Solução_Calculo_Stewart_6e

F.
134 ¤ CHAPTER 3 DIFFERENTIATION RULES
TX.10
63. (a) f(x) =x √ 5 − x ⇒
1
f 0 (x) =x
2 (5 − x)−1/2 (−1) + √ −x
5 − x =
2 √ 5 − x + √ 5 − x · 2 √ 5 − x
2 √ 5 − x = −x
2 √ 2(5 − x)
+
5 − x 2 √ 5 − x
=
−x +10− 2x
2 √ 5 − x
=
10 − 3x
2 √ 5 − x
(b) At (1, 2): f 0 (1) = 7 4 .
(c)
So an equation of the tangent line is y − 2= 7 4 (x − 1) or y = 7 4 x + 1 4 .
At (4, 4): f 0 (4) = − 2 2 = −1.
So an equation of the tangent line is y − 4=−1(x − 4) or y = −x +8.
(d)
The graphs look reasonable, since f 0 is positive where f has tangents with
positive slope, and f 0 is negative where f has tangents with negative slope.
65. y =sinx +cosx ⇒ y 0 =cosx − sin x =0 ⇔ cos x =sinx and 0 ≤ x ≤ 2π ⇔ x = π or 5π , so the points
4 4
are π
4 , √ 2 and 5π
4 , −√ 2 .
67. f(x) =(x − a)(x − b)(x − c) ⇒ f 0 (x) =(x − b)(x − c)+(x − a)(x − c)+(x − a)(x − b).
So f 0 (x)
f(x)
=
(x − b)(x − c)+(x − a)(x − c)+(x − a)(x − b)
(x − a)(x − b)(x − c)
= 1
x − a + 1
x − b + 1
x − c .
Or: f(x) =(x − a)(x − b)(x − c) ⇒ ln |f(x)| =ln|x − a| +ln|x − b| +ln|x − c| ⇒
f 0 (x)
f(x) = 1
x − a + 1
x − b + 1
x − c
69. (a) h(x) =f(x) g(x) ⇒ h 0 (x) =f(x) g 0 (x)+g(x) f 0 (x) ⇒
h 0 (2) = f(2) g 0 (2) + g(2) f 0 (2) = (3)(4) + (5)(−2) = 12 − 10 = 2
(b) F (x) =f(g(x)) ⇒ F 0 (x) =f 0 (g(x)) g 0 (x) ⇒ F 0 (2) = f 0 (g(2)) g 0 (2) = f 0 (5)(4) = 11 · 4=44
71. f(x) =x 2 g(x) ⇒ f 0 (x) =x 2 g 0 (x)+g(x)(2x) =x[xg 0 (x)+2g(x)]
73. f(x) =[g(x)] 2 ⇒ f 0 (x) =2[g(x)] · g 0 (x) =2g(x) g 0 (x)
75. f(x) =g(e x ) ⇒ f 0 (x) =g 0 (e x ) e x
77. f(x) =ln|g(x)| ⇒ f 0 (x) = 1
g(x) g0 (x) = g0 (x)
g(x)