Solução_Calculo_Stewart_6e

F.
102 ¤ CHAPTER 3 DIFFERENTIATION RULES
TX.10
49. y = e αx sin βx ⇒ y 0 = e αx · β cos βx +sinβx · αe αx = e αx (β cos βx + α sin βx) ⇒
y 00 = e αx (−β 2 sin βx + αβ cos βx)+(β cos βx + α sin βx) · αe αx
= e αx (−β 2 sin βx + αβ cos βx + αβ cos βx + α 2 sin βx) =e αx (α 2 sin βx − β 2 sin βx +2αβ cos βx)
= e αx (α 2 − β 2 )sinβx +2αβ cos βx
51. y =(1+2x) 10 ⇒ y 0 =10(1+2x) 9 · 2=20(1+2x) 9 .
At (0, 1), y 0 = 20(1 + 0) 9 =20, and an equation of the tangent line is y − 1=20(x − 0),ory =20x +1.
53. y =sin(sinx) ⇒ y 0 =cos(sinx) · cos x. At (π, 0), y 0 =cos(sinπ) · cos π =cos(0)· (−1) = 1(−1) = −1,andan
equation of the tangent line is y − 0=−1(x − π),ory = −x + π.
55. (a) y =
2
⇒ y 0 = (1 + e−x )(0) − 2(−e −x ) 2e −x
=
1+e −x (1 + e −x ) 2 (1 + e −x ) . 2
(b)
At (0, 1), y 0 =
2e 0
(1 + e 0 ) 2 = 2(1)
(1 + 1) 2 = 2 2 2 = 1 2 .Soanequationofthe
tangent line is y − 1= 1 2 (x − 0) or y = 1 2 x +1.
57. (a) f(x) =x √ 2 − x 2 = x(2 − x 2 ) 1/2 ⇒
f 0 (x) =x · 1
2 (2 − x2 ) −1/2 (−2x)+(2− x 2 ) 1/2 · 1=(2− x 2 ) −1/2 −x 2 +(2− x 2 ) = 2 − 2x2 √
2 − x
2
(b)
f 0 =0when f has a horizontal tangent line, f 0 is negative when f is
decreasing, and f 0 is positive when f is increasing.
59. For the tangent line to be horizontal, f 0 (x) =0. f(x) =2sinx +sin 2 x ⇒ f 0 (x) =2cosx +2sinx cos x =0 ⇔
2cosx(1 + sin x) =0 ⇔ cos x =0or sin x = −1, sox = π 2 +2nπ or 3π 2
+2nπ, wheren is any integer. Now
f π
2
=3and f
3π
2
where n is any integer.
= −1, so the points on the curve with a horizontal tangent are
π
2 +2nπ, 3 and 3π
2 +2nπ, −1 ,
61. F (x) =f(g(x)) ⇒ F 0 (x) =f 0 (g(x)) · g 0 (x),soF 0 (5) = f 0 (g(5)) · g 0 (5) = f 0 (−2) · 6=4· 6=24
63. (a) h(x) =f(g(x)) ⇒ h 0 (x) =f 0 (g(x)) · g 0 (x),soh 0 (1) = f 0 (g(1)) · g 0 (1) = f 0 (2) · 6=5· 6=30.
(b) H(x) =g(f(x)) ⇒ H 0 (x) =g 0 (f(x)) · f 0 (x),soH 0 (1) = g 0 (f(1)) · f 0 (1) = g 0 (3) · 4=9· 4=36.
65. (a) u(x) =f(g(x)) ⇒ u 0 (x) =f 0 (g(x))g 0 (x). Sou 0 (1) = f 0 (g(1))g 0 (1) = f 0 (3)g 0 (1). Tofind f 0 (3), note that f is
linear from (2, 4) to (6, 3), so its slope is 3 − 4
6 − 2 = − 1 4 .Tofind g0 (1), note that g is linear from (0, 6) to (2, 0), so its slope
is 0 − 6
2 − 0 = −3. Thus, f 0 (3)g 0 (1) =
− 1 4 (−3) =
3
. 4
(b) v(x) =g(f(x)) ⇒ v 0 (x) =g 0 (f(x))f 0 (x). Sov 0 (1) = g 0 (f(1))f 0 (1) = g 0 (2)f 0 (1), which does not exist since
g 0 (2) does not exist.