Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 3 REVIEW ¤ 133
√ x +1(2− x)
5
41. y =
⇒ ln y = 1
(x +3) 7 2
√ x +1(2− x)
y 0 5
1
=
(x +3) 7 2(x +1) − 5
2 − x − 7
x +3
ln(x +1)+5ln(2− x) − 7ln(x +3) ⇒
y0
y = 1
2(x +1) + −5
2 − x − 7
or y 0 = (2 − x)4 (3x 2 − 55x − 52)
2 √ x +1(x +3) 8 .
x +3
⇒
43. y = x sinh(x 2 ) ⇒ y 0 = x cosh(x 2 ) · 2x +sinh(x 2 ) · 1=2x 2 cosh(x 2 )+sinh(x 2 )
45. y =ln(cosh3x) ⇒ y 0 =(1/ cosh 3x)(sinh 3x)(3) = 3 tanh 3x
47. y =cosh −1 (sinh x) ⇒ y 0 =
49. y =cos e √
tan 3x
⇒
1
(sinh x)2 − 1 · cosh x = cosh x
sinh 2 x − 1
y 0 = − sin e √
tan 3x
· e √ 0
tan 3x = − sin e √
tan 3x
e √ tan 3x · 1 (tan 2 3x)−1/2 · sec 2 (3x) · 3
−3sin e √
tan 3x
e √ tan 3x sec 2 (3x)
=
2 √ tan 3x
51. f(t) = √ 4t +1 ⇒ f 0 (t) = 1 2 (4t +1)−1/2 · 4=2(4t +1) −1/2 ⇒
f 00 (t) =2(− 1 2 )(4t +1)−3/2 · 4=−4/(4t +1) 3/2 ,sof 00 (2) = −4/9 3/2 = − 4
27 .
53. x 6 + y 6 =1 ⇒ 6x 5 +6y 5 y 0 =0 ⇒ y 0 = −x 5 /y 5 ⇒
y 00 = − y5 (5x 4 ) − x 5 (5y 4 y 0 )
= − 5x4 y 4 y − x(−x 5 /y 5 ) = − 5x4 (y 6 + x 6 )/y 5
= − 5x4
(y 5 ) 2 y 10
y 6
y 11
55. We firstshowitistrueforn =1: f(x) =xe x ⇒ f 0 (x) =xe x + e x =(x +1)e x . We now assume it is true
for n = k: f (k) (x) =(x + k)e x . With this assumption, we must show it is true for n = k +1:
f (k+1) (x) = d
f (k) (x) = d
dx
dx [(x + k)ex ]=(x + k)e x + e x =[(x + k)+1]e x =[x +(k +1)]e x .
Therefore, f (n) (x) =(x + n)e x by mathematical induction.
57. y =4sin 2 x ⇒ y 0 =4· 2sinx cos x. At π
6 , 1 , y 0 =8· 1
2 · √3
2 =2√ 3, so an equation of the tangent line
is y − 1=2 √ 3 x − π 6
,ory =2
√
3 x +1− π
√
3/3.
59. y = √ 1+4sinx ⇒ y 0 = 1 2 (1 + 4 sin x)−1/2 · 4cosx =
2cosx
√ 1+4sinx
.
At (0, 1), y 0 = 2 √
1
=2, so an equation of the tangent line is y − 1=2(x − 0),ory =2x +1.
61. y =(2+x)e −x ⇒ y 0 =(2+x)(−e −x )+e −x · 1=e −x [−(2 + x)+1]=e −x (−x − 1).
At (0, 2), y 0 =1(−1) = −1, so an equation of the tangent line is y − 2=−1(x − 0),ory = −x +2.
The slope of the normal line is 1, so an equation of the normal line is y − 2=1(x − 0),ory = x +2.