206 CHAPTER 4 ADDITIONAL DERIVATIVE TOPICS
6.
d
dx2y2 -
d
dx3x3 -
d
dx5 =
d
dx(0)
4yy' - 9x2 - 0 = 0
y' =
9x2
4y
dy
dx (1,2) =
9 ! 12
4 ! 2=9
8
(4-5)
7. y = 3x2 - 5
dy
dt =
d(3x2)
dt!d(5)
dt
dy
dt = 6x
dx
dt
x = 12;
dx
dt = 3
dy
dt = 6·12·3 = 216 (4-6)
8. 25p + x = 1,000 (A) x = 1,000 - 25p (B) x = f(p) = 1,000 - 25p
f'(p) = -25
E(p) = -
pf'(p)
f(p) =
25p
1,000 ! 25p=
p
40 ! p
(C) E(15) =
15
40 ! 15=
15
25=
3
5 = 0.6
Demand is inelastic and insensitive to small changes in price.
(D) Revenue: R(p) = pf(p) = 1,000p - 25p2
(E) From (B), E(25) =
25
40 ! 25=25
15=5
3 = 1.6
Demand is elastic; a price cut will increase revenue. (4-7)
9. y = 100e-0.1x
y’ = 100(-0.1)e-0.1x; y’(0) = 100(-0.1) = -10 (4-2)
10. n 1000 100,000 10,000,000 100,000,000
1 +
2
n
! "
# $
n
7.374312 7.388908 7.389055 7.389056
limn!"
1 +2
n
! "
# $
n ≈ 7.38906 (5 decimal places);
limn!"
1 +2
n
! "
# $
n = e2 (4-1)
11.
d
dz[(ln z)7 + ln z7] =
d
dz[ln z]7 +
d
dz7 ln z
= 7[ln z]6
d
dzln z + 7
d
dzln z
= 7[ln z]6
1
z+7
z
=
!
7(ln z)6 + 7
z =
!
7[(ln z)6 + 1]
z (4-4)
CHAPTER 4 REVIEW 207
12.
d
dxx6 ln x = x6
d
dxln x + (ln x)
d
dxx6
= x6
!
1
x
"
# $
%
& ' + (ln x)6x5 = x5(1 + 6 ln x) (4-3)
13.
d
dx
ex
x6
!
" # $
% & =
x6 d
dxex! e
x d
dxx6
(x6)2 =
x6ex ! 6x5ex
x12 =
xex ! 6ex
x7 =
ex(x ! 6)
x7
(4-3)
14. y = ln(2x3 - 3x)
y' =
1
2x3 ! 3x(6x2 - 3) =
6x2 ! 3
2x3 ! 3x
(4-4)
15. f(x) = ex3-x2
f'(x) = ex3-x2(3x2 - 2x)
= (3x2 - 2x)ex3-x2 (4-4)
16. y = e-2x ln 5x
dy
dx = e-2x
1
5x
! "
# $ (5) + (ln 5x)(e
-2x)(-2)
= e-2x
1
x! 2 ln 5x
" #
$ % =
1 ! 2x ln 5x
xe2x (4-4)
17. f(x) = 1 + e-x f'(x) = e-x(-1) = -e-x
An equation for the tangent line to the graph of f at x = 0 is:
y - y1 = m(x - x1),
where x1 = 0, y1 = f(0) = 1 + e0 = 2, and m = f'(0) = -e0 = -1.
Thus, y - 2 = -1(x - 0) or y = -x + 2. An equation for the tangent line to the graph of f at x = -1 is
y - y1 = m(x - x1), where x1 = -1, y1 = f(-1) = 1 + e, and m = f'(-1) = -e. Thus, y - (1 + e) = -e[x - (-1)] or y - 1 - e = -ex - e and y = -ex + 1. (4-4)
18. x2 - 3xy + 4y2 = 23 Differentiate implicitly: 2x - 3(xy' + y·1) + 8yy' = 0 2x - 3xy' - 3y + 8yy' = 0 8yy' - 3xy' = 3y - 2x (8y - 3x)y' = 3y - 2x
y' =
3y ! 2x
8y ! 3x
y' (!1,2)
=
3 ! 2 " 2("1)
8 ! 2 " 3("1)=
8
19 [Slope at (-1, 2)] (4-5)
208 CHAPTER 4 ADDITIONAL DERIVATIVE TOPICS
19. x3 - 2t2x + 8 = 0 3x2x' - (2t2x' + x·4t) + 0 = 0 3x2x' - 2t2x' - 4xt = 0 (3x2 - 2t2)x' = 4xt x' =
4xt
3x2 ! 2t2
x' (!2,2)
=
4 ! 2 ! ("2)
3(22) " 2("2)2=
"16
12 " 8=
"16
4 = -4 (4-5)
20. x - y2 = ey Differentiate implicitly: 1 - 2yy' = eyy' 1 = eyy' + 2yy' 1 = y'(e y + 2y) y' =
1
ey + 2y
y' (1,0)
=
1
e0 + 2 ! 0 = 1
(4-5)
21. ln y = x2 - y2 Differentiate implicitly:
y'
y = 2x - 2yy'
y'
!
1
y+ 2y
"
# $
%
& ' = 2x
y'
!
1 + 2y2
y
"
# $ $
%
& ' ' = 2x
y' =
2xy
1 + 2y2
y' (1,1)
=
2 ! 1 ! 1
1 + 2(1)2=
2
3 (4-5)
22. y2 - 4x2 = 12 Differentiate with respect to t: 2y
dy
dt - 8x
dx
dt = 0
Given:
dx
dt = -2 when x = 1 and y = 4. Therefore,
2·4
dy
dt - 8·1·(-2) = 0
8
dy
dt + 16 = 0
dy
dt = -2.
The y coordinate is decreasing at 2 units per second. (4-6)
23. From the figure, x2 + y2 = 172. Differentiate with respect to t:
2x
dx
dt + 2y
dy
dt = 0 or x
dx
dt + y
dy
dt =0
We are given
dx
dt = -0.5 feet per second. Therefore,
x(-0.5) + y
dy
dt = 0 or
dy
dt =
0.5x
y =
x
2y
x
y 17
CHAPTER 4 REVIEW 209
Now, when x = 8, we have: 82 + y2 = 172 y2 = 289 - 64 = 225 y = 15
Therefore,
dy
dt (8,15) =
8
2(15)=
4
15 ≈ 0.27 ft/sec. (5-5)
(4-6)
24. A = πR2. Given:
dA
dt = 24 square inches per minute.
Differentiate with respect to t:
dA
dt = 2πR
dR
dt
24 = 2πR
dR
dt
Therefore,
dR
dt =
24
2!R=
12
!R.
dR
dt R =12 =
12
! " 12=
1
! ≈ 0.318 inches per minute (4-6)
25. x = f(p) = 20(p - 15)2 0 ≤ p ≤ 15 f'(p) = 40(p - 15)
E(p) = -
pf'(p)
f(p) =
!
"40p(p " 15)
20(p " 15)2=
"2p
p " 15
Elastic: E(p) =
!2p
p ! 15 > 1
-2p < p - 15 (p - 15 < 0 reverses inequality) -3p < -15 p > 5; 5 < p < 15
Inelastic: E(p) =
!2p
p ! 15 < 1
-2p > p - 15 (p - 15 < 0 reverses inequality) -3p > -15 p < 5; 0 < p < 5 (4-7)
26. x = f(p) = 5(20 - p) 0 ≤ p ≤ 20 R(p) = pf(p) = 5p(20 - p) = 100p - 5p2 R'(p) = 100 - 10p = 10(10 - p)
Critical values: p = 10 Sign chart for R'(p):
R'(p)
R(p)
x
Increasing Decreasing
Demand: Inelastic Elastic
0 10 20
+ + + 0 - - -
Test Numbers
p R'(p)
5 50(+)
15 !50(!)
R(p)
p10 20
100
200
300
400
500
Inelastic Elastic
(4-7)
210 CHAPTER 4 ADDITIONAL DERIVATIVE TOPICS
27. y = w3, w = ln u, u = 4 - ex
(A) y = [ln(4 - ex)]3
(B)
dy
dx =
dy
dw!dw
du!du
dx
= 3w2 ·
1
u · (-ex) = 3[ln(4 - ex)]2
!
1
4 " ex
#
$ %
&
' ( ("e
x)
=
!
"3ex[ln(4 " ex)]2
4 " ex (4-4)
28. y = 5x2-1
y' = 5x2-1(ln 5)(2x) = 2x5x
2-1(ln 5) (4-4)
29.
d
dxlog5(x
2 - x) =
1
x2 ! x"
1
ln 5"d
dx(x2 - x) =
1
ln 5!2x " 1
x2 " x (4-4)
30.
d
dx
!
ln(x2 + x) =
d
dx[ln(x2 + x)]1/2 =
1
2[ln(x2 + x)]-1/2
d
dxln(x2 + x)
=
1
2[ln(x2 + x)]-1/2
1
x2 + x
d
dx(x
2+ x)
=
1
2[ln(x2 + x)]-1/2 ·
2x + 1
x2 + x =
!
2x + 1
2(x2 + x)[ln(x2 + x)]1 2
(4-4)
31. exy = x2 + y + 1 Differentiate implicitly:
d
dxexy =
d
dxx2 +
d
dxy +
d
dx1
exy(xy' + y) = 2x + y' xexyy' - y' = 2x - yexy
y' =
2x ! yexy
xexy ! 1
y' (0,0)
=
2 ! 0 " 0 ! e0
0 ! e0 " 1 = 0 (4-5)
32. A = πr2, r ≥ 0 Differentiate with respect to t:
dA
dt = 2πr
dr
dt = 6πr since
dr
dt = 3
The area increases at the rate 6πr. This is smallest when r = 0; there is no largest value. (4-6)
CHAPTER 4 REVIEW 211
33. y = x3 Differentiate with respect to t:
dy
dt = 3x2
dx
dt
Solving for
dx
dt, we get
dx
dt =
1
3x2 ·
dy
dt =
5
3x2 since
dy
dt = 5
To find where
dx
dt >
dy
dt, solve the inequality
5
3x2 > 5
1
3x2 > 1
3x2 < 1
-
!
1
3 < x <
!
1
3 or
!
" 3
3 < x <
!
3
3 (4-6)
34. (A) The compound interest formula is: A = P(1 + r)t. Thus, the time for P to double when r = 0.05 and interest is compounded annually can be found by solving
2P = P(1 + 0.05)t or 2 = (1.05)t for t. ln(1.05)t = ln 2 t ln(1.05) = ln 2
t =
ln 2
ln(1.05) ≈ 14.2 or 15 years
(B) The continuous compound interest formula is: A = Pert. Proceeding as above, we have
2P = Pe0.05t or e0.05t = 2. Therefore, 0.05t = ln 2 and
t =
ln 2
.05 ≈ 13.9 years (4-1)
35. A(t) = 100e0.1t A'(t) = 100(0.1)e0.1t = 10e0.1t A'(1) = 11.05 or $11.05 per year A'(10) = 27.18 or $27.18 per year (4-1)
36. R(x) = xp(x) = 1000xe-0.02x R'(x) = 1000[xDxe
-0.02x + e-0.02xDxx]
= 1000[x(-0.02)e-0.02x + e-0.02x] = (1000 - 20x)e-0.02x (4-4)
212 CHAPTER 4 ADDITIONAL DERIVATIVE TOPICS
37. x =
!
5000 " 2p3 = (5000 - 2p3)1/2 Differentiate implicitly with respect to x: 1 =
1
2(5000 - 2p3)-1/2(-6p2)
dp
dx
1 =
!3p2
(5000 ! 2p3)1 2dp
dx
dp
dx =
!(5000 ! 2p3)1 2
3p2 (4-5)
38. Given: R(x) = 36x -
x2
20 and
dx
dt = 10 when x = 250.
Differentiate with respect to t:
dR
dt = 36
dx
dt -
1
20(2x)
dx
dt = 36
dx
dt -
x
10
dx
dt
Thus,
dR
dt x =250 and
dx
dt=10
= 36(10) -
250
10(10)
= $110 per day (4-6)
39. p = 16.8 - 0.002x
x = f(p) =
16.8
0.002!
1
0.002p = 8,400 - 500p
f'(p) = -500
Elasticity of demand: E(p) =
!pf'(p)
f(p) =
500p
8,400 ! 500p =
5p
84 ! 5p
E(8) =
40
84 ! 40=40
44=10
11 < 1
Demand is inelastic, a (small) price increase will increase revenue. (4-7)
40. f(t) = 1,700t + 20,500 f'(t) = 1,700
Relative rate of change:
f'(t)
f(t) =
1,700
1,700t + 20,500
Relative rate of change at t = 30:
1,700
1,700(30) + 20,500 ≈ 0.02378 (4-7)
41. C(t) = 5e-0.3t C'(t) = 5e-0.3t(-0.3) = -1.5e-0.3t After one hour, the rate of change of concentration is C'(1) = -1.5e-0.3(1) = -1.5e-0.3 ≈ -1.111 mg/ml per hour. After five hours, the rate of change of concentration is C'(5) = -1.5e-0.3(5) = -1.5e-1.5 ≈ -0.335 mg/ml per hour. (4-4)
CHAPTER 4 REVIEW 213
42. Given: A = πR2 and
dA
dt = -45 mm2 per day (negative because the area is
decreasing). Differentiate with respect to t:
dA
dt = π2R
dR
dt
-45 = 2πR
dR
dt
dR
dt = -
45
2!R
dR
dt R =15 =
!45
2" # 15=
!3
2" ≈ -0.477 mm per day (4-6)
43. N(t) = 10(1 - e-0.4t) (A) N'(t) = -10e-0.4t(-0.4) = 4e-0.4t N'(1) = 4e-0.4(1) = 4e-0.4 ≈ 2.68. Thus, learning is increasing at the rate of 2.68 units per day after 1 day. N'(5) = 4e-0.4(5) = 4e-2 = 0.54 Thus, learning is increasing at the rate of 0.54 units per day after 5 days.
(B) We solve N’(t) = 0.25 = 4e-0.4t for t
e-0.4t =
!
0.25
4 = 0.625
-0.4t = ln(0.625)
t =
!
ln(0.625)
"0.4 ≈ 6.93
The rate of learning is less than 0.25 after 7 days. (4-4)
44. Given: T = 2
!
1 +1
x3 2
"
# $
%
& ' = 2 + 2x-3/2, and
dx
dt = 3 when x = 9.
Differentiate with respect to t:
dT
dt = 0 + 2
!
"3
2x"5 2
#
$ %
&
' ( dx
dt = -3x-5/2
dxdt
dT
dt x =9 and
dx
dt=3 = -3(9)-5/2(3) = -3 · 3-5 · 3 = -3-3 =
!1
27
≈ -0.037 minutes per operation per hour (4-6)