Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 3.4 THE CHAIN RULE ¤ 103
(c) w(x) =g(g(x)) ⇒ w 0 (x) =g 0 (g(x))g 0 (x). Sow 0 (1) = g 0 (g(1))g 0 (1) = g 0 (3)g 0 (1). Tofind g 0 (3), note that g is
linear from (2, 0) to (5, 2),soitsslopeis 2 − 0
5 − 2 = 2 3 . Thus, g0 (3)g 0 (1) = 2
3
(−3) = −2.
67. (a) F (x) =f(e x ) ⇒ F 0 (x) =f 0 (e x ) d
dx (ex )=f 0 (e x )e x
(b) G(x) =e f(x) ⇒ G 0 (x) =e f(x) d
dx f(x) =ef(x) f 0 (x)
69. r(x) =f(g(h(x))) ⇒ r 0 (x) =f 0 (g(h(x))) · g 0 (h(x)) · h 0 (x),so
r 0 (1) = f 0 (g(h(1))) · g 0 (h(1)) · h 0 (1) = f 0 (g(2)) · g 0 (2) · 4=f 0 (3) · 5 · 4=6· 5 · 4=120
71. F (x) =f(3f(4f(x))) ⇒
F 0 (x)=f 0 (3f(4f(x))) ·
d
dx (3f(4f(x))) = f 0 (3f(4f(x))) · 3f 0 d
(4f(x)) ·
dx (4f(x))
= f 0 (3f(4f(x))) · 3f 0 (4f(x)) · 4f 0 (x), so
F 0 (0) = f 0 (3f(4f(0))) · 3f 0 (4f(0)) · 4f 0 (0) = f 0 (3f(4 · 0)) · 3f 0 (4 · 0) · 4 · 2=f 0 (3 · 0) · 3 · 2 · 4 · 2=2· 3 · 2 · 4 · 2=96.
73. y = Ae −x + Bxe −x ⇒
y 0 = A(−e −x )+B[x(−e −x )+e −x · 1] = −Ae −x + Be −x − Bxe −x =(B − A)e −x − Bxe −x
⇒
y 00 =(B − A)(−e −x ) − B[x(−e −x )+e −x · 1] = (A − B)e −x − Be −x + Bxe −x =(A − 2B)e −x + Bxe −x ,
so
y 00 +2y 0 + y =(A − 2B)e −x + Bxe −x +2[(B − A)e −x − Bxe −x ]+Ae −x + Bxe −x
=[(A − 2B)+2(B − A)+A]e −x +[B − 2B + B]xe −x =0.
75. The use of D, D 2 , ..., D n is just a derivative notation (see text page 157). In general, Df(2x) =2f 0 (2x),
D 2 f(2x) =4f 00 (2x), ..., D n f(2x) =2 n f (n) (2x). Sincef(x) =cosx and 50 = 4(12) + 2, wehave
f (50) (x) =f (2) (x) =− cos x,soD 50 cos 2x = −2 50 cos 2x.
77. s(t) =10+ 1 sin(10πt) ⇒ the velocity after t seconds is v(t) 4 =s0 (t) = 1 cos(10πt)(10π) = 5π cos(10πt) cm/s.
4 2
79. (a) B(t) =4.0+0.35 sin 2πt ⇒
dB
5.4 dt = 0.35 cos 2πt 2π
= 0.7π 2πt
cos
5.4 5.4 5.4 5.4 = 7π 2πt
cos
54 5.4
(b) At t =1, dB
dt = 7π 2π
cos
54 5.4 ≈ 0.16.
81. s(t) =2e −1.5t sin 2πt ⇒
v(t) =s 0 (t) =2[e −1.5t (cos 2πt)(2π)+(sin2πt)e −1.5t (−1.5)] = 2e −1.5t (2π cos 2πt − 1.5sin2πt)