Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 3 REVIEW ¤ 135
79. h(x) =
f(x) g(x)
f(x)+g(x)
⇒
h 0 (x) = [f(x)+g(x)] [f(x) g0 (x)+g(x) f 0 (x)] − f(x) g(x)[f 0 (x)+g 0 (x)]
[f(x)+g(x)] 2
= [f(x)]2 g 0 (x)+f(x) g(x) f 0 (x)+f(x) g(x) g 0 (x)+[g(x)] 2 f 0 (x) − f(x) g(x) f 0 (x) − f(x) g(x) g 0 (x)
[f(x)+g(x)] 2
= f 0 (x)[g(x)] 2 + g 0 (x)[f(x)] 2
[f(x)+g(x)] 2
81. Using the Chain Rule repeatedly, h(x) =f(g(sin 4x)) ⇒
h 0 (x) =f 0 (g(sin 4x)) ·
83. y =[ln(x +4)] 2 ⇒ y 0 =2[ln(x +4)] 1 ·
d
dx (g(sin 4x)) = f 0 (g(sin 4x)) · g 0 d
(sin 4x) ·
dx (sin 4x) =f 0 (g(sin 4x))g 0 (sin 4x)(cos 4x)(4).
1
+4)
· 1=2ln(x and y 0 =0 ⇔ ln(x +4)=0 ⇔
x +4 x +4
x +4=e 0 ⇒ x +4=1 ⇔ x = −3, so the tangent is horizontal at the point (−3, 0).
85. y = f(x) =ax 2 + bx + c ⇒ f 0 (x) =2ax + b. We know that f 0 (−1) = 6 and f 0 (5) = −2, so−2a + b =6and
10a + b = −2. Subtracting the first equation from the second gives 12a = −8 ⇒ a = − 2 . Substituting − 2 for a in the
3 3
first equation gives b = 14 3 .Nowf(1) = 4 ⇒ 4=a + b + c,soc =4+2 3 − 14
3 =0and hence, f(x) =− 2 3 x2 + 14
3 x.
87. s(t) =Ae −ct cos(ωt + δ) ⇒
v(t) =s 0 (t) =A{e −ct [−ω sin(ωt + δ)] + cos(ωt + δ)(−ce −ct )} = −Ae −ct [ω sin(ωt + δ)+c cos(ωt + δ)]
⇒
a(t) =v 0 (t) =−A{e −ct [ω 2 cos(ωt + δ) − cω sin(ωt + δ)] + [ω sin(ωt + δ)+c cos(ωt + δ)](−ce −ct )}
= −Ae −ct [ω 2 cos(ωt + δ) − cω sin(ωt + δ) − cω sin(ωt + δ) − c 2 cos(ωt + δ)]
= −Ae −ct [(ω 2 − c 2 )cos(ωt + δ) − 2cω sin(ωt + δ)] = Ae −ct [(c 2 − ω 2 )cos(ωt + δ)+2cω sin(ωt + δ)]
89. (a) y = t 3 − 12t +3 ⇒ v(t) =y 0 =3t 2 − 12 ⇒ a(t) =v 0 (t) =6t
(b) v(t) =3(t 2 − 4) > 0 when t>2,soitmovesupwardwhent>2 anddownwardwhen0 ≤ t2. The particle is slowing down when v and a have opposite
signs; that is, when 0