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ISM: Introductory Mathematical Analysis Section 11.3
389
89. 1 12 21( )f x x x x
x−
= + = +
312 21 1 1 1 1( )
2 2 2 2 2xf x x x
x x x x x− − −′ = − = − =
Thus 1 1 1( ) 0
2 2 2x x xf xx x x x x x− − −′− = − = .
90. (1 ) p cz b w bw= + −
(1 ) p
c c
dwdz b bdw dw
= + −
Rewriting the right side and factoring out 1 + b
gives (1 )(1 )1
p
c c
dwdz b bbdw dw b
+= + −
+,
(1 )1
p
c c
dwdz bbdw dw b
⎡ ⎤= + −⎢ ⎥+⎣ ⎦
.
91. 3 3y x x= − 2( ) 3 3y x x′ = −
( )22 3 2 3 9xy =
′ = − =
The tangent line at (2, 2) is given by y – 2 = 9(x – 2), or y = 9x – 16.
8
–2
–5 5
92. 1/ 33y x x= =
2 / 33 2
1 1( )3 3
y x xx
−′ = =
81
12xy =−′ =
The tangent line at (−8, −2) is given by 12 ( 8),
12y x+ = + or
1 4 .12 3
y x= −
5–15
1
–3
Principles in Practice 11.3
1. Here 5dPdp
= and ∆p = 25.5 – 25 = 0.5.
5(0.5) 2.5dPP pdp
∆ ≈ ∆ = =
The profit increases by 2.5 units when the price is changed from 25 to 25.5 per unit.
2. ( )216 16 16 16(2 ) 16 32dy d t t t tdt dt
= − = − = −
0.516 32(0.5) 16 16 0
t
dydt =
= − = − =
The graph of y(t) is shown. 5
00 1
When t = 0.5, the object is at the peak of its flight.
3. ( )2 24( ) 3 4 (2 ) 4 83
V r r r r r′ = π + π = π + π
When r = 2, 2( ) 4 (2) 8 (2) 32V r′ = π + π = π and
3 24 32 80( ) (2) 4 (2) 163 3 3
V r π= π + π = + π = π .
The relative rate of change of the volume when
r = 2 is 803
(2) 32 6 1.2(2) 5
VV′ π
= = =π
. Multiplying 1.2
by 100 gives the percentage rate of change: (1.2)(100) = 120%.
Problems 11.3
1. 2( ) 2 3s f t t t= = +
If 1,t∆ = then over [1, 2] we have
(2) (1) 14 5 9.2 1 1
s f ft
∆ − −= = =
∆ −
If 0.5,t∆ = then over [1, 1.5] we have
(1.5) (1) 9 5 8.1.5 1 0.5
s f ft
∆ − −= = =
∆ −
Continuing this way, we obtain the following table:
Chapter 11: Differentiation ISM: Introductory Mathematical Analysis
390
t∆ 1 0.5 0.2 0.1 0.01 0.001
st
∆∆ 9 8 7.4 7.2 7.02 7.002
We estimate the velocity when t = 1 to be 7 m/s. With differentiation we get 4 3,dsv tdt
= = +
14(1) 3 7 m/s.
t
dsdt =
= + =
2. ( ) 2 5y f x x= = + .
If ∆x = 1, then over [3, 4] we have
(4) (3) 13 11 0.28891
y f fx x
∆ − −= = ≈
∆ ∆
If ∆x = 0.5, then over [3, 3.5] we have
(3.5) (3) 12 11 0.29500.5
y f fx x
∆ − −= = ≈
∆ ∆
Continuing in this way we obtain the following table:
x∆ 1 0.5 0.2 0.1 0.01 0.001
yx
∆∆ 0.2889 0.2950 0.2988 0.3002 0.3014 0.3015
We estimate the rate of change to be 0.3015. 1Note: The actual rate of change is 0.3015.11
⎛ ⎞≈⎜ ⎟
⎝ ⎠
3. 2( ) 2 4s f t t t= = −
a. When t = 7, then 22(7 ) 4(7) 70s = − = m.
b. 2(7.5) (7) [2(7.5) 4(7.5)] 70 25
0.5 0.5s f ft
∆ − − −= = =
∆ m/s
c. 4 4.dsv tdt
= = − If t = 7, then v = 4(7) − 4 = 24 m/s
4. 1( ) 12
s f t t= = +
a. When t = 2, 1 (2) 1 22
s = + = m.
b. 12 (2.1) 1 2(2.1) (1) 0.5
0.1 0.1s f ft
⎡ ⎤+ −∆ − ⎣ ⎦= = =∆
m/s
c. 12
dsvdt
= = . If t = 2, then 12
v = m/s
ISM: Introductory Mathematical Analysis Section 11.3
391
5. 3( ) 2 6s f t t= = +
a. When t = 1, 32(1) 6 8s = + = m.
b.
3
(1.02) (1)0.02
2(1.02) 6 8
0.026.1208 m/s
s f ft
∆ −=
∆⎡ ⎤+ −⎣ ⎦=
=
c. 26dsv tdt
= = . If t = 1, then
26(1) 6v = = m/s
6. 2( ) 3 2 1s f t t t= = − + +
a. When t = 1, ( )23 1 2(1) 1 0s = − + + = m.
b.
2
(1.25) (1)0.25
3(1.25) 2(1.25) 1 04.75 m/s
0.25
s f ft
∆ −=
∆⎡ ⎤− + + −⎣ ⎦= = −
c. 6 2.dsv tdt
= = − + If t = 1, v = –4 m/s
7. 4 3( ) 2s f t t t t= = − +
a. When t = 2, ( )4 32 2 2 2 2s = − + = m.
b.
4 3
(2.1) (2)0.1
(2.1) 2(2.1) 2.1 210.261 m/s
0.1
s f ft
∆ −=
∆⎡ ⎤− + −⎣ ⎦= =
c. 3 24 6 1.dsv t tdt
= = − + If t = 2, then
( ) ( )3 24 2 6 2 1 9v = − + = m/s
8. 4 7 / 2( ) 3s f t t t= = −
a. When t = 0, 4 7 / 23 0 0 0.s = ⋅ = =
b. ( ) ( ) ( )4 7 / 21 11 4 44
1 14 4
3 0(0)
1 m/s64
f fst
⎡ ⎤⋅ − −− ⎢ ⎥∆ ⎣ ⎦= =∆
=
c. 3 5 / 2712 .2
dsv t tdt
= = − If t = 0, then
3 5 / 2712(0) (0) 0 m/s.2
v = − =
9. 3225
2dy xdx
= . If x = 9, 25 (27) 337.502
dydx
= = .
10. 2dA rdr
= π . If r = 3, 2 (3) 6dAdr
= π = π .
11. 0 0.27(1 0) 0.27e
dTdT
= + − =
12. 24dV rdr
= π
When 46.3 10 ,r −= ×
4 2
6
4 [6.3 10 ] 158.76
4.988 10 .
dVdr
− −8
−
= π × = π×10
≈ ×
13. c = 500 + 10q, 10dcdq
= . When q = 100,
10dcdq
= .
14. c = 5000 + 6q, 6dcdq
= . When q = 36, 6dcdq
= .
15. 0.1(2 ) 3 0.2 3dc q qdq
= + = + . When q = 5,
0.2(5) 3 4.dcdq
= + =
16. 0.2 3dc qdq
= + . When q = 3, 3.6dcdq
= .
17. 2 50dc qdq
= + . Evaluating when q = 15, 16 and
17 gives 80, 82 and 84, respectively.
Chapter 11: Differentiation ISM: Introductory Mathematical Analysis
392
18. 20.12 4.4dc q qdq
= − +
Evaluating when q = 5, 25, and 1000 gives 2.4, 54.4 and 119,004.4, respectively.
19. 5000.01 5c qq
= + +
20.01 5 500c cq q q= = + +
0.02 5dc qdq
= +
506
q
dcdq =
=
1007
q
dcdq =
=
20. 10002c
q= +
2 1000c cq q= = +
2dcdq
= for all q
21. 3 20.00002 0.01 6 20,000c cq q q q= = − + +
20.00006 0.02 6dc q qdq
= − +
If q = 100, then 4.6dcdq
= . If q = 500, then
11dcdq
= .
22. 3 20.002 0.5 60 7000c cq q q q= = − + +
20.006 60dc q qdq
= − +
If q = 15, then 46.35.dcdq
= If q = 25, then
38.75.dcdq
=
23. 0.8r q=
0.8drdq
= for all q.
24. 21 115 1530 30
r q q q q⎛ ⎞= − = −⎜ ⎟⎝ ⎠
11515
dr qdq
= −
For q = 5, 443
drdq
= ; for q =15, 14drdq
= ; for
q = 150, 5drdq
= .
25. 2 3250 45r q q q= + −
2250 90 3dr q qdq
= + − . Evaluating when
q = 5, 10 and 25 gives 625, 850 and 625, respectively.
26. 260 0.2r q q= −
60 0.4dr qdq
= −
Evaluating when q = 10 and 20 gives 56 and 52, respectively.
27. 6.750 0.000328(2 ) 6.750 0.000656dc q qdq
= − = −
20006.750 0.000656(2000) 5.438
q
dcdq =
= − =
10, 484.69 6.750 0.000328cc qq q
−= = + −
10, 484.69(2000) 6.750 0.000328(2000)2000
0.851655
c −= + −
=
28. 20.79 0.04284 0.0003dc q qdq
= − + −
700.7388
q
dcdq =
=
29. 0.93 5,000,000PR = 0.935,000,000P R−=
1.934,650,000dP RdR
−= −
30. 10,500dvdt
= − for all t.
ISM: Introductory Mathematical Analysis Section 11.3
393
31. a. 1.5dy xdx
= − −
61.5 6 7.5
x
dydx =
= − − = −
b. Setting –1.5 – x = –6 gives x = 4.5.
32. 2( ) 0.4 4 5c f q q q= = + +
0.8 4dc qdq
= +
If q = 2, then 5.6dcdq
= . Over the interval [2, 3],
(3) (2) 20.6 14.6 63 2 1
c f fq
∆ − −= = =
∆ −.
33. a. 1y′ =
b. 1
4yy x′=
+
c. (5) 1y′ =
d. 1 1 0.111
5 4 9= ≈
+
e. 11.1%
34. a. 3y′ = −
b. 3 3
7 3 3 7yy x x′ −= =
− −
c. (6) 3y′ = −
d. 3 3 0.2727
3(6) 7 11= ≈
−
e. 27.27%
35. a. 6y x′ =
b. 26
3 7y xy x′=
+
c. (2) 6(2) 12y′ = =
d. 12 12 0.632
12 7 19= ≈
+
e. 63.2%
36. a. 29y x′ = −
b. 2
39
5 3y xy x
′ −=
−
c. (1) 9y′ = −
d. 9 9 4.5
5 3 2−
= − = −−
e. −450%
37. a. 23y x′ = −
b. 2
33
8y xy x′ −=
−
c. (1) 3y′ = −
d. 3 3 0.429
8 1 7−
= − ≈ −−
e. –42.9%
38. a. 2 3y x′ = +
b . 22 3
3 4y xy x x′ +=
+ −
c. ( 1) 2( 1) 3 1y′ − = − + =
d. 1 1 0.167
1 3 4 6= − ≈ −
− −
e. –16.7%
39. 20.3 3.5 9c q q= + +
0.6 3.5dc qdq
= +
If q = 10, then 0.6(10) 3.5 9.5.dcdq
= + = If
q = 10, then c = 74 and
9.5(100) (100) 12.8%74
dcdq
c= ≈ .
Chapter 11: Differentiation ISM: Introductory Mathematical Analysis
394
40. 1100 100y xx
−= =
22
100100dy xdx x
−= − = −
If x = 10, 100 1100
dydx
= − = − and 1(100) (100) 10%
10yy′ −
= = − .
41. a. 30 0.6dr qdq
= −
b. If q = 10, 30 6 24 4 0.09
300 30 270 45rr′ −= = = ≈
−.
c. 9%
42. a. 10 0.4dq qdr
= −
b. If q = 25, 210 0.4(25) 0.
10(25) 0.2(25)rr′ −= =
−
c. 0%
43. 0.568
0.4320.864 0.432
2W tW tt
−′= =
44. a.
0.3
1.3
1.31855.241
1 1855.24
1.3I
I
RR I′= =
0.3
1.3
1.31101.292
2 1101.29
1.3I
I
RR I′= =
b. They are equal.
c. 1
1
1( )
n
nnC xf x n
f x xC x
−′= =
12
2
( )( )
n
nnC xg x n
g x xC x
−′= =
The rates are equal.
45. The cost of q = 20 bikes is 20(150) $3000qc = = . The marginal cost, $125, is the approximate cost of one
additional bike. Thus the approximate cost of producing 21 bikes is $3000 + $125 = $3125.
ISM: Introductory Mathematical Analysis Section 11.4
395
46. The relative rate of change of c is dcdq
c, which is given to be
1q
: 1
dcdq
c q= . Thus
dc c cdq q
= = , and the marginal cost
function dcdq
⎛ ⎞⎜ ⎟⎝ ⎠
and the average cost function ( )c are equal.
47. $5.07 per unit
48. 11,275 people per year
Principles in Practice 11.4
1. (2 0.15 ) (225 20 )dR dx xdx dx
= − + (225 20 ) (2 0.15 )dx xdx
+ + −
= (2 – 0.15x)(20) + (225 + 20x)(–0.15) = 40 – 3x – 33.75 – 3x = 6.25 – 6x
6.25 6dR xdx
= −
2. 2 31( )3
T x x x= −
( )T x′ 22x x= −
When the dosage is 1 milligram the sensitivity is 2(1) 2(1) 1 1T ′ = − = .
Problems 11.4
1. ( ) (4 1)(6) (6 3)(4)f x x x′ = + + + = 24x + 6 + 24x + 12 = 48x + 18 = 6(8x + 3)
2. ( ) (3 1)(7) (7 2)(3) 42 1f x x x x′ = − + + = −
3. 2 3 2 2 3 2 3 2 3 2( ) (5 3 )(3 4 ) ( 2 )( 3) 15 20 9 12 3 6 12 33 20s t t t t t t t t t t t t t t t′ = − − + − − = − − + − + = − + −
4. 2 2( ) (3 )(10 ) (5 2)(1) 15 30 2Q x x x x x x′ = + + − = + −
5. ( ) ( )2 2( ) 3 4 (2 5) 5 1 (6 )f r r r r r r′ = − − + − + 3 2 3 26 15 8 20 6 30 6r r r r r r= − − + + − + 3 212 45 2 20r r r= − − +
6. ( ) ( )2 2( ) 2 3 (6 4) 3 4 1 (4 )C I I I I I I′ = − − + − + 3 2 3 212 8 18 12 12 16 4I I I I I I= − − + + − + ( )3 22 12 12 7 6I I I= − − +
7. Without the product rule we have
( )2 2 4 2( ) 2 5 2 5f x x x x x= − = −
3( ) 8 10f x x x′ = −
8. Without the product rule we have
( )3 2 5 4 3( ) 3 2 2 3 6 6f x x x x x x x= − + = − +
4 3 2( ) 15 24 18f x x x x′ = − +
Chapter 11: Differentiation ISM: Introductory Mathematical Analysis
396
9. ( ) ( )2 23 2 (4 1) 2 3 (2 3)y x x x x x x′ = + − − + − − +
( )3 2 24 12 8 3 2x x x x x= + − − − + ( )3 2 24 2 6 6 3 9x x x x x+ − − + − −
3 28 15 20 7x x x= + − −
10. 2 2
2 2 3 2 2 3
3 2
( ) (3 5 2 )(1 8 ) (2 4 )( 5 4 )3 5 2 24 40 16 10 5 20 8 4 16
32 66 26 7
x x x x x x xx x x x x x x x x xx x x
φ′ = − + − + + − − +
= − + − + − − − + + + −
= − + − −
11. 2 2 3
4 3 2 4 3
4 3 2
( ) ( 3 7)(6 ) (2 4)(2 3)6 18 42 4 6 8 1210 24 42 8 12
f w w w w w ww w w w w ww w w w
′ = + − + − +
= + − + + − −
= + − − −
12. ( ) ( )2 2( ) 3 ( 1 2 ) 3 (3 2 )f x x x x x x x′ = − − − + − − −
2 3 2 2 33 5 2 9 3 3 6 2 2x x x x x x x x= − − + + − − − + + 3 24 6 12 9x x x= − − +
13. ( ) ( )2 21 9 6y x x′ = − − ( )33 6 5 (2 ) 4(8 2)x x x x+ − + − +
4 2 4 29 15 6 6 12 10 32 8x x x x x x= − + + − + − − 4 215 27 22 2x x x= − − −
14. ( ) ( )4 2( ) 4 5 3 8 5 (2) (2 2)(16 )h x x x x x⎡ ⎤′ = + − + +⎢ ⎥⎣ ⎦
4 2 220 3(16 10 32 32 )x x x x= + − + + 4 220 144 96 30x x x= + + −
15. 1/ 2 1/ 2
1/ 2 1/ 2 1/ 2
1/ 2 1/ 2
3 1( ) (5 2)(3) (3 1) 52 23 15 515 62 2 23[45 12 5 ]4
F p p p p
p p p
p p
−
−
−
⎡ ⎤⎛ ⎞′ = − + − ⋅⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦⎡ ⎤= − + −⎢ ⎥⎣ ⎦
= − −
16. 1/ 2 2 / 3 1/ 2 1/ 3 1/ 2 1/ 2
1/ 6 1/ 3 2 / 3 1/ 2 1/ 2 1/ 6 1/ 3 1/ 2
1/ 2 1/ 3 1/ 6 1/ 2 2 / 3
1 3 1( ) ( 5 2) ( 3 ) 53 2 2
1 5 2 3 15 1 33 5 153 3 3 2 2 2 21 ( 135 40 5 18 4 18)6
g x x x x x x x x
x x x x x x x x
x x x x x
− − −
− − − −
− − −
⎛ ⎞ ⎛ ⎞′ = + − − + − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= + − − − + + + − −
= − + + + − −
17. 273
y = ⋅ is a constant function, so 0y′ = .
18. 3 26 11 6y x x x= − + − 23 12 11y x x′ = − +
ISM: Introductory Mathematical Analysis Section 11.4
397
19. 3 26 47 31 28y x x x= + + − 218 94 31y x x′ = + +
20. 2 2
2
(4 1)(2) (2 3)(4) 8 2 8 12(4 1) (4 1)
14(4 1)
dy x x x xdx x x
x
+ − − + − += =
+ +
=+
21. 2 2
2
( 1)(5) (5 )(1) 5 5 5( )( 1) ( 1)
5( 1)
x x x xf xx x
x
− − − −′ = =
− −
= −−
22. 2
2 2
(5 )( 5) ( 5 )( 1)( )(5 )
25 5 5 25(5 ) (5 )
x xH xx
x xx x
− − − − −′ =−
− + −= = −
− −
23. 55
13 13( )33
f x xx
−−= = −
66
13 65( ) ( 5 )3 3
f x xx
−′ = − − =
24. ( )25( ) 27
f x x= −
5 10( ) (2 )7 7
f x x x′ = =
25. 2( 1)(1) ( 2)(1)
( 1)x xy
x− − +
′ =−
21 2
( 1)x x
x− − −
=−
23
( 1)x= −
−
26. ( )2
2
( 3)(6 5) 3 5 1 (1)( )
( 3)
w w w wh w
w
− + − + −′ =
−
2 2
26 13 15 3 5 1
( 3)w w w w
w− − − − +
=−
2
23 18 14
( 3)w w
w− −
=−
Chapter 11: Differentiation ISM: Introductory Mathematical Analysis
398
27. ( )
( )
2
22
4 ( 2) (6 2 )(2 )( )
4
z z zh z
z
− − − −′ =
−
( ) ( )2 2 2
2 22 2
2 8 12 4 2 12 8
4 4
z z z z z
z z
− + − + − += =
− −
( )( )
2
22
2 6 4
4
z z
z
− +=
−
28. 2 2
2 2
3 2 3 2
2 2
2
2 2
(3 5 3)(4 5) (2 5 2)(6 5)(3 5 3)
12 35 37 15 (12 40 13 10)(3 5 3)
5 24 25(3 5 3)
x x x x x xzx x
x x x x x xx x
x xx x
+ + + − + − +′ =+ +
+ + + − + + −=
+ +− + +
=+ +
29. ( ) ( )
( )
2 2
22
5 (16 2) 8 2 1 (2 5)
5
x x x x x xy
x x
− − − − + −′ =
−
( )( )
3 2 3 2
22
16 82 10 16 44 12 5
5
x x x x x x
x x
− + − − + −=
− ( )2
22
38 2 5
5
x x
x x
− − +=
−
30. ( )( ) ( )
( )
2 2 3 2
22
1 3 2 1 (2 )( )
1
x x x x x xf x
x
+ − − − +′ =
+
4 3 2 4 3
2 23 2 3 2 2 2 2
( 1)x x x x x x x
x− + − − + −
=+
( )( )
3
22
3 4
1
x x x
x
+ −=
+
31. ( ) ( )
( )
2 2
22
2 3 2 (2 4) 4 3 (4 3)
2 3 2
x x x x x xy
x x
− + − − − + −′ =
− +
( )( )
3 2 3 2
22
4 14 16 8 4 19 24 9
2 3 2
x x x x x x
x x
− + − − − + −=
− +
( )2
22
5 8 1
2 3 2
x x
x x
− +=
− +
ISM: Introductory Mathematical Analysis Section 11.4
399
32. The quotient rule can be used, or we can write
( )4
3 14 1( ) 43 3
zF z z zz
−+= = + ,
so ( )4
2 22
1 3 4( ) 3 43 3
zF z z zz
− −′ = − = .
33. ( ) ( )
( ) ( )
100 99 99
2 2100 100
7 (0) (1) 100 100( )7 7
x x xg xx x
+ −′ = = −
+ +
34. 559 9
22y x
x−−
= = −
6452
y x−′ =
35. 3 3
2 18 8( ) 8v vu v v vv v v
−−= = − = −
32
2 24 2( 4)( ) 2 8 2 vu v v v vv v
− +⎛ ⎞′ = + = + =⎜ ⎟
⎝ ⎠
36. 1 12 25 1 5
88xy x x
x−− ⎛ ⎞= = −⎜ ⎟
⎝ ⎠
312 2
1 3 32 2 2
1 1 5 1 1 5 58 2 2 16 16
xy x xx x x
− − ⎛ ⎞ +⎛ ⎞ ⎜ ⎟′ = + = + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
37. 5 2 13 3 3
13
2 2
33 1 3 1 3x x x xy x x x
x x
−− − − −= = = − −
2 1 4 23 3 3 3
1 43 3
2 1 2 15 53 3 3 3
y x x x xx x
− −′ = − + = − +
43
215 2 1
3
x x
x
− +=
38. ( )( ) ( )( )
( )
2.1 0.7 0.3 1.1
22.1
2 1 0.3 2 4.2
2 1
x x x xy
x
−+ − −′ =
+
1.4 0.7 1.4 1.1
2.1 20.6 0.3 4.2 8.4
(2 1)x x x x
x
−+ − +=
+
( )( )
1.8 2.1
20.7 2.1
0.3 1 28 12
2 1
x x
x x
+ −=
+
39. 2 2( 8)(0) (4)(1) (3 1)(2) (2 )(3)
( 8) (3 1)x x xy
x x− − + −
′ = − +− +
2 24 2
( 8) (3 1)x x= +
− +
40. 2 42
2 42
(3 5)(5) (5 1)(3)( ) 6 6(3 5)
286 6(3 5)
x xq x x xx
x xx
−
−
− − +′ = + +−
= − +−
41. 2[( 2)( 4)](1) ( 5)(2 2)
[( 2)( 4)]x x x xy
x x+ − − − −
′ =+ −
( )2 2
2
2 8 2 12 10
[( 2)( 4)]
x x x x
x x
− − − − +=
+ −
( )2
2
10 18
[( 2)( 4)]
x x
x x
− − +=
+ −
42. 2(9 1)(3 2) 27 15 2
4 5 4 5x x x xy
x x− + + −
= =− −
( )2
2
(4 5 )(54 15) 27 15 2 ( 5)
(4 5 )
x x x xy
x
− + − + − −′ =
−
2 2
2270 141 60 135 75 10
(4 5 )x x x x
x− + + + + −
=−
2135 216 50
(4 5 )x x
x− −
= −−
Chapter 11: Differentiation ISM: Introductory Mathematical Analysis
400
43. ( )s t′( )( ) ( )( )
( ) ( )
2 3 2 4 2
22 3
1 7 (2 3) 3 5 3 14
1 7
t t t t t t t t
t t
⎡ ⎤− + + − + − +⎢ ⎥⎣ ⎦=⎡ ⎤− +⎢ ⎥⎣ ⎦
( )( )6 5 4 3 2
22 3
3 12 6 21 14 21
1 7
t t t t t t
t t
− − + + − − −=
⎡ ⎤− +⎢ ⎥⎣ ⎦
44. 3 217( )
5 10 4f s
s s s=
− +
( )( )
( )
2 2
2 23 2 3 2
0 17 15 20 4 17 15 20 4( )
5 10 4 5 10 4
s s s sf s
s s s s s s
⎡ ⎤− − + − +⎣ ⎦′ = = −− + − +
45.
2( 1) 332( 1)13 3
2 2
x xx xx xy x x
x x
− −−−−
= − = −− −
3 22 23 3
( 1)( 2) 3 2x xx x
x x x x x x+ +
= + = +− − − +
3 2 2
2( 3 2 )(1) ( 2)(3 6 2)3
[ ( 1)( 2)]x x x x x xy
x x x− + − + − +
′ = +− −
3 2
22 3 12 43[ ( 1)( 2)]x x xx x x+ − +
= −− −
46.
2
2 25 2 5 2
3 3 32 22 2 4 2
1 33 12 3 12 3 125 5 7 10
xx x xy x x x
x x x x
+ −+ +
− −= − + = − + = − +
+ + + +
4 2 2 3 5 32 2
4 2 2 2 2 2( 7 10)(2 ) ( 3)(4 14 ) 2 12 6236 36
( 7 10) [( 2)( 5)]x x x x x x x x xy x x
x x x x+ + − − + − + +′ = − + = − +
+ + + +
47. 2 2( )(1) ( )( 1) 2( )
( ) ( )a x a x af x
a x a x− − + −
′ = =− −
48. Simplifying, 1 1
1 1( ) x a ax a xf xax a xx a
− −
− −+ +
= ⋅ =−−
2 2( )(1) ( )( 1) 2( )
( ) ( )a x a x af x
a x a x− − + −
′ = =− −
49. ( )( )2 34 2 5 7 4y x x x x= + − + +
( )( ) ( )2 2 34 2 5 3 7 7 4 (8 2)y x x x x x x′ = + − + + + + +
( 1) ( 3)(10) ( 4)( 6) 6y′ − = − + − − = −
ISM: Introductory Mathematical Analysis Section 11.4
401
50. 3
4 1xy
x=
+
4 2 3 3
4 2( 1)(3 ) ( )(4 )
( 1)x x x xy
x+ −
′ =+
2(2)(3) ( 1)( 4) 1( 1)
2(2)y − − −′ − = =
51. 6
1y
x=
−
2 2( 1)(0) (6)(1) 6
( 1) ( 1)xy
x x− −
′ = = −− −
26 3(3)
22y′ = − = −
The tangent line is 33 ( 3)2
y x− = − − , or 3 152 2
y x= − + .
52. 1 225 5xy x x
x− −+
= = +
2 32 3
1 1010y x xx x
− −′ = − − = − −
(1) 1 10 11y′ = − − = −
The tangent line is y − 6 = −11(x − 1) or y = −11x + 17.
53. ( )4 2(2 3) 2 5 4y x x x⎡ ⎤= + − +⎢ ⎥⎣ ⎦
( )3(2 3) 2 4 10y x x x⎡ ⎤′ = + −⎢ ⎥⎣ ⎦ ( )4 22 5 4 (2)x x⎡ ⎤+ − +⎢ ⎥⎣ ⎦
(0) (3)(0) [2(4)](2) 16y′ = + =
The tangent line is y – 24 = 16(x – 0), or y = 16x + 24.
54. 2 3 21 1
( 4) 4x xy
x x x x+ +
= =− −
( ) ( )( )
3 2 2
23 2
4 (1) ( 1) 3 8
4
x x x x xy
x x
− − + −′ =
−
2( 8)(1) (3)(–4) 4 1(2)
64 16( 8)y − −′ = = =
−
The tangent line is 3 1 ( 2)8 16
y x+ = − , or 1 1
16 2y x= − .
Chapter 11: Differentiation ISM: Introductory Mathematical Analysis
402
55. 2 6
xyx
=−
2 2(2 6)(1) (2) 6
(2 6) (2 6)x xy
x x− − −
′ = =− −
If x = 1, then 1 1
2 6 4y = = −
− and
26 6 3
16 8( 4)y − −′ = = = −
−.
Thus 3814
3 1.52
yy
−′= = =−
.
56. 11
xyx
−=
+
2 2(1 )( 1) (1 )(1) 2
(1 ) (1 )x xy
x x+ − − −
′ = = −+ +
When x = 5, then 1
1823
112
yy
−′= =−
.
57. 32
1s
t=
+. When t = 1, then s = 1 m.
3 2 2
3 2 3 2( 1)(0) 2(3 ) 6
( 1) ( 1)ds t t tvdt t t
+ −= = = −
+ +
If t = 1, then 6 1.54
v = − = − m/s.
58. 237
tst+
=+
2
2 2
2
2 2 2 2
( 7)(1) ( 3)(2 )( 7)
7 6 (7 )(1 )( 7) ( 7)
ds t t tvdt t
t t t tt t
+ − += =
+− − + −
= =+ +
v = 0 when t = −7 or t = 1. Since t is positive, we choose t = 1.
59. p = 50 − 0.01q 250 0.01r pq q q= = −
50 0.02dr qdq
= −
60. 500pq
=
500r pq= =
0drdq
=
61. 108 3
2p
q= −
+
108 32qr pq q
q= = −
+
2( 2)(108) (108 )(1) 3
( 2)dr q qdq q
+ −= −
+
2216 3
( 2)q= −
+
62. 75050
qpq+
=+
2 75050
q qr pqq+
= =+
( )2
2
( 50)(2 750) 750 (1)
( 50)
q q q qdrdq q
+ + − +=
+
2
2100 37,500( 50)
q qq
+ +=
+
63. 0.672dCdI
=
64. 0.712dCdI
=
65. 1/ 2 1/ 33 2C I I= + +
1/ 2 2 / 33 2
1 2 1 202 3 2 3
dC I IdI I I
− −= + + = +
When I = 1, then 1 2 7 .2 3 6
dCdI
= + =
3 2
1 21 12 3
dS dCdI dI I I
= − = − −
When I = 1, then 7 11 1 .6 6
dCdI
− = − = −
ISM: Introductory Mathematical Analysis Section 11.4
403
66. 3 14 6
dCdI I
= −
25
4360I
dCdI =
= , so 25
43 17160 60I
dSdI =
= − =
67. ( )( ) ( )
( )
38 12
2
4 1.2 0.2 16 0.8 0.2
4
I II I I I IdC
dI I
⎛ ⎞+ + − − + −⎜ ⎟⎝ ⎠=+
360.615
I
dCdI =
≈ , so 1 0.615 0.385dSdI
≈ − = when I = 36.
68. ( )( ) ( )
( )
310 12
2
5 0.75 0.4 20 0.5 0.4
5
I II I I I IdC
dI I
⎛ ⎞+ + − − + −⎜ ⎟⎝ ⎠=+
1000.393
I
dCdI =
≈ , so 1 0.393 0.607dSdI
≈ − = when I = 100.
69. Simplifying gives 1210 0.7 0.2C I I= + −
a. 12 0.10.7 0.1 0.7dC I
dI I−= − = −
0.11 0.3dS dCdI dI I
= − = +
25
0.10.3 0.325I
dSdI =
= + =
b. dCdIC
when I = 25 is 0.150.7
0.02610 0.7(25) 0.2(5)
−≈
+ −
70. Simplifying S gives
( )( )2 42 8 42 2
I II IS II I
+ −− −= = = −
+ +
Thus 1/ 21 1 .2 2
dS IdI I
−= =
150
1 0.040822 150I
dSdI =
= ≈⋅
and 150
1 0.04082 0.9592.I
dCdI =
≈ − ≈
71. 2 2
2 2 2( 2)(2 ) (1) 4 6 ( 4)6 6
( 2) ( 2) ( 2)dc q q q q q q qdq q q q
+ − + += ⋅ = ⋅ =
+ + +
Chapter 11: Differentiation ISM: Introductory Mathematical Analysis
404
72. We assume that ( ) 0d cdq
= . Thus 2
(1)0
dcdqq cdc d c
dq dq q q
⋅ −⎛ ⎞= = =⎜ ⎟
⎝ ⎠.
This implies that 0dcq cdq⋅ − = ,
dcq cdq⋅ = ,
dc c cdq q
= = , so the marginal cost function dcdq
⎛ ⎞⎜ ⎟⎝ ⎠
and the average
cost function ( )c are equal.
73. 900
10 45xy
x=
+
2(10 45 )(900) (900 )(45)
(10 45 )dy x xdx x
+ −=
+
22
(100)(900) (1800)(45) 910(100)x
dydx =
−= =
74. 0.05RT VA xV
=+
2( )(0.05) (0.05 )( )(RT)
( )d A xV V x
dV A xV+ −
=+
20.05
( )A
A xV=
+
Both numerator and denominator are always positive, so (RT) 0ddV
> . This rate of change means that if V
increases by one unit, RT increases.
75. 0.7355
1 0.02744xy
x=
+
2(1 0.02744 )(0.7355) (0.7355 )(0.02744)
(1 0.02744 )dy x xdx x
+ −=
+
20.7355
(1 0.02744 )x=
+
76. (1 ) (2 )( )
(2 )(1 ) (2 )a x b n xf x
a n x b n x+ − +
=+ + − +
For convenience let c = 2 + n.
Then (1 ) 1 (1 )( )(1 ) (1 )
a x bcx a x bcxf xac x bcx c a x bx
+ − + −= = ⋅
+ − + −.
21 [ (1 ) ]( ) [ (1 ) ]( )( )
[ (1 ) ]a x bx a bc a x bcx a bf x
c a x bx+ − − − + − −
′ = ⋅+ −
2 21 1 ( 1)
[ (1 ) ] [ (1 ) ]abc ab c ab
c ca x bx a x bx− + − +
= ⋅ = ⋅+ − + −
2 21 [ 1(2 ) 1] (1 )
2 [ (1 ) ] [ (1 ) ] (2 )n ab n ab
n a x bx a x bx n− + + − +
= ⋅ =+ + − + − +
( ) A Bxg xC Dx+
=+
ISM: Introductory Mathematical Analysis Section 11.5
405
2
2
2
( )( ) ( )( )( )( )
( )
( )
C Dx B A Bx Dg xC Dx
CB BDx AD BDxC Dx
BC ADC Dx
+ − +′ =+
+ + −=
+−
=+
Thus, ( )g x′ has the form given. When ( )g x′ is defined for ,CxD
⎛ ⎞≠⎜ ⎟⎝ ⎠
its sign is constant.
77. 2
(1)dcdqq cdc d c
dq dq q q
⋅ −⎛ ⎞= =⎜ ⎟
⎝ ⎠. When q = 20 we have
2 220(125) 20(150)
(20) 1150 120
dcdqq c
dcdq q
c c
⋅ − −
= = = −
78.
2
(3)(2 1)( 4) (3 1)(2)( 4) (3 1)(2 1)(1)
18 50 3
dy x x x x x xdx
x x
= − − + + − + + −
= − +
Principles in Practice 11.5
1. By the chain rule,
( )24 (6 ) (8 )(6) 48dy dy dx d dx t x xdt dx dt dx dt
= ⋅ = ⋅ = = .
Since x = 6t, 48(6 ) 288dy t tdt
= = .
Problems 11.5
1. (2 2)(2 1)dy dy du u xdx du dx
= ⋅ = − − ( )22 2 (2 1)x x x⎡ ⎤= − − −⎢ ⎥⎣ ⎦ ( )22 2 2 (2 1)x x x= − − − 3 24 6 2 2x x x= − − +
2. ( )( )2 26 8 7 3dy dy du u xdx du dx
= ⋅ = − − ( )( )6 4 2 22 3 42 147 4 7 3x x x x= − + − −
3. 3 3 32 2 2( 1)
(2 )dy dy dwdx dw dx w w x
⎛ ⎞= ⋅ = − − = =⎜ ⎟
−⎝ ⎠
4. 4 3
3/ 4 4 35 4 34
1 5 4(5 4 )4 4 ( 3)
dy dy dz x xz x xdx dz dx x x
− −= ⋅ = − =
⎛ ⎞− +⎜ ⎟⎝ ⎠
5. 2 22 2
( 1) ( 1) 2(3 ) 3 .( 1) ( 1)
dw dw du t tu udt du dt t t
⎡ ⎤ ⎡ ⎤+ − −= ⋅ = =⎢ ⎥ ⎢ ⎥
+ +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ If t = 1, then
1 1 0,1 1
u −= =
+ so 2
1
23(0) 04t
dwdt =
⎡ ⎤= =⎢ ⎥⎣ ⎦.
Chapter 11: Differentiation ISM: Introductory Mathematical Analysis
406
6. 12 (4 )
2dz dz du u sds du ds u
⎛ ⎞= ⋅ = +⎜ ⎟
⎝ ⎠. If s = –1, then
u = 1, so 1
5 ( 4) 102s
dzds =−
⎛ ⎞= − = −⎜ ⎟⎝ ⎠
7. (6 8)(4 )dy dy dw w xdx dw dx
= ⋅ = − . If x = 0, then
0dydx
= .
8. ( )29 2 7 (5)dy dy du u udx du dx
= ⋅ = − + . If x = 1, then
u = 3, so 1
(82)(5) 410x
dydx =
= =
9. 56(3 2) (3 2)dy x xdx
′ = + ⋅ +
5 56(3 2) (3) 18(3 2)x x= + = +
10. ( ) ( )32 24 4 4dy x xdx
′ = − ⋅ −
( ) ( )3 32 24 4 (2 ) 8 4x x x x= − = −
11. 3 4 3
3 4 2
2 3
5(3 2 ) (3 2 )
5(3 2 ) (6 )30 (3 2 )
dy x xdx
x xx x
′ = + ⋅ +
= +
= +
12. 2 3 2
2 3
2 3
4( ) ( )
4( ) (2 1)4(2 1)( )
dy x x x xdx
x x xx x x
′ = + +
= + +
= + +
13. ( ) ( )993 2 3 22 100 8 8dy x x x x x xdx
′ = ⋅ − + ⋅ − +
( ) ( )993 2 2200 8 3 16 1x x x x x= − + − +
( )( )992 3 2200 3 16 1 8x x x x x= − + − +
14. ( ) ( )
4242
2 1 1 2 12 2
xy x
+= = +
( )( )
32 2
32 3 2
1 4 2 1 (2 1)2
2(2 1) (4 ) 8 2 1
dy x xdx
x x x x
′ = ⋅ + +
= + = +
15. ( ) ( )42 23 2 2dy x xdx
−′ = − − ⋅ −
( ) ( )4 42 23 2 (2 ) 6 2x x x x− −
= − − = − −
16. 3 13 3
2 3 13
12(2 8 ) (2 8 )
12(6 8)(2 8 )
dy x x x xdx
x x x
−
−
′ = − − ⋅ −
= − − −
17. 2 12 / 7 2
2 12 / 7
52 ( 5 2) ( 5 2)7
10 (2 5)( 5 2)7
dy x x x xdx
x x x
−
−
⎛ ⎞′ = − + − ⋅ + −⎜ ⎟⎝ ⎠
= − + + −
18. ( ) ( )524 334 7 7 4
2y x x x
−⎛ ⎞′ = − − −⎜ ⎟⎝ ⎠
( )( )523 46 7 4 7x x x
−= − − −
19. ( )122 25 5y x x x x= − = −
( )1221 5 (10 1)
2y x x x
−′ = − −
( )1221 (10 1) 5
2x x x
−= − −
20. ( )122 23 7 3 7y x x= − = −
( ) ( )1 12 22 21 3 7 (6 ) 3 3 7
2y x x x x
− −′ = − = −
21. 144 2 1 (2 1)y x x= − = −
3 34 41 1(2 1) (2) (2 1)
4 2y x x− −′ = − = −
22. ( )133 2 28 1 8 1y x x= − = −
( ) ( )2 23 32 21 168 1 (16 ) 8 1
3 3y x x x x
− −′ = − = −
ISM: Introductory Mathematical Analysis Section 11.5
407
23. ( ) ( )2523 352 1 2 1y x x= + = +
( ) ( ) ( )3 35 53 2 2 32 122 1 3 1
5 5y x x x x
− −⎛ ⎞′ = + = +⎜ ⎟⎝ ⎠
24. 5 5 5 5 / 337 ( 3) 7( 3)y x x= − = −
5 2 / 3 4
4 5 2 / 3
57 ( 3) (5 )3
175 ( 3)3
y x x
x x
′ = ⋅ −
= −
25. ( ) 122
6 6 2 12 1
y x xx x
−= = − +
− +
( ) 226( 1) 2 1 (4 1)y x x x−
′ = − − + −
( ) 226(4 1) 2 1x x x−
= − − − +
26. ( ) 1443 3 2
2y x
x
−= = +
+
( ) ( ) ( )2 24 3 3 43( 1) 2 4 12 2y x x x x− −
′ = − + = − +
27.
( )( ) 22
22
1 33
y x xx x
−= = −
−
( ) 322 3 (2 3)y x x x−
′ = − − −
( ) 322(2 3) 3x x x−
= − − −
28. 44
1 (2 )(2 )
y xx
−= = ++
5 54(2 ) (1) 4(2 )y x x− −′ = − + = − +
29. 2 1/ 22
4 4(9 1)9 1
y xx
−= = ++
2 3/ 2
2 3/ 2
14 (9 1) (18 )2
36 (9 1)
y x x
x x
−
−
⎛ ⎞′ = − +⎜ ⎟⎝ ⎠
= − +
30.
( )( )
23
23
2
2
3 3 33
y x xx x
−= = −
−
( )53223 3 (6 1)
3y x x x
−⎛ ⎞′ = − − −⎜ ⎟⎝ ⎠
( )53–22(6 1) 3x x x= − − −
31. 133 3 37 7 (7 ) 7y x x x x= + = +
2 23 33 31 7(7 ) (7) 7(1) (7 ) 7
3 3y x x− −′ = + = +
32. 1 12 212 (2 ) (2 )
2y x x x
x−
= + = +
312 21 1(2 ) (2) (2 ) (2)
2 2y x x− −⎛ ⎞ ⎛ ⎞′ = + −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
312 2(2 ) (2 )x x− −
= −
33. 2 4 55( 4) (1) ( 4) (2 )y x x x x⎡ ⎤′ = − + −⎣ ⎦
4( 4) [5 2( 4)]x x x x= − + − 4( 4) (7 8)x x x= − −
34. 3 4
3 3
4( 4) (1) ( 4) (1)
( 4) (4 4) ( 4) (5 4)
y x x x
x x x x x
⎡ ⎤′ = + + +⎣ ⎦= + + + = + +
35. 122 24 5 1 4 (5 1)y x x x x= + = +
12
12
2
2
14 (5 1) (5) 5 1(8 )2
10 (5 1) 8 5 1
y x x x x
x x x x
−
−
⎛ ⎞′ = + + +⎜ ⎟⎝ ⎠
= + + +
Chapter 11: Differentiation ISM: Introductory Mathematical Analysis
408
36. 123 2 3 24 1 4 (1 )y x x x x= − = −
3 2 1/ 2 2 2
42 2
2
14 (1 ) ( 2 ) 1 (12 )2
4 12 11
y x x x x x
x x xx
−⎡ ⎤⎛ ⎞′ = − − + −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
= − + −−
37. ( ) ( )3 22 22 1 (5) (5 ) 3 2 1 (2 2)y x x x x x x⎡ ⎤′ = + − + + − +⎢ ⎥
⎣ ⎦
( ) ( )22 25 2 1 2 1 3 (2 2)x x x x x x⎡ ⎤= + − + − + +⎢ ⎥⎣ ⎦
( ) ( )22 25 2 1 7 8 1x x x x= + − + −
38. ( ) ( ) ( )3 42 3 2 34 1 3 1 (2 )y x x x x x⎡ ⎤′ = − + −⎢ ⎥
⎣ ⎦
( ) ( ) ( )( )3 33 3 3 3 32 1 6 1 2 7 1 1x x x x x x x⎡ ⎤= − + − = − −⎢ ⎥⎣ ⎦
39. 3 3 4 2(8 1) 4(2 1) (2) (2 1) 3(8 1) (8)y x x x x⎡ ⎤ ⎡ ⎤′ = − + + + −⎣ ⎦ ⎣ ⎦
2 38(8 1) (2 1) [(8 1) 3(2 1)]x x x x= − + − + + 2 38(8 1) (2 1) (14 2)x x x= − + +
2 316(8 1) (2 1) (7 1)x x x= − + +
40. 5 2 4
4
4
(3 2) [2(4 5)(4)] (4 5) [5(3 2) (3)](3 2) (4 5)[8(3 2) 15(4 5)](3 2) (4 5)(84 59)
y x x x xx x x xx x x
′ = + − + − +
= + − + + −
= + − −
41. 11
2
11
2
11
13
3 ( 2)(1) ( 3)(1)122 ( 2)
3 5122 ( 2)
60( 3)( 2)
x x xyx x
xx xx
x
⎡ ⎤− + − −⎛ ⎞′ = ⎢ ⎥⎜ ⎟+⎝ ⎠ +⎢ ⎥⎣ ⎦⎡ ⎤−⎛ ⎞= ⎢ ⎥⎜ ⎟+⎝ ⎠ +⎢ ⎥⎣ ⎦
−=
+
42. 3 3
2 52 ( 2)(2) 2 (1) 1284
2 ( 2) ( 2)x x x xy
x x x
⎡ ⎤+ −⎛ ⎞′ = =⎢ ⎥⎜ ⎟+⎝ ⎠ + +⎢ ⎥⎣ ⎦
43.
12
21 2 ( 3)(1) ( 2)(1)2 3 ( 3)
x x xyx x
− ⎡ ⎤− + − −⎛ ⎞′ = ⎢ ⎥⎜ ⎟+⎝ ⎠ +⎢ ⎥⎣ ⎦
12
2 25 2 5 3
3 22( 3) 2( 3)x xx xx x
−− +⎛ ⎞= =⎜ ⎟+ −⎝ ⎠+ +
ISM: Introductory Mathematical Analysis Section 11.5
409
44. ( ) ( )
( )
23 2 22
2 22
2 (16 ) 8 3 (2 )1 8 33 2 2
x x x xxyx x
− ⎡ ⎤+ − −⎛ ⎞ ⎢ ⎥−
′ = ⎜ ⎟ ⎢ ⎥⎜ ⎟+ ⎢ ⎥⎝ ⎠ +⎣ ⎦
( )
232
2 22
1 8 3 383 2 2
x xx x
−⎛ ⎞−
= ⎜ ⎟⎜ ⎟+⎝ ⎠ +
( ) ( )2 43 32 2
38
3 8 3 2
x
x x=
− +
45. ( ) ( )
( )
3 22 2
62
4 (2) (2 5) 3 4 (2 )
4
x x x xy
x
⎡ ⎤+ − − +⎢ ⎥
⎣ ⎦′ =+
( ) ( ){ }( )
22 2
62
4 4 (2) (2 5)[3(2 )]
4
x x x x
x
+ + − −=
+
( ) ( )2 2 2
4 42 2
2 8 12 30 10 30 8
4 4
x x x x x
x x
+ − + − + += =
+ +
( )( )
2
42
2 5 15 4
4
x x
x
− − −=
+
46. 2 3 4
2 2
3 2
2 2
3 2
2 2
(3 7)[4(4 2) (4)] (4 2) (6 )(3 7)
(4 2) [16(3 7) 6 (4 2)](3 7)
(4 2) (24 12 112)(3 7)
x x x xyx
x x x xx
x x xx
+ − − −′ =+
− + − −=
+
− + +=
+
47.
3 4 5 2
6
(3 1) 5(8 1) (8) (8 1) 3(3 1) (3)
(3 1)
x x x xy
x
⎡ ⎤ ⎡ ⎤− − − − −⎣ ⎦ ⎣ ⎦′ =−
2 4
6(3 1) (8 1) [(3 1)(40) (8 1)(9)]
(3 1)x x x x
x− − − − −
=−
4
4(8 1) (48 31)
(3 1)x x
x− −
=−
Chapter 11: Differentiation ISM: Introductory Mathematical Analysis
410
48. 2 2 1/ 33 ( 2) ( 2) [( 2) ( 2)]y x x x x= − + = − +
2 2 / 3 2
2 2 / 3
2 2 / 3
1/ 3 2 / 3
1[( 2) ( 2)] [(1)( 2) 2( 2)( 2)]31[( 2) ( 2)] ( 2)[ 2 2( 2)]31[( 2) ( 2)] ( 2)(3 2)31 ( 2) ( 2) (3 2)3
y x x x x x
x x x x x
x x x x
x x x
−
−
−
− −
′ = − + − + − +
= − + − − + +
= − + − +
= − + +
49. ( ) ( )( )122 4 2 46 5 2 5 6 5 2 5y x x x x
⎡ ⎤= + + = + +⎢ ⎥
⎣ ⎦
( ) ( ) ( ) ( )1 12 22 4 3 416 5 2 5 4 5 (10 )
2y x x x x x
−⎡ ⎤′ = + ⋅ + + +⎢ ⎥
⎣ ⎦
( )( ) ( ) ( )1 12 22 4 3 46 5 2 5 2 5 (10 )x x x x x
−⎡ ⎤= + + + +⎢ ⎥
⎣ ⎦
( )( ) ( ) ( )1 12 22 4 2 412 5 2 5 5 (5)x x x x x
−⎡ ⎤= + + + +⎢ ⎥
⎣ ⎦
Factoring out ( )124 5x
−+ gives
( ) ( )( ) ( )124 2 2 412 5 5 2 5 (5)y x x x x x
− ⎡ ⎤′ = + + + +⎢ ⎥⎣ ⎦
( ) ( )124 4 212 5 10 2 25x x x x
−= + + +
50. 23 4 (2)(7 1)(7) (7 1) (1)y x x x⎡ ⎤′ = − + + +⎣ ⎦
2 23 4 147 28 1 588 112 1x x x x⎡ ⎤= − + + = − − −⎣ ⎦
51. 2( 4)(1) ( 1)(1) 8 7 18 2 8
4 4( 4)t t ty
t+ − − −⎛ ⎞⎛ ⎞′ = + − ⋅⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠+
2 25 58 (8 7) 15 8
( 4) ( 4)t t
t t= + − − = − +
+ +
ISM: Introductory Mathematical Analysis Section 11.5
411
52. 3 4 3
2 2(2 6)(7 5) 14 10 42 30
(2 4) (2 4)x x x x xy
x x+ − − + −
= =+ +
2 3 2 4 3
4
3 2 4 3
4
4 3 3 2 4 3
3
4 3 2
(2 4) (56 30 42) (14 10 42 30)[2(2 4)(2)](2 4)
(2 4)[(2 4)(56 30 42) 4(14 10 42 30)](2 4)
112 60 84 224 120 168 56 40 168 120(2 4)
4(14 51 30 21 72)
x x x x x x xyx
x x x x x x xx
x x x x x x x xx
x x x x
+ − + − − + − +′ =+
+ + − + − − + −=
+− + + − + − − − +
=+
+ − − += 3(2 4)x +
53. 3 5 3 2 2 3 2 3 4 2
3 10( 5) [(2 1) (2)( 3)(1) ( 3) (3)(2 1) (2)] (2 1) ( 3) [5( 5) (3 )]
( 5)x x x x x x x x xy
x− + + + + + − + + −
′ =−
54. ( ) ( ) ( ) ( )1
22 22 2 21
2
2
(9 3) 2(2) 4 1 (8 ) 4 1 ( 2) 2 4 1 (9)
(9 3)
x x x x x x x xy
x
−⎡ ⎤− + − + − + − + −⎢ ⎥⎣ ⎦′ =
−
55. ( )32 23(5 6) (5) 4 1 (2 )dy dy du u x xdx du dx
⎡ ⎤⎡ ⎤= ⋅ = + +⎢ ⎥⎣ ⎦ ⎣ ⎦
When x = 0, then 0dydx
= .
56. (4 4)(6)(2)dz dz dy dx ydt dy dx dt
= ⋅ ⋅ = −
When t = 1, then x = 2 and y = 7. Thus 1
(24)(6)(2) 288t
dzdt =
= = .
57. ( )223 7 8 (2 7)y x x x′ = − − −
If x = 8, then slope 23(64 56 8) (16 7) 0y= ′ = − − − = .
58. 12( 1)y x= +
121 ( 1)
2y x −′ = +
If x = 8, 16
y′ = .
Chapter 11: Differentiation ISM: Introductory Mathematical Analysis
412
59. ( )232 8y x= −
( )( )
13
13
2
2
2 48 (2 )3
3 8
xy x xx
−′ = − =
−
If x = 3, then 12 43(1)
y′ = = . Thus the tangent line
is y – 1 = 4(x – 3), or y = 4x – 11.
60. 2 23( 3) (1) 3( 3)y x x′ = + = +
If x = −1, 23(2) 12.y′ = =
The tangent line is y − 8 = 12(x + 1) or y = 12x + 20.
61. ( )
121
22
( 1) (7 2) (7) 7 2(1)
( 1)
x x xy
x
−+ + − +
′ =+
( )7 12 7 2
2
( 1) 7 2
( 1)x
x x
x+
+ − +=
+
If x = 1, then ( )( )7 1
2 32 3 14 6
y−
′ = = − . The
tangent line is 3 1 ( 1)2 6
y x− = − − , or
1 56 3
y x= − + .
62. ( ) 323 3 1y x−
= − +
( ) 423( 3) 3 1 (6 )y x x−
′ = − − +
If x = 0, then 0y′ = . The tangent line is
y + 3 = 0(x – 0), or y = –3.
63. ( )32 9y x= + and ( )226 9y x x′ = + . When
x = 4, then 3(25)y = and 26(4)(25)y′ = , so 2
36(4)(25) 24(100) (100) (100) 96%
25(25)yy′
= = =
64. 2 31
( 1)y
x=
− and 2 4
6( 1)
xyx
′ = −−
When x = 2, 127
y = and 412 4 ,
273y′ = − = − so
4(100) 27(100) 400%27
yy′⎛ ⎞
= − ⋅ = −⎜ ⎟⎝ ⎠
65. q = 5m, p = –0.4q + 50; m = 6 dr dr dqdm dq dm
= ⋅
20.4 50 ,r pq q q= = − + 0.8 50,dr qdq
= − + . For
m = 6, then q = 30, so 6
24 50 26.m
drdq =
= − + =
Also, 5.dqdm
= Thus 6
(26)(5) 130.m
drdm =
= =
66. ( )21 20020
q m m= −
p = –0.1q + 70; m = 40 dr dr dqdm dq dm
= ⋅
20.1 70r pq q q= = − + , so 0.2 70dr qdq
= − + . If
m = 40, then q = 320, so
4064 70 6
m
drdq =
= − + = .
1 (200 2 )20
dq mdm
= − . When m = 40, 6dqdm
= .
Thus 40
(6)(6) 36m
drdm =
= = .
67. 2
2
10
9
mqm
=+
5253
pq
=+
; m = 4
dr dr dqdm dq dm
= ⋅
5253qr pq
q= =
+, so
2 2( 3)(1) (1) 1575525
( 3) ( 3)dr q qdq q q
+ −= ⋅ =
+ +.
ISM: Introductory Mathematical Analysis Section 11.5
413
If m = 4, then q = 32, so 4
1575 91225 7m
drdq =
= = .
( ) ( )1 12 22 2 21
22
9 (20 ) 10 9 (2 )
9
m m m m mdqdm m
−+ − ⋅ +
=+
( ) ( )122 2 3
2
9 20 9 10
9
m m m m
m
− ⎡ ⎤+ + −⎢ ⎥⎣ ⎦=+
( )32
3
2
10 180
9
m m
m
+=
+
When m = 4, then
32
10(64) 180(4) 1360 272125 25(25)
dqdm
+= = = . Thus
4
9 272 13.997 25m
drdm =
= ⋅ ≈ .
68. 2
100
19
mqm
=+
450010
pq
=+
; m = 9
dr dr dqdm dq dm
= ⋅
450010
qr pqq
= =+
, so 245,000
( 10)drdq q
=+
.
If m = 9, then q = 90, so 9
92m
drdq =
= .
( )322
1900
19
dqdm
m=
+. When m = 9, then
1910
dqdm
= .
Thus 9
9 19 8.552 10m
drdm =
= ⋅ = .
69. a. ( )122
2
10 20 (2 )2 20
dp qq qdq q
− −= − + =
+
b. 2 20
2100 20
qdpqdq
p q
−
+=
− +
2 220 100 20
q
q q= −
⎛ ⎞+ − +⎜ ⎟⎝ ⎠
2 2100 20 20
q
q q= −
+ − −
c. 2100 20r pq q q q= = − +
drdq
( )122 21100 20 (2 ) 20(1)
2q q q q
−⎡ ⎤= − ⋅ + + +⎢ ⎥
⎣ ⎦
22
2100 20
20
q qq
= − − ++
70. kpq
= ; q = f(m)
dr dr dqdm dq dm
= ⋅
r = pq = k, so 0drdq
= . Thus 0 0dr dqdm dm
= ⋅ = .
71. (12 0.4 )( 1.5)dc dc dq qdp dq dp
= ⋅ = + −
When p = 85, then q = 772.5, so
85481.5.
p
dcdp =
= −
72. 3250( ) 1
250f t
t⎛ ⎞= − ⎜ ⎟+⎝ ⎠
2
2250 250( ) 3
250 (250 )f t
t t
⎡ ⎤⎛ ⎞′ = − −⎢ ⎥⎜ ⎟+⎝ ⎠ +⎢ ⎥⎣ ⎦
2
2250 250(100) 3350 35025 1349 490
15 .4802
f ⎡ ⎤⎛ ⎞′ = − −⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦⎛ ⎞⎛ ⎞= − −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
=
Thus when t increases from 100 to 101, the proportion discharged increases by
approximately 15 .
4802
Chapter 11: Differentiation ISM: Introductory Mathematical Analysis
414
73. ( ) ( ) ( )
1 12 22 2 21
2
2
3 (10 ) 5 3 (2 )
3
q q q q qdcdq q
−⎡ ⎤+ − +⎢ ⎥
⎣ ⎦=+
Multiplying numerator and denominator by ( )122 3q + gives
( )( )
32
2 2
2
3 (10 ) 5 ( )
3
q q q qdcdq
q
+ −=
+ ( )( )
( )3 32 2
23
2 2
5 65 30
3 3
q qq q
q q
++= =
+ +.
74. a. 680 4360dS EdE
= − . If E = 16, 6520dSdE
= .
b. Solving 680 4360 5000E − = gives 680 9360, 13.8.E E= ≈
75. ( )2 8 74 10 (2 ) 10dV dV dr r tdt dr dt
− −⎡ ⎤= ⋅ = π +⎣ ⎦ . When t = 10, then ( )8 2 710 10 10 (10)r − −= + 6 6 610 10 2(10)− − −= + = .
Thus 26 8 7
104 2(10) 10 (2)(10) 10
t
dVdt
− − −
=
⎡ ⎤ ⎡ ⎤= π +⎣ ⎦ ⎣ ⎦ ( )12 7 194 4(10) 3 10 48 (10)− − −⎡ ⎤⎡ ⎤= π = π⎢ ⎥⎣ ⎦ ⎣ ⎦
76. a. 1 12 21 (2 ) (2 ) (2 )
2dp VI V V VIdI
ρ ρ ρ ρ− −= =
b.
12
12
(2 ) 12(2 )
dpdI V VIp IVI
ρ ρ
ρ
−
= =
77. a. 3 3( ) 0.001416 0.01356 1.696 34.9xd I x x xdx
= − + + −
If x = 65, ( ) 256.238.xd Idx
= −
b. If x = 65, ( ) 256.238 0.01578
16, 236.484
dxdx
x
I
I−
≈ ≈ −
If x = 65, the percentage rate of change is ( ) 25,623.8 1.578%.
16,236.484
dxdx
x
I
I−
⋅ = = −
78. (P + a)(v + b) = k kv b
P a+ =
+
kv bP a
= −+
1( )v k P a b−= + −
22( 1)( )
( )dv kk P adP P a
−= − + = −+
ISM: Introductory Mathematical Analysis Section 11.5
415
79. By the chain rule, dc dc dqdp dq dp
= ⋅ . We are given that 1100 100q pp
−= = , so 22
100100dq pdp p
− −= − = . Thus
2100dc dc
dp dq p
⎡ ⎤−= ⎢ ⎥
⎢ ⎥⎣ ⎦. When q = 200, then
100 1200 2
p = = and we are given that 0.01dcdq
= . Therefore
( )212
1000.01 4dcdp
⎡ ⎤−⎢ ⎥= = −⎢ ⎥⎢ ⎥⎣ ⎦
.
80. a. When m = 12, then q = 3000, so r = 1500.
Thus 1500 1 $0.503000 2
rpq
= = = = .
b. ( )
121
21000 3 (50) 50 (1000 3 ) (3)
1000 3
q q qdrdq q
−+ − +
=+
3000
2750 1110,000 40q
drdq =
= =
c. dr dr dqdm dq dm
= ⋅ . From part (b) we know drdq
. Now,
312 23(2 ) (2 1) (2) (2 1) (2)
2dq m m mdm
⎛ ⎞= + + +⎜ ⎟⎝ ⎠
, so 12
610m
dqdm =
= .
Thus 12
11 67161040 4m
drdm =
= ⋅ = .
81. ( ) ( )dy dy dx f x g tdt dx dt
= ⋅ = ′ ′ . We are given that g(2) = 3, so x = 3 when t = 2. Thus
2 (2) 2(3) (2) 10(4) 40
t x g t
dy dy dx f gdt dx dt= = =
= ⋅ = ′ ′ = = .
82. a. 2
324 5 19 19 19lim lim 0 018 18 1835q q
cqq→∞ →∞
⎛ ⎞⎜ ⎟= + + = + + =⎜ ⎟+⎝ ⎠
b. 2
324 1951835
qc cq qq
= = + ++
( )( )122 21
22
35(324) 324 35 (2 ) 191835
q q q qdcdq q
−+ − +
= ++
173
q
dcdq =
=
c. From part (b) the increase in cost of the additional unit is approximately $300. Since the corresponding revenue increases by $275, the move should not be made.
Chapter 11: Differentiation ISM: Introductory Mathematical Analysis
416
83. 86,111.37
84. 5.25
Chapter 11 Review Problems
1. 2( ) 2f x x= −
( )2 2
0 0
2 ( ) 2( ) ( )( ) lim limh h
x h xf x h f xf xh h→ →
⎡ ⎤− + − −+ − ⎣ ⎦′ = =
( )2 2 2 2
0 0
2 2 2 2lim limh h
x hx h x hx hh h→ →
⎡ ⎤− − − − − − −⎣ ⎦= =
0 0
(2 )lim lim (2 ) 2h h
h x h x h xh→ →
− += = − + = −
2. 2( ) 2 3 1f x x x= − +
0
( ) ( )( ) limh
f x h f xf xh→
+ −′ =
( )2 2
0
2( ) 3( ) 1 2 3 1limh
x h x h x x
h→
⎡ ⎤+ − + + − − +⎣ ⎦=
( )2 2 2
0
2 4 2 3 3 1 2 3 1limh
x hx h x h x x
h→
⎡ ⎤+ + − − + − − +⎣ ⎦=
2
0 0
4 2 3 (4 2 3)lim limh h
hx h h h x hh h→ →
+ − + −= =
0lim (4 2 3) 4 3h
x h x→
= + − = −
3. ( ) 3f x x=
0 0
3( ) 3( ) ( )( ) lim limh h
x h xf x h f xf xh h→ →
+ −+ −′ = =
0
3( ) 3 3( ) 3lim
3( ) 3h
x h x x h xh x h x→
+ − + += ⋅
+ +
( ) ( )0 0
3( ) 3 3lim lim3( ) 3 3( ) 3h h
x h x hh x h x h x h x→ →
+ −= =
+ + + +
0
3lim3( ) 3h x h x→
=+ +
3 3 33 3 2 3 2x x x x
= = =+