Solução_Calculo_Stewart_6e

F.
128 ¤ CHAPTER 3 DIFFERENTIATION RULES
TX.10
(e) Let y =coth −1 x.Thencoth y = x ⇒ −csch 2 y dy
dy
=1 ⇒
dx dx = − 1
csch 2 y = 1
1 − coth 2 y = 1
1 − x 2
by Exercise 13.
31. f(x) =x sinh x − cosh x ⇒ f 0 (x) =x (sinh x) 0 +sinhx · 1 − sinh x = x cosh x
33. h(x) =ln(coshx) ⇒ h 0 (x) = 1
cosh x (cosh x)0 = sinh x
cosh x =tanhx
35. y = e cosh 3x ⇒ y 0 = e cosh 3x · sinh 3x · 3=3e cosh 3x sinh 3x
37. f(t) =sech 2 (e t )=[sech(e t )] 2 ⇒
f 0 (t) =2[sech(e t )] [sech(e t )] 0 =2sech(e t ) − sech(e t )tanh(e t ) · e t = −2e t sech 2 (e t )tanh(e t )
39. y = arctan(tanh x) ⇒ y 0 =
1
1+(tanhx) (tanh 2 x)0 =
sech2 x
1+tanh 2 x
41. G(x) = 1 − cosh x
1+coshx
⇒
G 0 (x) =
(1 + cosh x)(− sinh x) − (1 − cosh x)(sinh x) − sinh x − sinh x cosh x − sinh x +sinhx cosh x
=
(1 + cosh x) 2 (1 + cosh x) 2
= −2sinhx
(1 + cosh x) 2
43. y =tanh −1√ x ⇒ y 0 1
= √ 2 · 1
1
2 x−1/2 =
1 − x 2 √ x (1 − x)
45. y = x sinh −1 (x/3) − √ 9+x 2 ⇒
x
y 0 =sinh −1 1/3
+ x
3
−
1+(x/3)
2
47. y =coth −1√ x 2 +1 ⇒ y 0 1
=
1 − (x 2 +1)
2x
x
2 √ 9+x 2 =sinh−1 +
3
49. As the depth d of the water gets large, the fraction 2πd
L
approaches 1. Thus, v =
x
√
9+x
2 −
2x
2 √ x 2 +1 = − 1
x √ x 2 +1
gL 2πd gL gL
2π tanh ≈
L 2π (1) = 2π .
x
x
√
9+x
2 =sinh−1 3
2πd
gets large, and from Figure 3 or Exercise 23(a), tanh L
51. (a) y =20cosh(x/20) − 15 ⇒ y 0 = 20 sinh(x/20) · 1 = sinh(x/20). Since the right pole is positioned at x =7,
20
we have y 0 (7) = sinh 7
20 ≈ 0.3572.
(b) If α is the angle between the tangent line and the x-axis, then tan α = slope of the line =sinh 7 20 ,so
α =tan −1 sinh 7
20
≈ 0.343 rad ≈ 19.66 ◦ .Thus,theanglebetweenthelineandthepoleisθ =90 ◦ − α ≈ 70.34 ◦ .
53. (a) y = A sinh mx + B cosh mx ⇒ y 0 = mA cosh mx + mB sinh mx ⇒
y 00 = m 2 A sinh mx + m 2 B cosh mx = m 2 (A sinh mx + B cosh mx) =m 2 y