446
CHAPTER 4
APPLICATIONS OF THE DERIVATIVE
Then f (x) = (x 2 2x 2)ex = 0 gives x = 1 + 3 as a candidate for an extrema.
Moreover, f (x) = (x 2 4x)ex = 0 gives x = 4 as a candidate for a point of inflection.
3
1+
3,
x
0, 1 +
f
+
0
f
M
1+
3
x
(0, 4)
S E C T I O N 3.8
Implicit Differentiation
5. (x 2 + y 2 )3/2
solution Assuming that y depends on x, then
3/2 3
1/2
d 2
=
2x + 2yy = 3 x + yy x 2 + y 2 .
x + y2
x2 + y2
dx
2
6. tan(xy)
solution Assuming that y depends on x, then
d
(tan (xy) = xy + y s
316
CHAPTER 3
DIFFERENTIATION
46. Show that (tan1 x) = cos2 (tan1 x) and then use Figure 9 to prove that (tan1 x) = (x 2 + 1)1 .
1 + x2
x
y
1
FIGURE 9 Right triangle with y = tan1 x.
solution
Let y = tan1 x. Then x = tan y and
1 = sec2 y
From Figure 9, co
Chapter Review Exercises
551
so the critical points are
3
+ n
4
3
for all integers n. Because g ( ) 0 for all , it follows that g
+ n is neither a local maximum nor a local
4
minimum for all integers n.
=
30. h( ) = 2 cos 2 + cos 4
solution
Let h( ) = 2 c
S E C T I O N 4.6
Interval
signs of f and f
(, 35 )
+
( 35 , 1)
(1, 95 )
( 95 , )
+
Graph Sketching and Asymptotes
481
+
+
The graph has an inflection point at x = 35 , a local maximum at x = 1 (at which the graph has a cusp), and a local
minimum at x =
226
CHAPTER 3
DIFFERENTIATION
49. Compute the derivatives, where c is a constant.
d 3
d
(a)
(b)
ct
(5z + 4cz2 )
dt
dz
(c)
d
(9c2 y 3 24c)
dy
solution
d 3
(a)
ct = 3ct 2 .
dt
d
(b)
(5z + 4cz2 ) = 5 + 8cz.
dz
d
(c)
(9c2 y 3 24c) = 27c2 y 2 .
dy
50. Find the
586
CHAPTER 5
THE INTEGRAL
51. lim MN ,
N
f (x) = x,
[0, 2]
solution Let f (x) = x on [0, 2]. Let N > 0 be an integer and set a = 0, b = 2, and x = (b a)/N =
2
1
2k1
N . Also, let xk = 0 + (k 2 )x = N , k = 1, 2, . . . N , be the midpoints of the N subint
S E C T I O N 4.3
11. y = ex ,
The Mean Value Theorem and Monotonicity
421
[0, 1]
solution Let f (x) = ex , a = 0, and b = 1. Then f (x) = ex . Because f is continuous on the closed interval [0, 1]
and differentiable on the open interval (0, 1), by the MV
416
CHAPTER 4
APPLICATIONS OF THE DERIVATIVE
Squaring both sides of this last equation gives
cos2 i =
n2
cos2 r,
4
while squaring both sides of Snells Law gives
sin2 i = n2 sin2 r
or 1 cos2 i = n2 (1 cos2 r).
Solving this equation for cos2 r gives
cos2 r
S E C T I O N 5.2
The Definite Integral
611
In Exercises 3342, use properties of the integral and the formulas in the summary to calculate the integrals.
4
33.
(6t 3) dt
0
4
4
4
1
solution
(6t 3) dt = 6
t dt 3
1 dt = 6 (4)2 3(4 0) = 36.
2
0
0
0
2
34.
296
CHAPTER 3
DIFFERENTIATION
36. y = cos3 (12 )
solution After two applications of the chain rule,
y = 3 cos2 (12 )( sin(12 )(12) = 36 cos2 (12 ) sin(12 ).
1
x
Let f (x) = sec x 1 . Then
sec (1/x) tan (1/x)
f (x) = sec x 1 tan x 1 x 2 =
.
x2
37. y = s
S E C T I O N 4.3
The Mean Value Theorem and Monotonicity
431
solution The graph of h(x) is shown below at the left. Because h(x) is negative for x < 1 and for 0 < x < 1, it
follows that f (x) is decreasing for x < 1 and for 0 < x < 1. Similarly, f (x) is
The Derivative as a Function
S E C T I O N 3.2
18. f (x) =
solution
1
12 .
3
x,
a=8
Let f (x) =
3
x = x 1/3 . Then f (x) = 13 (x 1/31 ) = 13 x 2/3 . In particular, f (8) =
1
3
1
4
221
=
f (8) = 2, so the tangent line at x = 8 is
y = f (8)(x 8) + f (8) =
606
CHAPTER 5
THE INTEGRAL
solution
Let f (x) be given by Figure 14.
&2
(a) The definite integral 0 f (x) dx is the signed area of a semicircle of radius 1 which lies below the x-axis.
Therefore,
2
1
f (x) dx = (1)2 = .
2
2
0
&6
(b) The definite integral
S E C T I O N 4.2
Extreme Values
401
20. f (x) = sec1 x ln x
solution
Let f (x) = sec1 x ln x. Then
1
1
f (x) =
.
x x2 1 x
not exist at x = 0
This derivative is equal to zero when x 2 1 = 1, or when
x = 2. Moreover, the derivative does
and at x = 1. Am
326
CHAPTER 3
DIFFERENTIATION
so
dx
4y 3 2y
y(2y 2 1)
=
=
.
dy
2x
x
2
dx
= 0 when y = 0 and when y =
. Substituting y = 0 into the equation y 4 + 1 = y 2 + x 2
Thus,
dy
2
2
3
2
2
gives 1 = x , so x = 1. Substituting y =
, gives x = 3/4, so x =
. Thus,
5 THE INTEGRAL
5.1 Approximating and Computing Area
Preliminary Questions
1. What are the right and left endpoints if [2, 5] is divided into six subintervals?
1
solution If the interval [2, 5] is divided into six subintervals, the length of each subinterv
S E C T I O N 4.6
and
1
x4 x 2 6x + 8
lim
Graph Sketching and Asymptotes
=
1
x4+ x 2 6x + 8
and
lim
491
= .
y
5
3
1
x
2
4
5
6
5
64. y =
x3 + 1
x
x3 + 1
= x 2 + x 1 . Then f (x) = 2x x 2 , so that f is decreasing for x < 0 and for
solution Let f (x) =
x
0
466
CHAPTER 4
APPLICATIONS OF THE DERIVATIVE
Further Insights and Challenges
76. Show that LHpitals Rule applies to lim
x
x
x2 + 1
but that it does not help. Then evaluate the limit directly.
Both the numerator f (x) = x and the denominator g(x) = x 2 +
The Definite Integral
S E C T I O N 5.2
In other words, RN = bk+1 SN . Therefore,
b
k
k+1
k+1
k+1
x dx = lim RN = lim b SN = b
lim SN = b
N
0
N
N
1
621
x k dx.
0
86. Verify for 0 b 1 by interpreting in terms of area:
b
1
1
1 x 2 dx = b 1 b2 + sin1 b
2
476
CHAPTER 4
APPLICATIONS OF THE DERIVATIVE
local minima of f (x), and f ( 1+8 33 ) is a local maximum. Further sign analyses reveal that f (x) changes from positive
to negative at x = 0 and from negative to positive at x = 32 , so that there are points
286
CHAPTER 3
DIFFERENTIATION
45. Find the values of x between 0 and 2 where the tangent line to the graph of y = sin x cos x is horizontal.
solution
Let y = sin x cos x. Then
y = (sin x)( sin x) + (cos x)(cos x) = cos2 x sin2 x.
When y = 0, we have sin x
S E C T I O N 3.3
=
1
h0 g(a + h)g(a)
lim
f (a + h) f (a)
h0
h
g(a) lim
Product and Quotient Rules
g(a + h) g(a)
f (a) lim
h0
h
g(a)f (a) f (a)g (a)
g(a)f
(a)
f
(a)g
(a)
=
(g(a)2
(g(a)2
f
gf f g
In other words, p =
=
.
g
g2
59. Derivative of
566
CHAPTER 4
APPLICATIONS OF THE DERIVATIVE
and
y =
8(x 4 + 16) 8x 4x 3
128 24x 4
=
.
(x 4 + 16)2
(x 4 + 16)2
Solving y = 0 yields x = 0 as the only critical point. Because y (0) = 12 > 0, we conclude the function has
a local minimum at x = 0. Moreover,
S E C T I O N 3.5
Higher Derivatives
271
solution Let f (x) = 4ex x 3 . Then f (x) = 4ex 3x 2 , f (x) = 4ex 6x, f (x) = 4ex 6, and
f (3) = 4e3 6.
t
24. f (1), f (t) =
t +1
t
solution Let f (t) =
. Then
t +1
f (t) =
(t + 1)(1) (t)(1)
1
1
=
= 2
(t + 1)2
(t
S E C T I O N 3.7
The Chain Rule
301
77. The average molecular velocity v of a gas in a certain container is given by v(T ) = 29 T m/s, where
T is the temperature
in kelvins. The temperature is related to the pressure (in atmospheres) by T = 200P .
dv
.
246
CHAPTER 3
DIFFERENTIATION
39. (f g) (4) and (f/g) (4)
solution
Let h = f g and H = f/g. Then h = f g + gf and H =
gf fg
.
g2
Finally,
h (4) = f (4)g (4) + g(4)f (4) = (10)(1) + (5)(2) = 20,
and
H (4) =
g(4)f (4) f (4)g (4)
(5)(2) (10)(1)
=
= 0.
2
(g(