6.
d
d
d
d
2y2 3x3 5=
(0)
dx
dx
dx
dx
4yy' - 9x2 - 0 = 0
9x 2
y' =
4y
9 ! 12
9
dy
=
=
4! 2
8
dx (1,2)
(4-5)
7. y = 3x2 - 5
d(3x 2) d(5)
dy
=
!
dt
dt
dt
dy
dx
= 6x
dt
dt
dx
x = 12;
= 3
dt
dy
= 6·12·3 = 216
dt
(4-6)
8. 25p + x = 1,000
(A) x = 1,000 - 25p
(B) x = f(p) = 1,000 - 25p
f'(p) = -25
pf '(p)
25p
p
=
E(p) = =
f(p)
1,000 ! 25p
40 ! p
15
15
3
(C) E(15) =
= 0.6
=
=
40 ! 15
25
5
Demand is inelastic and insensitive to small changes in price.
(D) Revenue: R(p) = pf(p) = 1,000p - 25p2
25
25
5
(E) From (B), E(25) =
= 1.6
=
=
40 ! 25
15
3
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
10.
n
1000
100,000
10,000,000
100,000,000
2# n
!
1
+
"
n$
7.374312
7.388908
7.389055
7.389056
2 n
!
lim " 1 + #$ ≈ 7.38906 (5 decimal places);
n
n! "
2 n
!
lim " 1 + #$ = e2
n
n! "
11.
206
(4-2)
d
d
d
[(ln z)7 + ln z7] =
[ln z]7 +
7 ln z
dz
dz
dz
d
d
= 7[ln z]6
ln z + 7
ln z
dz
dz
1
7
= 7[ln z]6
+
z
z
6
7(ln z) + 7
7[(ln z)6 + 1]
=
=
z
z
CHAPTER 4
!
!
ADDITIONAL
DERIVATIVE TOPICS
(4-1)
(4-4)
12.
d 6
d
d 6
x ln x = x6
ln x + (ln x)
x
dx
dx
dx
"1%
= x6 $ ' + (ln x)6x5 = x5(1 + 6 ln x)
#x &
6 d
x
x d
x
e
!
e
x6
x
!
x 6e x ! 6x5ex
d # e $&
dx
dx
13.
=
=
!
(x6)2
x12
dx " x6 %
=
(4-3)
xe x ! 6ex
ex(x ! 6)
=
x7
x7
(4-3)
3
2
14. y = ln(2x3 - 3x)
15. f(x) = ex -x
2
3
2
6x ! 3
1
f'(x) = ex -x (3x2 - 2x)
y' =
(6x2 - 3) =
3
3
2x ! 3x
3
2
2x ! 3x
= (3x2 - 2x)ex -x
(4-4)
16. y = e-2x ln 5x
dy
! 1
= e-2x " #$ (5) + (ln 5x)(e-2x)(-2)
5x
dx
1 ! 2x ln 5x
1
"
= e-2x # ! 2 ln 5x $% =
xe2x
x
(4-4)
(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' =
2x - 3xy' - 3y + 8yy' =
8yy' - 3xy' =
(8y - 3x)y' =
0
0
3y 3y 3y !
y' =
8y !
3!2
y'
=
(!1,2)
8! 2
2x
2x
2x
3x
" 2("1)
8
=
" 3("1)
19
[Slope at (-1, 2)]
CHAPTER 4 REVIEW
(4-5)
207
19.
x3 - 2t2 x + 8
3x2x' - (2t2x' + x·4t) + 0
3x2x' - 2t2x' - 4xt
(3x2 - 2t2)x'
=
=
=
=
0
0
0
4xt
4xt
x' =
2
3x ! 2t2
4 ! 2 ! ("2)
"16
"16
=
=
x'
=
= -4
2
2
(!2,2)
3(2 ) " 2("2)
12 " 8
4
(4-5)
20. x - y2 = e y
Differentiate implicitly:
1 - 2yy' = e y y'
1 = e y y' + 2yy'
1 = y'(e y + 2y)
1
y' = y
e + 2y
1
y'
= 0
= 1
(1,0)
e + 2!0
21. ln y = x2 - y2
Differentiate implicitly:
y'
= 2x - 2yy'
y
"1
%
y' $ + 2y ' = 2x
#y
&
"1 + 2y 2 %
'' = 2x
y' $$
y
#
&
!
2xy
y' =
(4-5)
1 + 2y 2
2 !1 ! 1
2
!
=
y'
=
2
(1,1)
1 + 2(1)
3
(4-5)
22. y2 - 4x2 = 12
Differentiate with respect to t:
dy
dx
2y
- 8x
= 0
dt
dt
dx
Given:
= -2 when x = 1 and y = 4. Therefore,
dt
dy
2·4
- 8·1·(-2) = 0
dt
dy
8
+ 16 = 0
dt
dy
= -2.
dt
The y coordinate is decreasing at 2 units per second.
23. From the figure, x2 + y2 = 172.
Differentiate with respect to t:
dx
dy
dx
dy
2x
+ 2y
= 0 or x
+ y
=0
dt
dt
dt
dt
dx
We are given
= -0.5 feet per second.
dt
0.5x
dy
dy
x(-0.5) + y
= 0 or
=
y
dt
dt
208
CHAPTER 4
ADDITIONAL DERIVATIVE TOPICS
(4-6)
17
y
Therefore,
x
=
x
2y
82 + y2 = 172
y2 = 289 - 64 = 225
y = 15
8
4
dy
=
Therefore,
=
≈ 0.27 ft/sec.
2(15)
15
dt (8,15)
Now, when x = 8, we have:
(5-5)
(4-6)
dA
= 24 square inches per minute.
dt
Differentiate with respect to t:
24. A = πR2.
Given:
dA
dR
= 2πR
dt
dt
dR
24 = 2πR
dt
dR
24
12
=
.
=
dt
2!R
!R
dR
12
1
=
≈ 0.318 inches per minute
=
dt R =12
! " 12
!
Therefore,
(4-6)
25. x = f(p) = 20(p - 15)2
0 ≤ p ≤ 15
f'(p) = 40(p - 15)
pf '(p)
"40p(p " 15)
"2p
E(p) = =
=
2
f(p)
p " 15
20(p " 15)
!2p
Elastic: E(p) =
> 1
p ! 15
-2p < p - 15 (p - 15 < 0 reverses inequality)
!
-3p < -15
p > 5;
5 < p < 15
!2p
Inelastic: E(p) =
< 1
p ! 15
-2p > p - 15 (p - 15 < 0 reverses inequality)
-3p > -15
p < 5;
0 < p < 5
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):
Test Numbers
R'(p)
+ + + 0 - - p
R'(p)
x
5
50(+)
0
10
20
R(p)
15 !50(!)
Increasing
Demand: Inelastic
Decreasing
Elastic
(4-7)
R(p)
500
400
300
200
Inelastic
Elastic
100
10
20
p
(4-7)
CHAPTER 4 REVIEW
209
27. y = w3, w = ln u, u = 4 - ex
(A) y = [ln(4 - ex)]3
(B)
dy
dy dw du
=
!
!
dx
dw du dx
# 1 & x
1
= 3w2 ·
· (-ex) = 3[ln(4 - ex)]2 %
(("e )
u
$ 4 " ex '
"3ex [ln(4 " ex )]2
=
4 " ex
!
(4-4)
2
28. y = 5x -1
29.
30.
!
2
2
y' = 5x -1(ln 5)(2x)! = 2x5x -1(ln 5)
(4-4)
1
2x " 1
1
1
d
d
! 2
log5(x2 - x) = 2
(x2 - x) =
"
"
ln 5 x " x
x ! x ln 5 dx
dx
(4-4)
d
d
1
d
[ln(x2 + x)]1/2 =
[ln(x2 + x)]-1/2
ln(x2 + x)
ln(x 2 + x) =
dx
dx
2
dx
1
d
1
=
[ln(x2 + x)]-1/2 2
(x 2 + x)
x + x dx
2
2x + 1
1
2x + 1
=
[ln(x2 + x)]-1/2 · 2
=
2
x + x
2
2(x + x)[ln(x 2 + x)]1 2
(4-4)
31. exy = x2 + y +
Differentiate
d xy
e =
dx
exy(xy' + y) =
xexyy' - y' =
1
implicitly:
d 2
d
d
x +
y +
1
dx
dx
dx
2x + y'
2x - yexy
2x ! ye xy
y' =
xe xy ! 1
2 ! 0 " 0 ! e0
y'
=
= 0
(0,0)
0 ! e0 " 1
!
(4-5)
32. A = πr 2 , r ≥ 0
Differentiate with respect to t:
dA
dr
dr
= 2πr
= 6πr since
= 3
dt
dt
dt
The area increases at the rate 6πr. This is smallest when r = 0; there
is no largest value.
(4-6)
210
CHAPTER 4
ADDITIONAL DERIVATIVE TOPICS
33. y = x 3
Differentiate with respect to t:
dy
dx
= 3x 2
dt
dt
dx
Solving for
, we get
dt
1
5
dx
dy
dy
=
·
=
since
= 5
2
2
3x
3x
dt
dt
dt
dx
dy
To find where
>
, solve the inequality
dt
dt
5
> 5
3x 2
1
> 1
3x 2
3x 2 < 1
1
1
" 3
3
< x <
or
< x <
3
3
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
ln 2
t =
≈ 14.2 or 15 years
ln(1.05)
(B) The continuous compound interest formula is: A = Pert.
as above, we have
Proceeding
2P = Pe0.05t or e0.05t = 2.
Therefore, 0.05t = ln 2 and
ln 2
t =
≈ 13.9 years
.05
(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)
CHAPTER 4 REVIEW
211
37. x = 5000 " 2p3 = (5000 - 2p3)1/2
Differentiate implicitly with respect to x:
1
dp
1 =
(5000 - 2p3)-1/2(-6p2)
2
dx
!
2
!3p
dp
1 =
3
1
2
(5000 ! 2p ) dx
!(5000 ! 2 p3)1 2
dp
=
3p 2
dx
x2
dx
and
= 10 when x = 250.
20
dt
Differentiate with respect to t:
dR
dx
1
dx
dx
x dx
= 36
(2x)
= 36
dt
dt
20
dt
dt
10 dt
dR
250
Thus,
= 36(10) (10)
dt x = 250 and dx =10
10
dt
= $110 per day
(4-5)
38. Given: R(x) = 36x -
(4-6)
39. p = 16.8 - 0.002x
16.8
1
x = f(p) =
p = 8,400 - 500p
!
0.002
0.002
f'(p) = -500
! pf '(p)
5p
500p
Elasticity of demand: E(p) =
=
=
f(p)
84 ! 5p
8,400 ! 500p
40
40
10
E(8) =
< 1
=
=
84 ! 40
44
11
Demand is inelastic, a (small) price increase will increase revenue.
(4-7)
40.
f(t) = 1,700t + 20,500
f'(t) = 1,700
1,700
f '(t)
=
1,700t + 20,500
f(t)
1,700
Relative rate of change at t = 30:
≈ 0.02378
1,700(30) + 20,500
Relative rate of change:
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.
212
CHAPTER 4
ADDITIONAL DERIVATIVE TOPICS
(4-7)
(4-4)
42. Given: A = πR2 and
dA
= -45 mm2 per day (negative because the area is
dt
decreasing).
Differentiate with respect to t:
dA
dR
= π2R
dt
dt
dR
-45 = 2πR
dt
dR
45
= dt
2!R
dR
!45
!3
=
≈ -0.477 mm per day
=
dt R =15
2" # 15
2"
(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
0.25
e-0.4t =
= 0.625
4
-0.4t = ln(0.625)
ln(0.625)
t=
≈ 6.93
"0.4
!
The rate of learning is less than 0.25 after 7 days.
(4-4)
"
dx
1 %
44. Given: T = 2 $1!+ 3 2 ' = 2 + 2x-3/2, and
= 3 when x = 9.
dt
#
x &
Differentiate with respect to t:
dx
# 3
& dx
dT
= 0 + 2 %" x "5 2 (
= -3x-5/2 dt
$ 2
' dt
dt
!
dT
!1
= -3(9)-5/2(3) = -3 · 3-5 · 3 = -3-3 =
dx
=3
dt x = 9 and
27
dt
≈ -0.037 minutes per operation per hour
!
(4-6)
CHAPTER 4 REVIEW
213