Section 3.2 The Product and Quotient Rules
44. If g(x) = exx, find g(n)(x).
Solution:
186 ¤ CHAPTER 3 DIFFERENTIATION RULES 41. () = 2
1 + ⇒ 0() =(1 + )(2) − 2(1)
(1 + )2 = 2 + 22− 2
(1 + )2 = 2+ 2
2+ 2 + 1 ⇒
00() = (2+ 2 + 1)(2 + 2) − (2+ 2)(2 + 2)
(2+ 2 + 1)2 = (2 + 2)(2+ 2 + 1 − 2− 2) [( + 1)2]2
= 2( + 1)(1)
( + 1)4 = 2 ( + 1)3, so 00(1) = 2
(1 + 1)3 = 2 8 = 1
4.
42. () =
⇒ 0() = · 1 − ·
()2 = (1 − )
()2 = 1 −
⇒
00() = · (−1) − (1 − )
()2 = [−1 − (1 − )]
()2 = − 2
⇒
000() = · 1 − ( − 2)
()2 = [1 − ( − 2)]
()2 = 3 −
⇒
(4)() = · (−1) − (3 − )
()2 = [−1 − (3 − )]
()2 = − 4
. The pattern suggests that ()() = ( − )(−1)
. (We could use mathematical induction to prove this formula.) 43. We are given that (5) = 1, 0(5) = 6, (5) = −3, and 0(5) = 2.
(a) ()0(5) = (5)0(5) + (5)0(5) = (1)(2) + (−3)(6) = 2 − 18 = −16
(b)
0
(5) =(5)0(5) − (5)0(5)
[(5)]2 = (−3)(6) − (1)(2)
(−3)2 = −20 9
(c)
0
(5) = (5)0(5) − (5)0(5)
[ (5)]2 = (1)(2) − (−3)(6) (1)2 = 20 44. We are given that (4) = 2, (4) = 5, 0(4) = 6, and 0(4) = −3.
(a) () = 3() + 8() ⇒ 0() = 30() + 80(), so
0(4) = 30(4) + 80(4) = 3(6) + 8(−3) = 18 − 24 = −6.
(b) () = () () ⇒ 0() = () 0() + () 0(), so
0(4) = (4) 0(4) + (4) 0(4) = 2(−3) + 5(6) = −6 + 30 = 24.
(c) () = ()
() ⇒ 0() = () 0() − () 0() [()]2 , so
0(4) = (4) 0(4) − (4) 0(4)
[(4)]2 = 5(6) − 2(−3)
52 = 30 + 6 25 = 36
25. (d) () = ()
() + () ⇒
0(4) = [ (4) + (4)] 0(4) − (4) [0(4) + 0(4)]
[ (4) + (4)]2 = (2 + 5) (−3) − 5 [6 + (−3)]
(2 + 5)2 = −21 − 15
72 = −36 49
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50. If f (2) = 10 and f0(x) = x2f (x) for all x, find f00(2).
Solution:
SECTION 3.2 THE PRODUCT AND QUOTIENT RULES ¤ 187 45. () = () ⇒ 0() = 0() + ()= [0() + ()]. 0(0) = 0[0(0) + (0)] = 1(5 + 2) = 7
46.
()
= 0() − () · 1
2 ⇒
()
=2
= 20(2) − (2)
22 = 2(−3) − (4)
4 = −10
4 = −25 47. () = () ⇒ 0() = 0() + () · 1. Now (3) = 3(3) = 3 · 4 = 12 and
0(3) = 30(3) + (3) = 3(−2) + 4 = −2. Thus, an equation of the tangent line to the graph of at the point where = 3 is − 12 = −2( − 3), or = −2 + 18.
48. 0() = 2 () ⇒ 00() = 20() + () · 2. Now 0(2) = 22 (2) = 4(10) = 40, so
00(2) = 22(40) + 10(4) = 200.
49. (a) From the graphs of and , we obtain the following values: (1) = 2 since the point (1 2) is on the graph of ;
(1) = 1since the point (1 1) is on the graph of ; 0(1) = 2since the slope of the line segment between (0 0) and (2 4)is 4 − 0
2 − 0 = 2; 0(1) = −1 since the slope of the line segment between (−2 4) and (2 0) is 0 − 4
2 − (−2) = −1.
Now () = ()(), so 0(1) = (1)0(1) + (1) 0(1) = 2 · (−1) + 1 · 2 = 0.
(b) () = ()(), so 0(5) = (5)0(5) − (5)0(5) [(5)]2 = 2
−13
− 3 · 23
22 = −83
4 = −2 3 50. (a) () = () (), so 0(2) = (2) 0(2) + (2) 0(2) = 3 ·24 + 2 · 0 = 32
(b) () = ()(), so 0(7) = (7) 0(7) − (7) 0(7)
[(7)]2 = 1 · 14− 5 ·
−23
12 = 1
4+10 3 = 43
12 51. (a) = () ⇒ 0= 0() + () · 1 = 0() + ()
(b) =
() ⇒ 0= () · 1 − 0()
[()]2 = () − 0() [()]2 (c) = ()
⇒ 0= 0() − () · 1
()2 = 0() − ()
2 52. (a) = 2 () ⇒ 0= 20() + ()(2)
(b) = ()
2 ⇒ 0= 20() − ()(2)
(2)2 = 0() − 2()
3
(c) = 2
() ⇒ 0= ()(2) − 20() [ ()]2 (d) = 1 + ()
√ ⇒
0=
√ [0() + ()] − [1 + ()] 1 2√
(√
)2
= 320() + 12 () −12−12−1212 ()
· 212
212 = () + 220() − 1 232
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58. Use the method of Hint to compute Q0(0), where
Q(x) = 1 + x + x2+ xex 1 − x + x2− xex
Hint: Instead of finding Q0(x) first, let f (x) be the numerator and g(x) the denominator of Q(x) and compute Q0(x) from f (0), f0(0), g(0), and g0(0).
Solution:
188 ¤ CHAPTER 3 DIFFERENTIATION RULES 53. If = () =
+ 1, then 0() = ( + 1)(1) − (1)
( + 1)2 = 1
( + 1)2. When = , the equation of the tangent line is
−
+ 1 = 1
( + 1)2( − ). This line passes through (1 2) when 2 −
+ 1 = 1
( + 1)2(1 − ) ⇔ 2( + 1)2− ( + 1) = 1 − ⇔ 22+ 4 + 2 − 2− − 1 + = 0 ⇔ 2+ 4 + 1 = 0.
The quadratic formula gives the roots of this equation as = −4 ±
42− 4(1)(1)
2(1) = −4 ±√
12
2 = −2 ±√3, so there are two such tangent lines. Since
−2 ±√ 3
= −2 ±√ 3
−2 ±√
3 + 1 = −2 ±√ 3
−1 ±√
3·−1 ∓√ 3
−1 ∓√ 3
= 2 ± 2√ 3 ∓√
3 − 3
1 − 3 = −1 ±√ 3
−2 = 1 ∓√ 3 2 , the lines touch the curve at
−2 +√
31−2√3
≈ (−027 −037) and
−2 −√
31 +2√3
≈ (−373 137).
54. = − 1
+ 1 ⇒ 0= ( + 1)(1) − ( − 1)(1)
( + 1)2 = 2
( + 1)2. If the tangent intersects the curve when = , then its slope is 2( + 1)2. But if the tangent is parallel to
− 2 = 2, that is, = 12 − 1, then its slope is12. Thus, 2
( + 1)2 = 1
2 ⇒
( + 1)2= 4 ⇒ + 1 = ±2 ⇒ = 1 or −3. When = 1, = 0 and the equation of the tangent is − 0 = 12( − 1) or =12 − 12. When = −3, = 2 and the equation of the tangent is − 2 = 12( + 3)or = 12 +72.
55. =
⇒ 0= 0− 0
2 . For () = − 33+ 55, 0() = 1 − 92+ 254, and for () = 1 + 33+ 66+ 99, 0() = 92+ 365+ 818.
Thus, 0(0) = (0)0(0) − (0)0(0)
[(0)]2 = 1 · 1 − 0 · 0
12 = 1
1 = 1.
56. =
⇒ 0= 0− 0
2 . For () = 1 + + 2+ , 0() = 1 + 2 + + , and for () = 1 − + 2− , 0() = −1 + 2 − − .
Thus, 0(0) =(0)0(0) − (0)0(0)
[(0)]2 = 1 · 2 − 1 · (−2)
12 = 4
1 = 4.
57. If () denotes the population at time and () the average annual income, then () = ()() is the total personal income. The rate at which () is rising is given by 0() = ()0() + ()0() ⇒
0(1999) = (1999)0(1999) + (1999)0(1999) = (961,400)($1400yr) + ($30,593)(9200yr)
= $1,345,960,000yr + $281,455,600yr = $1,627,415,600yr So the total personal income was rising by about $1.627 billion per year in 1999.
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