x 3
13.
limx0
5 5
limx0
5x x 5x
x
fx limx0
f x x f x
x
f x 5x 15.
lims0
23
s
s
23
lims0
3
23
s s 3 23ss
hs lims0
hs s hs
s
hs 3 23
s
-
54 Chapter 2 Differentiation
17.
limx0
4xx 2x2 x
x lim
x0 4x 2x 1 4x 1
limx0
2x2 4xx 2x2 x x 1 2x2 x 1
x
limx0
2x x2 x x 1 2x2 x 1
x
fx limx0
f x x f x
x
f x 2x2 x 1
19.
limx0
3x2 3xx x2 12 3x2 12
limx0
3x2x 3xx2 x3 12x
x
limx0
x3 3x2x 3xx2 x3 12x 12x x3 12x
x
limx0
x x3 12x x x3 12x
x
fx limx0
f x x f x
x
f x x3 12x
21.
1
x 12
limx0
1
x x 1x 1
limx0
x
xx x 1x 1
limx0
x 1 x x 1xx x 1x 1
limx0
1x x 1
1
x 1x
fx limx0
f x x f x
x
f x 1x 1
23.
1
x 1 x 1
1
2x 1
limx0
1
x x 1 x 1
limx0
x x 1 x 1
xx x 1 x 1
limx0
x x 1 x 1
x x x 1 x 1x x 1 x 1
fx limx0
f x x f x
x
f x x 1
-
25. (a)
At the slope of the tangent line isThe equation of the tangent line is
y 4x 3.
y 5 4x 8
y 5 4x 2
m 22 4.2, 5,
limx0
2x x 2x
limx0
2xx x2
x
limx0
x x2 1 x2 1
x
fx limx0
f x x f x
x
f x x2 1 17. (b)
(2, 5)
5 5
2
8
27. (a)
At the slope of the tangent is The equation of the tangent line is
y 12x 16.
y 8 12x 2
m 322 12.2, 8,
limx0
3x2 3xx x2 3x2
limx0
3x2x 3xx2 x3
x
limx0
x x3 x3
x
fx limx0
f x x f x
x
f x x3 18. (b)
5 5
4
(2, 8)
10
29. (a)
At the slope of the tangent line is
The equation of the tangent line is
y 12
x 12
.
y 1 12
x 1
m 1
21
12
.
1, 1,
limx0
1
x x x
1
2x
limx0
x x x
xx x x
limx0
x x x
xx x xx x x
fx limx0
f x x f x
x
f x x 18. (b)
5
1
1
3
(1, 1)
Section 2.1 The Derivative and the Tangent Line Problem 55
-
56 Chapter 2 Differentiation
31. (a)
At the slope of the tangent line is
The equation of the tangent line is
y 34
x 2
y 5 34
x 4
m 1 4
16
34
4, 5,
x2 4
x2 1
4x2
limx0
x2 xx4
xx x
limx0
x2x xx2 4x
xxx x
limx0
x3 2x2x xx2 x3 x2x 4x
xxx x
limx0
xx xx x 4x x2x x 4x x
xxx x
limx0
x x 4
x x x 4x
x
fx limx0
f x x f x
x
f x 4x
(b)
12
6
12
10
(4, 5)
33. From Exercise 27 we know that Since theslope of the given line is 3, we have
Therefore, at the points and the tangentlines are parallel to These lines haveequations
and
y 3x 2. y 3x 2
y 1 3x 1y 1 3x 1
3x y 1 0.1, 11, 1
x 1.
3x2 3
fx 3x2. 35. Using the limit definition of derivative,
Since the slope of the given line is , we have
Therefore, at the point the tangent line is parallel toThe equation of this line is
y 12
x 32
.
y 1 12
x 12
y 1 12
x 1
x 2y 6 0.1, 1
x 1.
1
2xx
12
12
fx 12xx
.
37. because the tangent line passes through
g5 2 05 9
2
4
12
5, 2g5 2 39. f x x fx 1 b
-
Section 2.1 The Derivative and the Tangent Line Problem 57
41. matches (a)
decreasing slope as x
f x x fx 43.
Answers will vary.
Sample answer: y x
x1
1
2
3
4
2
1
3
4
2 134 2 3 4
y
45. (a) If and is odd, then
(b) If and is even, then fc f c 3ffc 3
fc fc 3ffc 3
47. Let be a point of tangency on the graph of f. By the limit definition for the derivative, The slope of theline through and equals the derivative of f at
Therefore, the points of tangency are and and the corresponding slopes are 2 and . The equations of the tangentlines are
y 2x 1 y 2x 9
y 5 2x 2 y 5 2x 2
23, 3,1, 3
0 x0 1x0 3 x0 1, 3
0 x02 4x0 3
5 4x0 x02 8 8x0 2x02 5 y0 2 x04 2x0
5 y02 x0
4 2x0
x1 2 3 62
2
4
3
5
7
6
1
(1, 3)(3, 3)
(2, 5)
yx0:x0, y02, 5fx 4 2x.x0, y0
49. (a)
(b)
(c) Because g is decreasing (falling) at
(d) Because g is increasing (rising) at
(e) Because and are both positive, is greater than , and
(f) No, it is not possible. All you can say is that g is decreasing (falling) at x 2.
g6 g4 > 0.g4g6g6g4
x 4.g4 73 ,
x 1.g1 83 ,
g3 0
g0 3
51.
By the limit definition of the derivative we have fx 34 x2.
2
2
2
2
f x 14 x3
x 0 0.5 1 1.5 2
0 2
3 0 3271634
316
316
34
2716fx
2732
14
132
132
14
27322f x
0.511.52
-
58 Chapter 2 Differentiation
53.
The graph of is approximately the graph of fx.gx
3
1
2 4
g
f
2x 0.01 x 0.012 2x x2 100
gx f x 0.01 f x0.01
55.
Exact: f2 0f2 3.99 42.1 2
0.1
f 2.1 2.14 2.1 3.99f 2 24 2 4,
57. and
As is nearly horizontal and thus f 0.x , f
5
5
2 5
f
f
fx 12x32
.f x 1x
59.
(a)
(b) As the line approaches the tangent line to at 2, 3.fx 0,
x 0.1: Sx 1910x 2 3 1910
x 45
x 0.5: Sx 32x 2 3 32
x
5
1
2 7
f
S0.1
S1
S0.5
x 1: Sx x 2 3 x 1
4 2 x 32 3
xx 2 3 1 x 1
2
xx 2 3 x 2x 2 3
Sx x f 2 x f 2
xx 2 f 2
f x 4 x 32
61.
f2 limx2
f x f 2
x 2 lim
x2 x2 1 3
x 2 lim
x 2 x 2x 2
x 2 lim
x2 x 2 4
f x x2 1, c 2
63.
f2 limx2
f x f 2
x 2 lim
x 2 x2x 2
x 2 lim
x2 x3 2x2 1 1
x 2 lim
x2 x2 4
f x x3 2x2 1, c 2
65.
Does not exist.
As
As x 0, x
x
1x
x 0, x
x
1x
g0 limx0
gx g0
x 0 lim
x0 x
x.
gx x, c 0 67.
Does not exist.
limx6
1
x 613
limx6
x 623 0
x 6
f6 limx6
f x f 6
x 6
f x x 623, c 6
-
Section 2.1 The Derivative and the Tangent Line Problem 59
69.
Does not exist.
limx5
x 5x 5
limx5
x 5 0x 5
h5 limx5
hx h5
x 5
h x x 5, c 5
73. is differentiable everywhere except at (Discontinuity)
x 1.f x 75. is differentiable everywhere except at (Sharp turn in the graph)
x 3.f x
77. is differentiable on the interval (At the tangent line is vertical)x 1
1, .f x 79. is differentiable everywhere except at (Discontinuity)
x 0.f x
81.
The derivative from the left is
The derivative from the right is
The one-sided limits are not equal. Therefore, f is not differentiable at x 1.
limx1
f x f 1
x 1 lim
x1 x 1 0
x 1 1.
limx1
f x f 1
x 1 lim
x1 x 1 0
x 1 1.
f x x 1 83.
The derivative from the left is
The derivative from the right is
These one-sided limits are equal. Therefore, f is differentiable at x 1. f1 0
limx1
x 1 0.
limx1
f x f 1
x 1 lim
x1 x 12 0
x 1
limx1
x 12 0.
limx1
f x f 1
x 1 lim
x1 x 13 0
x 1
f x x 13,x 12, x 1x > 1
85. Note that f is continuous at
The derivative from the left is
The derivative from the right is
The one-sided limits are equal. Therefore, f is differentiable at x 2. f2 4
limx2
f x f 2
x 2 lim
x2 4x 3 5
x 2 lim
x2 4 4.
limx2
f x f 2
x 2 lim
x2 x2 1 5
x 2 lim
x2 x 2 4.
f x x2 1,4x 3, x 2x > 2x 2.
71. is differentiable everywhere except at (Sharp turn in the graph.)
x 3.f x
87. (a) The distance from to the line is
.
(b)
The function d is not differentiable at This corresponds to the linewhich passes through the point 3, 1.y x 4,
m 1.
5
1
4 4
m3 11 4m2 1
3m 3m2 1
d Ax1 By1 CA2 B2
x
2
3
1 2 3 4
1
ymx y 4 03, 1
-
60 Chapter 2 Differentiation
Section 2.2 Basic Differentiation Rules and Rates of Change
89. False. the slope is limx0
f 2 x f 2
x.
93.
Using the Squeeze Theorem, we have Thus, and f is continuous atUsing the alternative form of the derivative we have
Since this limit does not exist (it oscillates between and 1), the function is not differentiable at
Using the Squeeze Theorem again we have Thus, and f is continu-ous at Using the alternative form of the derivative again we have
Therefore, g is differentiable at x 0, g0 0.
limx0
f x f 0
x 0 lim
x0 x2 sin1x 0
x 0 lim
x0 x sin
1x
0.
x 0.limx0
x2 sin1x 0 f 0x2 x2 sin1x x2, x 0.
gx x2 sin1x,0, x 0x 0x 0.1
limx0
f x f 0
x 0 lim
x0 x sin1x 0
x 0 lim
x0 sin1x.
x 0.limx0
x sin1x 0 f 0x x sin1x x, x 0.
f x x sin1x,0, x 0x 0
91. False. If the derivative from the left of a point does not equal the derivative from the right of a point, then the derivative doesnot exist at that point. For example, if then the derivative from the left at is and the derivative from theright at is At the derivative does not exist.x 0,1.x 0
1x 0f x x,
1. (a) (b)
y1 32
y 32 x12
y x32
y1 12
y 12 x12
y x12 (c) (d)
y1 3
y 3x2
y x3
y1 2
y 2x
y x2
3.
y 0
y 8 5.
y 6x5
y x6 7.
y 7x8 7x8
y 1x7
x7 9.
y 15
x45 1
5x45
y 5x x15
11.
fx 1
f x x 1 13.
fx 4t 3
f t 2t2 3t 6 15.
gx 2x 12x2gx x2 4x3 17.
st 3t2 2
st t3 2t 4
19.
y
2 cos sin
y
2 sin cos 21.
y 2x 12
sin x
y x2 12
cos x 23.
y 1x2
3 cos x
y 1x
3 sin x
-
31.
f1 6
fx 6x3 6x3
f x 3x2
3x2, 1, 3 33.
f0 0
fx 215
x2
f x 12
75
x3, 0, 12 35.
y0 4
y 8x 4
4x2 4x 1
y 2x 12, 0, 1
37.
f0 41 1 3
f 4 cos 1
f 4 sin , 0, 0 39.
fx 2x 6x3 2x 6x3
f x x2 5 3x2 41.
gt 2t 12t4 2t 12t4
gt t2 4t3
t2 4t3
43.
fx 1 8x3
x3 8
x3
f x x3 3x2 4
x2 x 3 4x2 45.
y 3x2 1
y xx2 1 x3 x
47.
fx 12
x12 2x23 1
2x
2
x23
f x x 6 3x x12 6x13 49.
h(s 45
s45 23
s13 4
5s15
23s13
hs s45 s23
51.
fx 3x12 5 sin x 3x
5 sin x
f x 6x 5 cos x 6x12 5 cos x
53. (a)
At .
Tangent line:
(b) 3
1
2 2(1, 0)
2x y 2 0
y 0 2x 1
y 413 61 21, 0:
y 4x3 6x
y x4 3x2 2
Function Rewrite Derivative Simplify
25. y 5x 3
y 5x3y 52
x2y 5
2x2
27. y 98x4
y 98
x4y 38
x3y 3
2x3
29. y 1
2x32y
12
x32y x12
y xx
55. (a)
At
Tangent line:
(b)
7
1
2
5
(1, 2)
3x 2y 7 0
y 32
x 72
y 2 32
x 1
f1 32
1, 2,
fx 32
x74 3
2x74
f x 24x3 2x34
Section 2.2 Basic Differentiation Rules and Rates of Change 61
-
62 Chapter 2 Differentiation
57.
Horizontal tangents: , 2, 142, 140, 2,
y 0 x 0, 2
4xx 2x 2
4xx2 4
y 4x3 16x
y x4 8x2 2 59.
cannot equal zero.
Therefore, there are no horizontal tangents.
y 2x3 2x3
y 1x2
x2
61.
At .
Horizontal tangent: ,
y x ,
cos x 1 x
y 1 cos x 0
0 x < 2 y x sin x, 63.
Hence, and
For and for x 3, k 10.x 3, k 2
x2 2x 4x 4x 9 x2 9 x 3.k 2x 4
2x k 4 Equate derivatives
x2 kx 4x 9 Equate functions
65.
Hence, and
34
x2
x
34
x 3 34
x 34
x 3 32
x 3 x 2 k 3.k 34
x2
kx2
34
Equate derivatives
kx
34
x 3 Equate functions
67. (a) The slope appears to be steepest between A and B. (c)
(b) The average rate of change between A and B is greater than the instantaneous rate of change at B.
x
f
CA
B
ED
y
69. gx f x 6 gx fx 71.
If f is linear then its derivative is a constant function.
fx a
f x ax b
3
3
1
21123
2
x
f
f
y
-
Section 2.2 Basic Differentiation Rules and Rates of Change 63
73. Let and be the points of tangency on and respectively. The derivatives of these functions are
and
Since and
and
Thus, the tangent line through and is
and
Thus, the tangent line through and is
y 1 3 12 1x 1 y 2x 1.1, 12, 3
y1 1x1 1x2 2 y2 3,
y 0 4 02 1x 1 y 4x 4.2, 41, 0
y1 4x1 2x2 1 y2 0,
x2 1 or 2
2x2 2x2 1 05
4
3
1
1
2
2
2 3x
), 4)2
, )) 01
y 2x22 6x2 4 0
2x22 12x2 14 4x2
2 18x2 18
x22 6x2 5 x22 6x2 9 2x2 62x2 3
x22 6x2 5 x2 32
x2 x2 3 2x2 6
m y2 y1x2 x1
x22 6x2 5 x12
x2 x1 2x2 6.
5
4
3
1
1
2
2 3x
32) ,
1)1,)
)
yy2 x22 6x2 5,y1 x1
2
x1 x2 3
m 2x1 2x2 6
y 2x 6 m 2x2 6.y 2x m 2x1
y x2 6x 5,y x2x2, y2x1, y1
75.
The point is on the graph of f.
Tangent line:
0 x 4y 4
4y 8 x 4
y 2 0 2
4 4x 4
4, 2
x 4, y 2
4 x 2x
4 x 2xx
4 x 2xy
1
2x
0 y4 x
fx 12
x12 1
2x
f x x, 4, 0 77.
1.243.33
0.77
3.64
f1 1
-
79. (a) One possible secant is between and :
(b)
is an approximation of the tangent line
(c) As you move further away from the accuracy of the approximation T gets worse.
(d)
2
2 12T
f
20
4, 8,
Tx.Sx
Tx 3x 4 8 3x 4
fx 32
x12 f4 32
2 3
y Sx 2.981x3.924
y 8 2.981x 4
y 8 8 7.7019
4 3.9x 4
2
2 12
(4, 8)
204, 83.9, 7.7019
0.5 0.1 0 0.1 0.5 1 2 3
1 2.828 5.196 6.548 7.702 8 8.302 9.546 11.180 14.697 18.520
2 5 6.5 7.7 8 8.3 9.5 11 14 171T4 x
f 4 x
123x
81. False. Let and Thenbut f x gx.fx gx 2x,
x2 4.gx f x x2 83. False. If then 2 is a constant.dydx 0.y 2,
85. True. If gx 3f x, then gx 3fx.
87.
Instantaneous rate of change is the constant 2. Average rate of change:
(These are the same because f is a line of slope 2.)
f 2 f 12 1
22 7 21 7
1 2
ft 2
f t 2t 7, 1, 2 89.
Instantaneous rate of change:
Average rate of change:
f 2 f 12 1
12 1
2 1
12
2, 12 f2 14
1, 1 f1 1
fx 1x2
f x 1x, 1, 2
64 Chapter 2 Differentiation
-
Section 2.2 Basic Differentiation Rules and Rates of Change 65
91. (a)
(b)
(c)
When .
When .
(d)
(e)
81362 295.242 ftsec
v13624 321362
4
t2 136216
t 1362
4 9.226 sec
16t2 1362 0
v2 64 ftsect 2:
v1 32 ftsect 1:
vt st 32t
s2 s12 1
1298 1346 48 ftsec
vt 32t
st 16t2 1362 93.
v10 9.810 120 22 msec
v5 9.85 120 71 msec
vt 9.8t 120
4.9t2 120t
st 4.9t2 v0t s0
97.
1 mimin2 min 2 mi
v 60 mph 1mimin
0 mimin2 min 0 mi
v 0 mph 0 mimin
23 mimin6 min 4 mi
Time (in minutes)
Dis
tanc
e (i
n m
iles)
t2 4 6 8 10
2
4
6
8
10
(10, 6)
(8, 4)
(6, 4)
(0, 0)
sv 40 mph 23 mimin95.
(The velocity has been converted to miles per hour)
t
Time (in minutes)2 4 6 8 10
10
20
30
40
50
60
Vel
ocity
(in
mph
)
v
99. (a) Using a graphing utility, you obtain
(c)
(e)
For .
For .
For .T100 1.315v 100,
T80 1.081v 80,
T40 0.612v 40,
dTdv
0.01172v 0.1431
T R B 0.00586v2 0.1431v 0.44
R 0.167v 0.02.
(b) Using a graphing utility, you obtain
(d)
(f) For increasing speeds, the total stopping distance increases.
1000
0
T
B
R
60
B 0.00586v2 0.0239v 0.46.
101.
When
square meters per meter change in s.dAds
8
s 4 m,
dAds
2sA s2, 103.
When dCdQ
$1.93.Q 350,
C351 C350 5083.095 5085 $1.91
dCdQ
1,008,000
Q2 6.3
C 1,008,000
Q 6.3Q
105. (a) is the rate of change of the amount of gasoline sold when the price is $1.47 per gallon.
(b) is usually negative. As prices go up, sales go down.f1.47
f1.47
-
107.
Since the parabola passes through and we have
.
Thus, From the tangent line we know that the derivative is 1 at the point
Therefore, y 2x2 3x 1.
b a 1 3
a 2
1 a 1
1 2a1 a 1
y 2ax a 1
1, 0.y x 1,y ax2 a 1x 1.
0 a12 b1 1 b a 11, 0:
1 a02 b0 c c 10, 1:
1, 0,0, 1
y ax2 bx c
109.
Tangent lines through
The points of tangency are and At the slope is At the slope is
Tangent lines:
9x y 0 9x 4y 27 0
y 9x y 94 x 274
y 0 9x 0 and y 818 94 x 32
y32 94 .32 , 818 y0 9.0, 032 , 818 .0, 0 x 0 or x 32
0 2x3 3x2 x22x 3
x3 9x 9 3x3 3x2 9x 9
y 9 3x2 9x 1
1, 9:
y 3x2 9
y x3 9x
111.
f must be continuous at to be differentiable at
For f to be differentiable at the left derivative must equal the right derivative.
b 8a 4 43
a 13
12a 4
3a22 22
x 2,
fx 3ax2,2x, x < 2x > 2
limx2
f x limx2
x2 b 4 b
limx2
f x limx2
ax3 8a
x 2.x 2
f x ax3,x2 b, x 2x > 2
8a 4 b 8a 4 b
66 Chapter 2 Differentiation
-
Section 2.3 The Product and Quotient Rules and Higher-Order Derivatives
Section 2.3 The Product and Quotient Rules and Higher-Order Derivatives 67
113. Let
0 sin x1 sin x
limx0
cos xcos x 1
x lim
x0 sin xsin xx
limx0
cos x cos x sin x sin x cos x
x
fx limx0
f x x f x
x
f x cos x.
1.
4x3 6x2 2x 2
2x3 2x2 2x 2 2x3 4x2
gx x2 12x 2 x2 2x2x
gx x2 1x2 2x 3.
7t2 4
3t23
2t 43 t2 43t23
ht t132t t 2 413
t 23
ht 3tt2 4 t13t2 4
5.
3x2 cos x x 3 sin x
fx x 3sin x cos x3x2
f x x 3 cos x 7.
fx x2 11 x2x
x2 12 1 x2
x2 12
f x xx2 1
9.
1 8x 3
3x23x 3 12
x3 1 x9x2
3x23x 3 12
hx x 3 11
3x23 x133x2
x 3 12
hx 3x
x3 1
x13
x3 111.
gx x2cos x sin x2x
x22 x cos x 2 sin x
x 3
gx sin xx2
13.
f0 15
10x4 12x3 3x2 18x 15
fx x3 3x4x 3 2x2 3x 53x2 3
f x x3 3x2x2 3x 5 15.
f1 1 6 41 32 14
x2 6x 4
x 32
fx x 32x x2 41
x 32 2x2 6x x2 4
x 32
f x x2 4
x 3
17.
28
4 f4 22
4 22 cos x x sin x fx xsin x cos x1
f x x cos x
-
68 Chapter 2 Differentiation
Function Rewrite Derivative Simplify
19. y x2 2x
3y
13
x2 23
x y 23
x 23
y 2x 2
3
21. y 7
3x3y
73
x3 y 7x4 y 7x4
23. y 4x32
xy 4x, x > 0 y 2x12 y
2x
25.
2
x 12 , x 1
2x2 4x 2
x2 12 2x 12x2 12
fx x2 12 2x 3 2x x22x
x2 12
f x 3 2x x2
x2 1
27.
x2 6x 3
x 32
fx 1 x 34 4x1x 32 x2 6x 9 12
x 32
f x x1 4x 3 x 4x
x 329.
2x 5
2xx
2x 52x32
fx x12 52
x32 x32x 52
f x 2x 5x
2x12 5x12
31.
hs 6s5 12s2 6s2s3 2
hs s3 22 s6 4s3 4
33.
2x2 2x 3
x2 3x2 2x2 2x 3
x2x 32
fx x2 3x2 2x 12x 3
x2 3x2 2x2 6x 4x2 8x 3
x2 3x2
f x 2
1x
x 3
2x 1xx 3
2x 1x2 3x
35.
15x 4 48x 3 33x 2 32x 20
9x 4 36x3 41x 2 16x 20 6x 4 12x 3 8x 2 16x
9x2 4x2 4x 5 3x 4 3x 3 4x 2 4x 3x 4 15x3 4x2 20x
fx 9x2 4x 5x 1 3x3 4x1x 1 3x3 4xx 51
f x 3x3 4xx 5x 1
37.
4xc2
x2 c22
fx x2 c22x x2 c22x
x2 c22
f x x2 c2
x2 c239.
tt cos t 2 sin t
ft t2 cos t 2t sin t
f x t2 sin t
-
Section 2.3 The Product and Quotient Rules and Higher-Order Derivatives 69
41.
ft t sin t cos tt2
t sin t cos t
t2
f t cos tt
43.
fx 1 sec2 x tan2 x
f x x tan x
45.
gt 14
t34 8 sec t tan t 1
4t34 8 sec t tan t
gt 4t 8 sec t t14 8 sec t 47.
32
sec x tan x tan2 x 1
y 32
sec x tan x sec2 x 32
sec xtan x sec x
y 31 sin x
2 cos x
32
sec x tan x
51.
xx sec2 x 2 tan x
fx x2 sec2 x 2x tan x
f x x2 tan x49.
cos x cot2 x
cos xcsc2 x 1
cos xsin2 x
cos x
y csc x cot x cos x
y csc x sin x
53.
4x cos x 2 sin x x2 sin x
y 2x cos x 2 sin x x2sin x 2x cos x
y 2x sin x x2 cos x
57.
(form of answer may vary)g 1 sin cos sin 12
g
1 sin
55.
(form of answer may vary)gx 2x2 8x 1x 22
gx x 1x 22x 5
59.
y6 223
1 22 43
y 1 csc xcsc x cot x 1 csc xcsc x cot x
1 csc x2 2 csc x cot x1 csc x2
y 1 csc x1 csc x
61.
h sec tan 1 2
1
2
sec tt tan t 1
t2
ht tsec t tan t sec t1t2
ht sec tt
-
70 Chapter 2 Differentiation
53. (b)
10
10 10
10
(1, 3)
63. (a)
Tangent line: y 3 1x 1 y x 2
f1 1 slope at 1, 3.
4x3 6x2 6x 5
fx x3 3x 11 x 23x2 3
f x x3 3x 1x 2, 1, 3
65. (a)
.
Tangent line: y 2 1x 2 y x 4
f2 12 12 1 slope at 2, 2
fx x 11 x1x 12 1
x 12
f x xx 1
, 2, 2 51. (b)
3
3 6
6
(2, 2)
55. (b)
4
4
4( (, 1
67. (a)
.
Tangent line:
4x 2y 2 0
y 1 2x
2
y 1 2x 4
f4 2 slope at
4, 1
fx sec2 x
f x tan x, 4, 1
69.
.
Horizontal tangents are at and 2, 4.0, 0
fx 0 when x 0 or x 2
x2 2xx 12
xx 2x 12
fx x 12x x21
x 12
f x x2
x 171.
and differ by a constant.gf
gx 5x 4x 2 3x
x 2 2x 4x 2 f x 2
gx x 25 5x 41x 22 6
x 22
fx x 23 3x1x 22 6
x 22
73.
When
When
When
When
For general n, .fx x n1 x cos x n sin x
n 4: fx x3x cos x 4 sin x.
n 3: fx x2x cos x 3 sin x.
n 2: fx xx cos x 2 sin x.
n 1: fx x cos x sin x.
x n1 x cos x n sin x
fx x n cos x nx n1 sin x
f x x n sin x 75.
6t 1
2t cm2sec
3t12 12
t12
At 232t12 12
t12
Area At 2t 1t 2t32 t12
-
Section 2.3 The Product and Quotient Rules and Higher-Order Derivatives 71
77.
(a) When
(b) When
(c) When
As the order size increases, the cost per itemdecreases.
x 20: dCdx
$3.80.
x 15: dCdx
$10.37.
x 10: dCdx
$38.13.
dCdx
100400x3 30
x 302
C 100200x2 x
x 30, 1 x 79.
P2 31.55 bacteria per hour
2000 50 t2
50 t2 2
500200 4t2
50 t 2 2
Pt 50050 t24 4t2t
50 t22
Pt 5001 4t50 t2
81. (a)
(b)
(c)
ddx
cot x ddx
cos xsin x
sin xsin x cos xcos xsin x2
sin2 x cos2 xsin2 x
1
sin2x csc2 x
cot x cos xsin x
ddx
csc x ddx
1sin x
sin x0 1cos xsin x2
cos xsin x sin x
1
sin x
cos xsin x
csc x cot x
csc x 1
sin x
ddx
sec x ddx
1cos x
cos x0 1sin xcos x2
sin xcos x cos x
1
cos x
sin xcos x
sec x tan x
sec x 1
cos x
83.
f x 3x12 3x
fx 6x12 f x 4x32 85.
f x 2x 13
fx x 11 x1x 12 1
x 12
f x xx 1
87.
f x 3 sin x
fx 3 cos x
f x 3 sin x 89.
f x 2x
fx x2 91.
f 4x 12
2x12 1x
f x 2x
93.
One such function is f x x 22.
f 2 0
x
1
1
2
2
3
3 4
4
y 95.
0
22 4
f2 2g2 h2
fx 2gx hx
f x 2gx hx 97.
10
12 34
12
f2 h2g2 g2h2h22
fx hxgx gxhxhx2
f x gxhx
-
72 Chapter 2 Differentiation
99.
It appears that is cubic; so would be quadratic and would be linear.f
ff
2
2
1
112x
f
y
f
f
101.
The speed of the object is decreasing.
a3 6 msec
v3 27 msec
at 2t
vt 36 t2, 0 t 6
103.
1500
2t 152
at 2t 15100 100t22t 152
vt 100t2t 15
105.
(a)
(b)
Note: (read n factorial.)n! nn 1 . . . 3 2 1
n!
n 1!1! gn1xhx gnxhx
gxhnx n!1!n 1! gxh
n1x n!2!n 2! gxh
n2x . . .
nn 1n 2 . . . 21
n 1n 2 . . . 211 gn1xhx gnxhx
nn 1n 2 . . . 21
321n 3n 4 . . . 21 g xhn3x . . .
nn 1n 2 . . . 21
21n 2n 3 . . . 21 gxhn2xf nx gxhnx nn 1n 2
. . . 211n 1n 2 . . . 21 gxh
n1x
gxh4x 4gxh x 6gxhx 4g xhx g4xhx
g xhx g4xhx
f 4x gxh4x gxh x 3gxh x 3gxhx 3gxhx 3g xhx
gxh x 3gxhx 3gxhx g xhx
f x gxh x gxh x 2gxhx 2gxhx hxg x hxg x
gxhx 2gxhx hxgx
f x gxhx gxhx hxgx hxgx
fx gxhx hxgx
f x gxhx
(a)
(b)
(c) a20 1500220 152 0.5 ftsec2
a10 1500210 152 1.2 ftsec2
a5 150025 152 2.4 ftsec2
-
Section 2.4 The Chain Rule
Section 2.4 The Chain Rule 73
107.
(a)
14x
32
32 x
3 12
P2x 12
f ax a2 f ax a f a
P1x fax a f a 32 x
3 12
f x cos x f 3 cos
3
12
fx sin x f3 sin
3
32
f x cos x f 3 cos
3
12
(b) 2
2
P2 P1
2
f
(c) is a better approximation.P2 (d) The accuracy worsens as you move farther awayfrom x a 3.
109. False. If then
dydx
f xgx gxfx.
y f xgx, 111. True
0
f c0 gc0
hc f cgc gcfc
113. True
115.
does not exist since the left and right derivativesare not equal.f 0
f x 2, 2, if x > 0if x < 0
fx 2x, 2x, if x 0if x < 0 2x
f x xx x2, x2, if x 0if x < 0
y f uu gxy f gx
1. y u4u 6x 5y 6x 54
3. y uu x2 1y x2 1
5. y u3u csc xy csc3 x
7.
y 32x 722 62x 72 y 2x 73 9.
gx 124 9x39 1084 9x3 gx 34 9x4
11.
fx 23
9 x2132x 4x39 x213
f x 9 x223 13.
ft 12
1 t121 121 t
f t 1 t12
-
74 Chapter 2 Differentiation
15.
y 13
9x2 42318x 6x9x2 423
y 9x2 413 17.
x
44 x23
y 2144 x2342xy 24 x214
19.
y 12 x21 1x 22
y x 21 21.
ft 2t 33 2t 33
f t t 32
23.
dydx
12
x 232 12x 232
y x 212 25.
2xx 233x 2
2xx 232x x 2
fx x24x 231 x 242x
f x x2x 24
27.
1 2x2
1 x2
1 x212x2 1 x2
x21 x212 1 x212
y x121 x2122x 1 x2121 y x1 x2 x1 x212 29.
1
x2 132
x2 132x2 x2 1
x2x2 132 x2 112
y x12x2 1322x x2 1121
y x
x2 1 xx2 112
31.
2x 52 10x x2
x2 23
gx 2 x 5x2 2x2 2 x 52x
x2 22
gx x 5x2 22
33.
91 2v2
1 v4
fv 31 2v1 v 2
1 v2 1 2v1 v2
f v 1 2v1 v 3
35.
The zero of corresponds to the point on the graph ofy where the tangent line is horizontal.
2
1 5
y
y
2
y
y 1 3x2 4x32
2xx2 12
y x 1x2 1
37.
The zeros of correspond to the points on the graph ofg where the tangent lines are horizontal.
2
5 3
gg
24
g
gt 3tt2 3t 2
t2 2t 132
gt 3t2
t2 2t 1
-
Section 2.4 The Chain Rule 75
39.
has no zeros.
2
5 4
y
y
4
y
y x 1x2xx 1
y x 1x 41.
The zero of corresponds to the point on the graph ofwhere the tangent line is horizontal.
3
3 6
s
s
3
stst
st t1 t
st 22 t1 t
3
43.
The zeros of correspond to the points on the graph of y where the tangent lines are horizontal.y
x sin x cos x 1
x 2
dydx
x sin x cos x 1
x 2
3
5 5
y
y
3 y cos x 1
x
45. (a)
1 cycle in 0, 2
y0 1
y cos x
y sin x (b)
2 cycles in
The slope of at the origin is a.sin ax
0, 2
y0 2
y 2 cos 2x
y sin 2x
47.
dydx
3 sin 3x
y cos 3x 49.
gx 12 sec2 4x
gx 3 tan 4x
51.
y cos x22 2x 2 2x cos 2x2
y sin x2 sin 2 x 2
53.
Alternate solution:
hx 12
cos 4x4 2 cos 4x
hx 12
sin 4x
2 cos 4x.
2 cos2 2x 2 sin2 2x
hx sin 2x2 sin 2x cos 2x2 cos 2x
hx sin 2x cos 2x 55.
sin2 x 2 cos2 x
sin3 x
1 cos2 xsin3 x
fx sin2 xsin x cos x2 sin x cos x
sin4 x
f x cot xsin x
cos xsin2 x
-
76 Chapter 2 Differentiation
57.
y 8 sec x sec x tan x 8 sec2 x tan x
y 4 sec2 x 59.
sin 2 cos 2 12 sin 4
f 214sin 2cos 22 f 14 sin2 2
14 sin 22
61.
6 sec2 t 1 tan t 1 6 sin t 1cos3 t 1
ft 6 sec t 1 sec t 1 tan t 1
f x 3 sec2 t 1 63.
1
2x 2x cos2x2
dydx
12
x12 14
cos4x28x
x 14
sin4x2
y x 14
sin2x 2
65.
sin x coscos x
dydx
coscos x sin x
y sincos x 67.
s2 34
t 1
t 2 2t 8
st 12
t 2 2t 8122t 2
st t 2 2t 812, 2, 4
69.
f1 925
fx 3x3 423x2 9x2
x3 42
f x 3x3 4
3x3 41, 1, 35 71.
f0 5
ft t 13 3t 21t 12 5
t 12
f t 3t 2t 1
, 0, 2
73.
y0 0 6 sec32x tan2x
y 3 sec22x2 sec(2x tan2x
y 37 sec32x, 0, 36
75. (a)
Tangent line:
(b)
3
5 5
7
(3, 5)
y 5 95
x 3 9x 5y 2 0
f3 95
3x
3x2 2
fx 12
3x2 2126x
f x 3x2 2, 3, 5 77. (a)
Tangent line:
63. (b)
2
0 2
2
( , 0)
y 2x 2x y 2 0
f 2
fx 2 cos 2x
f x sin 2x, , 0
-
Section 2.4 The Chain Rule 77
79.
125x2 1x2 1f x 60x 4 72x2 12
12x5 24x3 12x
12xx 4 2x2 1
fx 6x2 122x
f x 2x2 13 81.
2cos x2 2x2 sin x2
f x 2x2xsin x2 2 cos x2 fx 2x cos x 2
f x sin x 2
83.
The zeros of correspond to the points where the graphof f has horizontal tangents.
f
3
3
2
1
22
2
3
x
f
y
f
85.
The zeros of correspond to the points where the graphof f has horizontal tangents.
f
3
2
213x
f
y
f
87.
gx f3x3 gx 3 f3x
gx f 3x
89. (a)
f5 32 63 24
fx gxhx gxhx
f x gxhx 73. (b)
Need to find f5.g3
f5 g32 2g3
fx ghxhx
f x ghx
73. (c)
f5 36 3232 129
43
fx hxgx gxhxhx2
f x gxhx 73. (d)
f5 3326 162
f x 3gx 2gx
f x gx3
91. (a)
When , f 1.461.v 30
132,400
331 v 2
f 1132,400331 v21
f 132,400331 v1
93.
The maximum angular displacement is (since
When radians persecond.
ddt 1.6 sin 24 1.4489t 3,
d
dt
0.28 sin 8t 1.6 sin 8t
1 cos 8t 1.
0.2
0.2 cos 8t 95.
Since r is constant, we have and
4.224 102 0.04224.
dSdt
1.76 10521.2 102105
drdt 0
dSdt
C2R dRdt 2r drdt
S CR2 r 2
(b)
When .f 1.016v 30,
132,400331 v 2
f 1132,400331 v21
f 132,400331 v1
-
78 Chapter 2 Differentiation
97. (a)
(b)
The function is quadratic, not linear. The cost function levels off at the end of the day, perhaps due to fatigue.dCdt
294.696t2 2317.44t 30.492
dCdt
604.9116t2 38.624t 0.5082
601.6372t3 19.3120t2 0.5082t 0.6161 1350
C 60x 1350
x 1.6372t3 19.3120t2 0.5082t 0.6161
99.
(a)
(b)
(c)
f 2k1x 1k1 2k1 cos x
f 2kx 1k 2k sin x
f x 2 f x 2 sin x 2sin x 0
f 4 4 sin x
f x 3 cos x
f x 2 sin x
fx cos x
f x sin x 101. (a)
Note that and
Also, Thus,
(b)
Note that and
Thus, s4 12
54
58
.
f4 54
.
f 4 52
, g52 6 46 2
12
s4 gf 4f 4
sx gf xf x
r1 0g1 0.
f4 5 06 2
54
g1 4
r1 fg1g1
rx fgxgx
103.
ag
x x n2xx n
2x n2xx n
2x n
2x2 nx
dgdx
12
x2 nx122x n
x2 nx
g xx n 105.
gx 2 2x 32x 3, x 32
gx 2x 3
107.
hx x sin x xx cos x, x 0 hx xcos x
-
Section 2.5 Implicit Differentiation 79
111. False. If then y 12 1 x121.y 1 x12, 113. True
109. (a)
(b)
(c) is a better approximation than
(d) The accuracy worsens as you move away from x c 1.
P1P2
P1
30
0
2
f
P2
P2x 12
4x 12 f1x 1 f 1
8x 12
2x 1 1
P1x f1x 1 f 1
2x 1 1.
f 1 8
21 4
f x 2
sec2 x4
tan x4
4
f1 4
2 2
f x 4
sec2 x4
f 1 1f x tan x4
Section 2.5 Implicit Differentiation
1.
y xy
2x 2yy 0
x2 y2 36 3.
y x12
y12 yx
12
x12 12
y12y 0
x12 y12 9
5.
y y 3x2
2y x
2y xy y 3x23x2 xy y 2yy 0
x3 xy y2 4 7.
y 1 3x2y3
3x3y2 1
3x3y2 1y 1 3x2y33x3y2y 3x2y3 y 1 0
x3y3 y x 0
9.
y1 6xy 3x2 2y2
4xy 3x2
4xy 3x2y 6xy 3x2 2y23x2 3x2y 6xy 4xyy 2y2 0
x3 3x2 2xy2 12 11.
y cos x
4 sin 2y
cos x 4sin 2yy 0
sin x 2cos 2y 1
-
80 Chapter 2 Differentiation
13.
y cos x tan y 1
x sec2 y
cos x xsec2 yy 1 tan y1
sin x x1 tan y 15.
y y cosxy
1 x cosxy
y x cosxyy y cosxy
y xy y cosxy
y sinxy
17. (a)
y 16 x2
y2 16 x2x2 y2 16 (b)
x
y x= 16 2
y x= 16 2
2 626
2
6
6
y
(c) Explicitly:
x
16 x2
x
16 x2
xy
dydx
12
16 x2122x
(d) Implicitly:
y xy
2x 2yy 0
19. (a)
y 3416 x2
y2 1
16144 9x2 9
1616 x2
16y2 144 9x2 (b)
x6 2 2 6
6
4
2
4
6y x= 16 2
y x= 16 2
3
3
4
4
y
(c) Explicitly:
3x
416 x2
3x443y
9x16y
dydx
38
16 x2122x
(d) Implicitly:
y 9x16y
18x 32yy 0
21.
At 4, 1: y 14
y yx
xy y
xy y1 0
xy 4 23.
At is undefined.y2, 0,
y 8x
yx2 42
2yy 16x
x2 42
2yy x2 42x x2 42x
x2 42
y2 x2 4x2 4
-
Section 2.5 Implicit Differentiation 81
25.
At .8, 1: y 12
y x13
y13 3yx
23
x13 23
y13y 0
x23 y23 5 27.
At .0, 0: y 0
x2
x2 1
tan2x y
tan2x y 1 sin2x y
y 1 sec2x y
sec2x y
1 y sec2x y 1
tanx y x
29.
At
Or, you could just solve for y: y 8
x2 4
2, 1: y 3264
12
16x
x2 42
2x8x2 4
x2 4
y 2xyx2 4
x2 4y y2x 0
x2 4y 8 31.
At .1, 1: y 0
y 2xy x3 xy2
x2y y3 x2
4yx2y y3 x2 42xy x3 xy2
4x2yy 4y3y 4x2y 8xy 4x3 4xy2
4x3 4x2yy 4xy2 4y3y 4x2y 8xy
2x2 y22x 2yy 4x2y y8x
x2 y22 4x2y
33.
y 1
1 x2
sec2 y 1 tan2 y 1 x2
y 1
sec2 y cos2 y,
2 < y <
2
ysec2 y 1
tan y x 35.
y y1 xy
y2
y xxyy2
y2 x2
y3
36y3
y xy
2x 2yy 0
x2 y2 36
37.
y y2 x2
y3
16y3
y2 y3y x2
1 yy xy2
0
1 yy y2 0
x yy 0
y xy
2x 2yy 0
x2 y2 16 39.
3y4x2
3x4y
2x3 3y2x 6y
4x2
y 2x3y 3y2
4x2
y 3x2
2y
3x2
2y
xyxy
3y2x
x3
y2
3y2x
2yy 3x2
y2 x3
-
82 Chapter 2 Differentiation82 Chapter 2 Differentiation
41.
At
x 3y 12 0
y 13
x 4
Tangent line: y 1 13
x 9
9, 1, y 13
y yx
12
x12 12
y12y 0
14
1
1
9
(9, 1)
x y 4
43.
At
Tangent line:
Normal line: .
At
Tangent line:
Normal line: .y 4 43
x 3 4x 3y 0
y 4 34
x 3 3x 4y 25 0
6
9 9
( 3, 4)
63, 4:
y 3 34
x 4 3x 4y 0
y 3 43
x 4 4x 3y 25 0
6
9 9
(4, 3)
64, 3:
y xy
x2 y2 25
45.
Let be a point on the circle. If then the tangent line is horizontal, the normal line is vertical and, hence, passesthrough the origin. If then the equation of the normal line is
which passes through the origin.
y y0x0
x
y y0 y0x0
x x0
x0 0,x0 0,x0, y0
yx
slope of normal line
y xy
slope of tangent line
2x 2yy 0
x2 y2 r 2
-
Section 2.5 Implicit Differentiation 83
47.
Horizontal tangents occur when
Horizontal tangents: .
Vertical tangents occur when
Vertical tangents: .0, 5, 8, 5
25xx 8 0 x 0, 8
25x2 400 200x 800 400 0
y 5:
4, 0, 4, 10 yy 10 0 y 0, 10
2516 16y2 2004 160y 400 0
x 4:
y 200 50x160 32y
50x 32yy 200 160y 0
x2 4 8 610
2
10
6
4(0, 5)
( 4, 10)
( 4, 0)
( 8, 5)
y25x2 16y2 200x 160y 400 0
49. Find the points of intersection by letting in the equation
and
The curves intersect at
Ellipse: Parabola:
At the slopes are:
.
At the slopes are:
.
Tangents are perpendicular.
y 1y 1
1, 2,
y 1y 1
1, 2,
y 2y
y 2xy
2yy 44x 2yy 06 6
4
y x= 42
2 + = 6x y22
4
(1, 2)
(1, 2)
1, 2.
x 3x 1 02x2 4x 6
2x2 y2 6.y2 4x
51. and
Point of intersection:
At the slopes are:
.
Tangents are perpendicular.
y 1y 1
0, 0,
y sec y
1 y cos yy 1
x sin y:y x:
0, 06 6
4 x y+ = 0
x y= sin
(0, 0)
4x sin yy x
53.
At any point of intersection the product of theslopes is The curves are orthogonal.yxxy 1.
x, y
y yx y x
y
xy y 0 2x 2yy 0
2
3 3
C = 4
K = 2
2
2
3 3C = 1
K = 1
2 xy C x2 y2 K
-
84 Chapter 2 Differentiation84 Chapter 2 Differentiation
55.
(a)
y 12x3
4y
3x3
y
4yy 12x3
4yy 12x3 0
2y2 3x4 0
(b)
ydydt
3x3dxdt
4ydydt
12x3dxdt
0
57.
(a)
y 3 cos x
sin y
sin yy 3 cos x 0
cos y 3 sin x 1
(b)
sinydydt
3 cosxdxdt
sinydydt
3 cos xdxdt
0
59. A function is in explicit form if is written as a functionof For example, An implicit equationis not in the form For example, x2 y2 5.y f x.
y x3.x: y f x.y
61. (a)
y 4x2 14x4 y2 4x2
14
x4
4y2 16x2 x41010
10
10 x4 44x2 y2
(b)
Note that
Hence, there are four values of x:
To find the slope, .
For and the line is
For and the line is
For and the line is
For and the line is
CONTINUED
y4 137 7x 1 7 3 137 7x 87 23.
x 1 7, y 13 7 7,1010
10
y4y2y3
y1
10y3 137 7x 1 7 3 137 7x 23 87 .
x 1 7, y 13 7 7,y2
137 7x 1 7 3 137 7x 23 87.
x 1 7, y 13 7 7,y1
137 7x 1 7 3 137 7x 87 23.
x 1 7, y 13 7 7,
2yy 8x x3 y x8 x2
23
17, 17, 17, 1 7
x2 8 28 8 27 1 72.
x2 16 256 144
2 8 28
x4 16x2 36 0
36 16x2 x4
y 3 9 4x 2 14
x4
-
Section 2.6 Related Rates
Section 2.6 Related Rates 85
1.
(a) When and
(b) When and
dxdt
225 2 20.
dydt 2,x 25
dydt
1
243 3
4.
dxdt 3,x 4
dxdt
2xdydt
dydt
12xdxdt
y x 3.
(a) When and
(b) When and
dxdt
14
6 32
.
dydt 6,x 1, y 4,
dydt
12
810 5
8.
dxdt 10,x 8, y 12,
dxdt
xydydt
dydt
yxdxdt
xdydt
ydxdt
0
xy 4
61. CONTINUED
(c) Equating and
If then and the lines intersect at 877 , 5.y 5x 877 , x
877
167 14x
7x 7 7 7x 7 77 7x 7 7 7x 7 77
7 7x 1 7 7 7x 1 7
13
7 7x 1 7 3 13
7 7x 1 7 3
y4,y3
63. Let where p and q are nonzero integers and First consider the case where The derivative ofis given by
where Observe that
As the denominator approaches That is,
Now consider From the Chain Rule,
1q
xp1q1pxp1 pq
xpqpp1 pq
xpq1nxn1 n pq.fx 1q
xp1q1 ddx
xp
f x xpq xp1q.
ddx
x1q 1qx11q
1q
x1q1.
qx11q.t x,
1
t11q t12qx1q . . . t1qx12q x11q .
t1q x1q
t1q x1qt11q t12qx1q . . . t1qx12q x11q
f t f xt x
t1q x1q
t x
t1q x1q
t1qq x1qq
t x x.
ddx
x1q limx0
f x x f x
x lim
tx
f t f xt x
f x x1qp 1.q > 0.f x xn xpq,
-
86 Chapter 2 Differentiation
5.
(a) When
(b) When
(c) When
dydt
212 4 cmsec.
x 1,
dydt
202 0 cmsec.
x 0,
dydt
212 4 cmsec.
x 1,
dydt
2x dxdt
dxdt
2
y x2 1 7.
(a) When
(b) When
(c) When
dydt
122 2 cmsec.
x 0,
dydt
222 4 cmsec.
x 4,
dydt
222 8 cmsec.
x 3,
dydt
sec2 x dxdt
dxdt
2
y tan x
9. (a)
(b)dy
dt positive
dx
dt negative
dx
dt negative
dy
dt positive 11. Yes, changes at a constant rate:
No, the rate is a multiple of dxdt
.dydt
dydt
a dxdt
.y
13.
dDdt
12
x4 3x2 1124x3 6xdxdt
2x3 3x
x4 3x2 1
dxdt
4x3 6x
x4 3x2 1
dxdt
2
D x2 y2 x2 x2 12 x4 3x2 1
15.
(a) When
(b) When
dAdt
2 243 144 cm2min.
r 24,
dAdt
2 63 36 cm2min.
r 6,
dAdt
2r drdt
drdt
3
A r 2 17. (a)
(b) where
When
When
(c) If is constant, is proportional to cos.dAdtddt
3,
dAdt
s2
2 12
12
s2
8
6,
dAdt
s2
2 32 12 3s2
8
ddt
12
radmin.dAdt
s2
2 cos
ddt
b
s s
h
s2
2 2 sin
2 cos
2 s2
2 sin
A 12
bh 12 2s sin
2s cos
2
cos
2
hs h s cos
2
sin
2
12bs
b 2s sin 2
-
Section 2.6 Related Rates 87
19.
(a) When
(b) When r 60, drdt
1
4 60 2 800 1
18 cmmin.
r 30, drdt
1
4 302 800 2
9 cmmin.
drdt
1
4r2 dVdt
14r 2
800
dVdt
4r 2 drdt
V 43
r 3, dVdt
800 21.
(a) When
(b) When
dsdt
12103 360 cm2sec.
x 10,
dsdt
1213 36 cm2sec.
x 1,
dsdt
12x dxdt
dxdt
3
s 6x2
23.
When h 15, dhdt
410
9 152 8
405 ftmin.
dVdt
94
h2 dhdt
dhdt
4dVdt
9h2
dVdt
10
34
h3h
r
V 13
r 2h 13
94h2 h since 2r 3h
25.
(a) Total volume of pool
Volume of 1m. of water
(see similar triangle diagram)
% pool filled
(b) Since for you have
dVdt
36h dhdt
14
dhdt
1
144h
11441
1144
mmin.
V 12
bh6 3bh 36hh 18h2
0 h 2, b 6h,
18144 100% 12.5%
2
h = 1
12
b = 6
12
166 18 m3
12
2126 1612 144 m3
1
1
3
6
12
-
88 Chapter 2 Differentiation
27.
(a) When (b)
When
When From part (a) we have
and
Thus,
(c)
Using and , we have ddt
24252 124 2
7242
712
112
radsec.cos 2425
x 7, y 24, dxdt
2, dydt
7
12
ddt
cos2 1y dxdt
xy2
dydt
sec2 ddt
1y
dxdt
xy2
dydt 25y
x
tan xy
52724
21.96 ft2sec.
dAdt
127
712 242
dydt
712
.
x 7, y 24, dxdt
2,x 24, y 7, dydt
224
7
487
ftsec.
dAdt
12x
dydt
y dxdtx 15, y 400 20,
dydt
215
20
32
ftsec.
A 12
xyx 7, y 576 24, dydt
27
24
712
ftsec.
dydt
xy
dxdt
2x
y since
dxdt
2.
2x dxdt
2y dydt
025
y
x
x2 y2 252
29. When and
Also,
.
Thus,
(horizontal)
(vertical).dydt
xy
dxdt
63
6
315
15
msec
dxdtx x
12xy s
dsdt
dxdt
sy
12x
dsdt
126
12630.2 1
53
315
msec
x dxdt
y 12xy dxdt s
dsdt
2x dxdt
2y dydt
0 dydt
xy
dxdt
x2 y2 122
x dxdt
y 12dydt
s dsdt
2x dxdt
212 y1dydt
2s dsdt
x2 12 y2 s2
108 36 12.
s x2 12 y2
12
( , )x y
s
x
12 y
y
y 6, x 122 62 63,
-
Section 2.6 Related Rates 89
31. (a)
When and and
(b) t 250750
13
hr 20 min
dsdt
150450 200600
250 750 mph.
y 200, s 250x 150
dsdt
xdxdt ydydt
s
2s dsdt
2x dxdt
2y dydt
dydt
600 s
100 200
200
100
x
y
Distance (in miles)
Dista
nce (in
mile
s) dxdt
450
s2 x2 y2
33.
When
dsdt
30
301028 28
10 8.85 ftsec.
s 902 302 3010
x 30,
2s dsdt
2x dxdt
dsdt
xs
dxdt
dxdt
28
x 30
s
x
90 ft
30 ft
Home
1st
2nd
3rd
s2 902 x2
35. (a)
(b)d y x
dt
dydt
dxdt
253
5 103
ftsec
dydt
53
dxdt
53
5 253
ftsec
dxdt
5
y 53
x15
6
yx
156
y
y x 15y 15x 6y
-
90 Chapter 2 Differentiation
37.
(a) Period:
(b) When
Lowest point:
(c) When and
.
Thus,
.
Speed 5120 5120 msec
15 112
32
24
15
5
120
dydt
14
154
12 cos6
2x dxdt
2y dydt
0 dydt
xy
dxdt
x2 y2 1
dxdt
12
6 cos t6
12 cos
t6
t 1x 14
, y 1 142
15
4
0, 32 x
12
, y 12 122
32
m.
26
12 seconds
xt 12
sin t6
, x 2 y 2 139. Since the evaporation rate is proportional to the surface
area, However, since we have
Therefore,
k4r 2 4r 2 drdt
k drdt
.
dVdt
4r 2 drdt
.
V 43r 3,dVdt k4r2.
41.
1.3p dVdt
V dpdt
V 0.31.3p dVdt V dpdt 0
1.3 pV 0.3 dVdt
V1.3 dpdt
0
pV1.3 k
43.
When and Thus,
ddt
1
30 123
120
radsec.
cos 22.y 30, 4
ddt
1
30 cos2
dydt
sec2 ddt
1
30 dydt
dydt
3 msec.
30
y
x
y tan y
30
-
Section 2.6 Related Rates 91
45.
(a) When
(b) When
(c) When 75, ddt
120 sin2 75 111.96 radhr 1.87 radmin.
60, ddt
12034 90 radhr 32
radmin.
30, ddt
1204
30 radhr 12
radmin.
52
L215
dxdt
sin2 15600 120 sin2
ddt
cos2 5x2dxdt
x2
L25x2
dxdt
sec2 ddt
5x2
dxdt
dxdt
600 mihrx
y = 5L
tan yx, y 5
47.
(a)
(b)
(c) is greatest when
is least when
(d) For
For 60, dxdt
600 sin60 600 32
3003 cmsec
30, dxdt
600 sin30 600 12
300 cmsec.
or n 180. ndxdtor 90 n 180 2 nsin 1 dxdt 600 sin
2000
0
2000
4
dxdt
30 sin ddt
30 sin 20 600 sin
sin ddt
1
30 dxdt
cos x
30x
30
x
P
ddt
10 revsec2 radrev 20 radsec
49.
ddt
1
25 cos2 ,
4
4
2 50 sec2 ddt
dxdt
50 sec2 ddt
tan x
50 x 50 tan
-
Review Exercises for Chapter 2
51. acceleration of the top of the ladder
When and (see Exercise 27). Since is constant,
d 2ydt 2
1
2470 22 7
122 1244
49144
124
625144 0.1808 ftsec2
d 2xdt 2
0.dxdt
dxdt
2x 7, y 24, dydt
7
12,
d 2ydt 2
1yxd 2xdt 2
dxdt2
dydt2
Second derivative: x d 2xdt 2
dxdt
dxdt
yd 2ydt 2
dydt
dydt
0
x dxdt
y dydt
0
First derivative: 2x dxdt
2y dydt
0
d 2ydt 2
x2 y2 25;
53. (a) Using a graphing utility, you obtain
(b)
(c) then and
Thus,dmdt
1.76215.5 29.101.2 2.15 million.
dsdt
1.2.s 15.5If t s 1995,
dmdt
dmds
dsdt
1.762s 29.10dsdt
ms 0.881s2 29.10s 206.2
1.
limx0
2xx x2 2x
x lim
x0 2x x 2 2x 2
limx0
x2 2xx x2 2x 2x 3 x2 2x 3
x
limx0
x x2 2x x 3 x2 2x 3
x
fx limx0
f x x f x
x
f x x2 2x 3
3.
limx0
1
x x x
1
2x
limx0
x x x
xx x x
limx0
x x x
xx x xx x x
limx0
x x 1 x 1
x
fx limx0
f x x f x
x
f x x 1 5. f is differentiable for all x 1.
92 Chapter 2 Differentiation
-
Review Exercises for Chapter 2 93
7.
(a) Continuous at
(b) Not differentiable at because of the sharp turnin the graph.
x1
23
2
3
45
6
7
1 2 3 4 5 6
y
x 2
x 2.
f x 4 x 2 9. Using the limit definition, you obtain
At x 1, g1 43
16
32
gx 43
x 16
.
11. (a) Using the limit defintion,
At The tangent line is
(b) 2
4
40
(1, 2)
y 3x 1
y 2 3x 1
x 1, f1 3.
fx 3x2. 13.
limx2
x2 x 2 8
limx2
x 2x2 x 2
x 2
limx2
x3 x2 4
x 2
limx2
x2x 1 4
x 2
g2 limx2
gx g2
x 2
15.
x
1
1
2
1
ff
y 17.
y 0
y 25 19.
fx 8x7f x x8
21.
ht 12t 3ht 3t 4
25.
hx 3x12 x23 3x
13x2
hx 6x 3 3x 6x12 3x13
23.
fx 3x2 6x 3xx 2
f x x3 3x2
27.
gx 43
t3 43t3
gt 23
t2
29.
f 2 3 cos
f 2 3 sin 31.
f 3 sin cos 4
f 3 cos sin 4
-
94 Chapter 2 Differentiation
33.
(a) When vibrations/sec/lb.
(b) When vibrations/sec/lb.F9 3313T 9,
F4 50T 4,
Ft 100T
F 200T 35.
The building is approximately 1354 feet high (or 415 m).
s0 1354.24
s9.2 169.22 s0 0
st 16t2 s0
37. (a)
Total horizontal distance: 50
(b)
implies x 50.0 x1 x50
0 x 0.02x2
15
10
5
x604020
y (c) Ball reaches maximum height when
(d)
(e) y25 0
y50 1
y30 0.2
y25 0
y10 0.6
y0 1
y 1 0.04x
y x 0.02x2
x 25.
39.
(a)
(c) for
x 32 232 1 1212 14t 32 .vt 0
at vt 2
vt xt 2t 3
xt t2 3t 2 t 2t 1(b) for
(d) for
The speed is 1 when the position is 0.
v2 22 3 1
v1 21 3 1
t 1, 2.xt 0
t < 32 .vt < 0
41.
26x3 9x2 16x 7
fx 3x2 72x 2 x2 2x 36x
f x 3x2 7x2 2x 3 43.
hx 12x
sin x x cos x
hx x sin x x12 sin x
45.
2x3 1
x3
fx 2 2x3 21 1x3f x 2x x2 47.
x2 1x2 12
fx x2 12x 1 x2 x 12x
x2 12
f x x2 x 1x2 1
49.
fx 4 3x226x 6x4 3x22
f x 4 3x21
53.
y 3x2 sec x tan x 6x sec x
y 3x2 sec x
51.
y cos x 2x x2sin x
cos2 x
2x cos x x2 sin xcos2 x
y x2
cos x
55.
y x sec2 x tan x
y x tan x
-
Review Exercises for Chapter 2 95
57.
y x sin x cos x cos x x sin x
y x cos x sin x
61.
f 6 sec sec tan 6 sec2 tan
f 3 sec2
f 3 tan
59.
g t 6t
gt 3t2 3
gt t3 3t 2
63.
0
y y 2 sin x 3 cos x 2 sin x 3 cos x
y 2 sin x 3 cos x
y 2 cos x 3 sin x
y 2 sin x 3 cos x
65.
3x2
21 x3
fx 12
1 x3123x2
f x 1 x312 67.
2x 3x2 6x 1
x2 13
hx 2 x 3x2 1x2 11 x 32x
x2 12
hx x 3x2 12
69.
ss2 1328s3 3s 25
ss2 1323ss2 1 5s3 5
fs s2 1523s2 s3 552s2 1322sf s s2 152s3 5 71.
y 9 sin3x 1
y 3 cos3x 1
73.
csc 2x cot 2x
y 12
csc 2x cot 2x2
y 12
csc 2x 75.
12
1 cos 2x sin2 x
y 12
14
cos 2x2
y x2
sin 2x
4
77.
cos3 xsin x
cos xsin x1 sin2 x
y sin12 x cos x sin52 x cos x
y 23
sin32 x 27
sin72x 79.
y x 2 cos x sin x
x 22
y sin xx 2
81.
The zeros of correspond to the points on the graph of fwhere the tangent line is horizontal.
0.1
0.1 1.3
f
f
0.1
f
ft tt 147t 2
f t t2t 15 83.
does not equal zero for any value of x in the domain.The graph of g has no horizontal tangent lines.
2
2 7
g
g
4
g
gx x 2x 132
gx 2xx 112
-
96 Chapter 2 Differentiation
85.
does not equal zero for any x in the domain. Thegraph of f has no horizontal tangent lines.
1
2 7
f
5
f
f
ft 56t 116
f t t 112t 113 t 156 87.
does not equal zero for any x in the domain. The graphhas no horizontal tangent lines.
4
20 2
y
y
5
y
y sec21 x
21 x
y tan1 x
89.
y 4 4 sin 2x
y 4x 2 cos 2x
y 2x2 sin 2x 91.
2 csc2 x cot x
f 2 csc xcsc x cot x
fx csc2 x
f x cot x 93.
f t 2t 21 t4
ft t 11 t3
f t t1 t2
95.
g 18 sec2 3 tan 3 sin 1
g 3 sec2 3 cos 1
g tan 3 sin 1
97.
(a) When
.
(c) When
.T 14005 2
25 30 102 3.240 deghr
t 5,
T 14001 21 4 102 18.667 deghr
t 1,
T 1400t 2
t2 4t 102
T 700t2 4t 101
99.
y 2x 3y
3x y2
3x y2y 2x 3y
2x 3xy 3y 3y2y 0
x2 3xy y3 10
(b) When
(d) When
T 140010 2
100 40 102 0.747 deghr.
t 10,
T 14003 29 12 102 7.284 deghr.
t 3,
101.
y 2xy y
2x
2y
2xy x
2yx yy
2xy xx
2xy x
2yy
2xy y
2x
x x2yy y y
2x
y12x12 x12y x12
y12y y12 0 yx xy 16
-
Review Exercises for Chapter 2 97
105.
At
Tangent line:
Normal line:
2x y 0
y 4 2x 2
x 2y 10 0
y 4 12
x 2
y 12
2, 4:
y xy
2x 2yy 0
6
9 9
6
(2, 4)
x2 y2 20103.
y y sin x sin ycos x x cos y
yx cos y cos x y sin x sin y
x cos yy sin y y sin x y cos x
x sin y y cos x
107.
(a) When units/sec.
(b) When units/sec.
(c) When units/sec.dxdt
8x 4,
dxdt
4x 1,
dxdt
22x 12
,
dydt
1
2x dxdt
dxdt
2xdydt
4x
dydt
2 unitssec
y x
109.
Width of water at depth h:
When dhdt
2
25 mmin.h 1,
dhdt
2dVdt54 h
2
h
s2
212
12
dVdt
52
4 hdhdt
V 522
4 h2 h
54
8 hh
w 2 2s 2 214h 4 h
2
dVdt
1
s 14
h
sh
122
111.
38.34 msec
dxdt
3 dsdt
39.8 54.9
xt 3st
tan 30 1
3
stxt
t 5
4.9
4.9t2 25
s 35 60 4.9t2
st 9.8t
x t( )
s t( )
30
st 60 4.9t2