Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 3 REVIEW ¤ 137
105. A = x 2 + 1 π 1
2 2 x2 =
1+ π 8 x
2
⇒ dA = 2+ π 4
xdx.Whenx =60
and dx =0.1, dA = 2+ π 4
approximately 12 + 3π 2 ≈ 16.7 cm2 .
√
4
16 + h − 2 d
107. lim
=
h→0 h dx
60(0.1) = 12 +
3π
2
, so the maximum error is
4√
x
x =16
= 1 4 x−3/4 x
=16
=
1
4 √ 4
16 3 = 1
32
√ √ √ √ √ √
1+tanx − 1+sinx 1+tanx − 1+sinx 1+tanx + 1+sinx
109. lim
=lim
x→0 x 3
x→0 x √ 3 1+tanx + √ 1+sinx
(1 + tan x) − (1 + sin x)
=lim
x→0 x 3√ 1+tanx + √ 1+sinx =lim sin x (1/ cos x − 1)
x→0 x 3√ 1+tanx + √ 1+sinx · cos x
cos x
sin x (1 − cos x)
=lim
x→0 x 3√ 1+tanx + √ 1+sinx cos x · 1+cosx
1+cosx
sin x · sin 2 x
=lim
x→0 x 3√ 1+tanx + √ 1+sinx cos x (1 + cos x)
= lim
x→0
=1 3 ·
3
sin x
1
lim √ √
x x→0 1+tanx + 1+sinx cos x (1 + cos x)
1
√
1+
√
1
· 1 · (1 + 1)
= 1 4
111.
d
dx [f(2x)] = x2 ⇒ f 0 (2x) · 2=x 2 ⇒ f 0 (2x) = 1 2 x2 .Lett =2x. Thenf 0 (t) = 1 1
2 2 t2 = 1 8 t2 ,sof 0 (x) = 1 8 x2 .