Solução_Calculo_Stewart_6e

F.
126 ¤ CHAPTER 3 DIFFERENTIATION RULES
TX.10
3.11 Hyperbolic Functions
1. (a) sinh 0 = 1 2 (e0 − e 0 )=0 (b) cosh 0 = 1 2 (e0 + e 0 )= 1 2
(1 + 1) = 1
3. (a) sinh(ln 2) = eln 2 − e −ln 2
2
= eln 2 − (e ln 2 ) −1
2
= 2 − 2−1
2
= 2 − 1 2
2
= 3 4
(b) sinh 2 = 1 2 (e2 − e −2 ) ≈ 3.62686
5. (a) sech 0 = 1
cosh 0 = 1 1 =1 (b) cosh−1 1=0because cosh 0 = 1.
7. sinh(−x) = 1 2 [e−x − e −(−x) ]= 1 2 (e−x − e x )=− 1 2 (e−x − e x )=− sinh x
9. cosh x +sinhx = 1 2 (ex + e −x )+ 1 2 (ex − e −x )= 1 2 (2ex )=e x
11. sinh x cosh y +coshx sinh y = 1
2 (ex − e −x ) 1
2 (ey + e −y ) + 1
2 (ex + e −x ) 1
2 (ey − e −y )
= 1 4 [(ex+y + e x−y − e −x+y − e −x−y )+(e x+y − e x−y + e −x+y − e −x−y )]
= 1 4 (2ex+y − 2e −x−y )= 1 2 [ex+y − e −(x+y) ]=sinh(x + y)
13. Divide both sides of the identity cosh 2 x − sinh 2 x =1by sinh 2 x:
cosh 2 x
sinh 2 x − sinh2 x
sinh 2 x = 1
sinh 2 x ⇔ coth2 x − 1=csch 2 x.
15. Putting y = x in the result from Exercise 11, we have
sinh 2x =sinh(x + x) =sinhx cosh x +coshx sinh x =2sinhx cosh x.
17. tanh(ln x) =
sinh(ln x)
cosh(ln x) = (eln x − e − ln x )/2
(e ln x + e − ln x )/2 = x − (eln x ) −1 x − x−1 x − 1/x
= =
x +(e ln x )
−1
x + x−1 x +1/x = (x2 − 1)/x
(x 2 +1)/x = x2 − 1
x 2 +1
19. By Exercise 9, (cosh x +sinhx) n =(e x ) n = e nx =coshnx +sinhnx.
21. sech x = 1
cosh x ⇒ sech x = 1
5/3 = 3 5 .
cosh 2 x − sinh 2 x =1 ⇒ sinh 2 x =cosh 2 x − 1=
5 2
− 1= 16 ⇒ sinh x = 4 [because x>0].
3
9 3
csch x = 1
sinh x ⇒ csch x = 1
4/3 = 3 4 .
tanh x = sinh x
cosh x
⇒ tanh x =
4/3
5/3 = 4 5 .
coth x = 1
tanh x ⇒ coth x = 1
4/5 = 5 4 .
e x − e −x
23. (a) lim tanh x = lim
x→∞ x→∞ e x + e
(b) lim tanh x =
x→−∞
lim
x→−∞
−x ·
e−x
= lim
e−x e x − e −x ex
·
e x + e−x e = x
x→∞
lim
x→−∞
1 − e −2x
1+e −2x = 1 − 0
1+0 =1
e 2x − 1
e 2x +1 = 0 − 1
0+1 = −1