= (1 + 3x)(-2) + (5 - 2x)(3) = -2 - 6x + 15 - 6x = 13 - 12x An equation for the tangent line at x = 2 is:
y - y1 = m(x - x1) where x1 = 2, y1 = f(x1) = f(2) = 7, and m = f'(x1) = f'(2) = -11. Thus, we have:
y - 7 = -11(x - 2) or y = -11x + 29
51. f(x) = x - 83x - 4
First find f'(x):
f'(x) = (3x - 4)(x - 8)' - (x - 8)(3x - 4)'
(3x - 4)2
= (3x - 4)(1) - (x - 8)(3)
(3x - 4)2 = 20
(3x - 4)2
An equation for the tangent line at x = 2 is: y - y1 = m(x - x1) where x1 = 2, y1 = f(x1) = f(2) = -3, and m = f'(x1) = f'(2) = 5. Thus, we have: y - (-3) = 5(x - 2) or y = 5x – 13
After 10 months, the total sales are 60,000 CD's and the sales are INCREASING at the rate of 4,000 CD's per month.
(C) The total sales after 11 months will be approximately 64,000 CD's.
85. x =
4,000
0.1p + 1, 10 ≤ p ≤ 70
(A)
dx
dp =
(0.1p + 1)(0) ! 4,000(0.1)
(0.1p + 1)2 =
!400
(0.1p + 1)2
(B) x(40) =
4,000
0.1(40) + 1=4,000
5 = 800;
dx
dp =
!400
[0.1(40) + 1]2=
!400
25 = -16
At a price level of $40, the demand is 800 CD players and the demand is DECREASING at the rate of 16 CD players per dollar.
(C) At a price of $41, the demand will be approximately 784 CD players.
87. C(t) =
0.14t
t2 + 1
(A) C'(t) =
(t2 + 1)(0.14t)'!(0.14t)(t2 + 1)'
(t2 + 1)2
=
(t2 + 1)(0.14) ! (0.14t)(2t)
(t2 + 1)2 =
0.14 ! 0.14t2
(t2 + 1)2 =
0.14(1 ! t2)
(t2 + 1)2
EXERCISE 4-4 175
(B) C'(0.5) =
!
0.14(1 " [0.5]2)
([0.5]2 + 1)2 =
!
0.14(1 " 0.25)
(1.25)2 = 0.0672
Interpretation: At t = 0.5 hours, the concentration is increasing at the rate of 0.0672 mg/cm3 per hour.
C'(3) =
!
0.14(1 " 32)
(32 + 1)2 =
!
0.14("8)
100 = -0.0112
Interpretation: At t = 3 hours, the concentration is decreasing at the rate of 0.0112 mg/cm3 per hour.
89. N(x) =
!
100x + 200
x + 32
(A) N'(x) =
!
(x + 32)(100x + 200 " ) # (100x + 200)(x + 32 " )
(x + 32)2
=
!
(x + 32)(100) " (100x + 200)(1)
(x + 32)2
=
!
100x + 3200 " 100x " 200
(x + 32)2=
3000
(x + 32)2
(B) N'(4) =
!
3000
(36)2 =
!
3000
1296 ≈ 2.31; N'(68) =
!
3000
(100)2 =
!
3000
10,000 =
!
3
10 = 0.30
EXERCISE 4-4 Things to remember: 1. COMPOSITE FUNCTIONS
A function m is a COMPOSITE of functions f and g if m(x) = f[g(x)] The domain of m is the set of all numbers x such that x is in the domain of g and g(x) is in the domain of f.
2. GENERAL POWER RULE
If u(x) is a differentiable function, n is any real number, and y = f(x) = [u(x)]n then f’(x) = n[u(x)]n-1u’(x)
This rule is often written more compactly as
y’ = nun-1u’ or
!
d
dxun = nun-1
!
d
dx, u = u(x)
176 CHAPTER 4 ADDITIONAL DERIVATIVE TOPICS
3. THE CHAIN RULE: GENERAL FORM If y = f(u) and u = g(x) define the composite function
y = m(x) = f[g(x)], then
dy
dx =
dy
du
du
dx provided that
dy
du and
du
dx exist.
Or, equivalently, m'(x) = f'[g(x)]g'(x) provided that f'[g(x)] and g'(x) exist.
4. GENERAL DERIVATIVE RULES
(a)
d
dx[f(x)]n = n[f(x)]n-1f'(x)
(b)
d
dxln[f(x)] =
1
f(x)f'(x)
(c)
d
dxef(x) = ef(x)f'(x)
1. f(u) = u3, g(x) = 3x2 + 2 f[g(x)] = (3x2 + 2)3
3. f(u) = eu, g(x) = -x2 f[g(x)] = e-x
2
5. Let u = g(x) = 3x2 - x + 5 and f(u) = u4. Then y = f(u) = u4.
7. Let u = g(x) = 1 + x + x2 and f(u) = eu. Then y = f(u) = eu.
An equation for the tangent line to the graph of f at x = 2 is: y - y1 = m(x
- x1) where x1 = 2, y1 = f(x1) = f(2) = 16, and m = f'(x1) = f'(2) = -16. Thus, y - 16 = -16(x - 2) or y = -16x + 48.
65. f(x) =
x
(2x ! 5)3
f'(x) =
(2x ! 5)3(1) ! x(3)(2x ! 5)2(2)
[(2x ! 5)3]2
=
(2x ! 5)3 ! 6x(2x ! 5)2
(2x ! 5)6 =
(2x ! 5) ! 6x
(2x ! 5)4 =
!4x ! 5
(2x ! 5)4
An equation for the tangent line to the graph of f at x = 3 is: y - y1 = m(x - x1) where x1 = 3, y1 = f(x1) = f(3) = 3, and m = f'(x1) = f'(3) = -17. Thus, y - 3 = -17(x - 3) or y = -17x + 54.
level of 24 calculators, total costs are INCREASING at the rate of $12.50 per calculator; also, the cost of producing the 25th calculator is approximately $12.50.
C'(42) =
1
[2(42) + 16]1 2 =
1
(100)1 2=
1
10 or $10.00; at a production
level of 42 calculators, total costs are INCREASING at the rate of $10.00 per calculator; also the cost of producing the 43rd calculator is approximately $10.00.
95. x = 80
!
p + 25 - 400 = 80(p + 25)1/2 - 400, 20 ≤ p ≤ 100
(A)
dx
dp = 80
!
1
2
"
# $ %
& ' (p + 25)-1/2(1) =
40
(p + 25)1 2
EXERCISE 4-4 183
(B) At p = 75, x = 80
!
75 + 25 - 400 = 400 and
dx
dp =
40
(75 + 25)1 2 =
40
(100)1 2 = 4.
At a price of $75, the supply is 400 speakers, and the supply is INCREASING at a rate of 4 speakers per dollar.
97. C(t) = 4.35e-t =
4.35
et, 0 ≤ t ≤ 5
(A) C'(t) =
!4.35et
e2t =
!4.35
et = -4.35e-t
C'(1) = -4.35e-1 ≈ -1.60 C'(4) = -4.35e-4 ≈ -0.08
Thus, after one hour, the concentration is decreasing at the rate of 1.60 mg/ml per hour; after four hours, the concentration is decreasing at the rate of 0.08 mg/ml per hour.
(B) C'(t) = -4.35e-t < 0 on (0, 5) Thus, C is decreasing on (0, 5); there are no local extrema.
C"(t) =
4.35et
e2t=4.35
et = 4.35e-t > 0 on (0, 5)
Thus, the graph of C is concave upward on (0, 5). The graph of C is shown at the right.
!
t C(t)0 4.351 1.604 0.085 0.03
99. P(x) = 40 + 25 ln(x + 1) 0 ≤ x ≤ 65
P'(x) = 25
!
1
x + 1
"
# $
%
& ' (1) =
25
x + 1
P'(10) =
25
11 ≈ 2.27
P'(30) =
25
31 ≈ 0.81
P'(60) =
25
61 ≈ 0.41
Thus, the rate of change of pressure at the end of 10 years is 2.27 millimeters of mercury per year; at the end of 30 years the rate of change is 0.81 millimeters of mercury per year; at the end of 60 years the rate of change is 0.41 millimeters of mercury per year.
184 CHAPTER 4 ADDITIONAL DERIVATIVE TOPICS
101. T = f(n) = 2n
!
n " 2 = 2n(n - 2)1/2
(A) f'(n) = 2n[(n - 2)1/2]' + (n - 2)1/2(2n) '
= 2n
!
1
2
"
# $ %
& ' (n ( 2)-1/2(1) + (n - 2)1/2(2)
=
!
n
(n " 2)1 2 + 2(n - 2)1/2
=
!
n + 2(n " 2)
(n " 2)1 2=
3n " 4
(n " 2)1 2
(B) f'(11) =
!
29
3 = 9.67; when the list contains 11 items, the learning
time is increasing at the rate of 9.67 minutes per item;
f'(27) =
!
77
5 = 15.4; when the list contains 27 items, the learning
time is increasing at the rate of 15.4 minutes per item.
