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Unformatted text preview: (a) cosh2 x0 = 1 + sinh2 x0 = 1 + (2)2 = 5, cosh x0 = (b) sinh2 x0 = cosh2 x0 − 1 = 9 25 3 −1= , sinh x0 = (because x0 > 0) 16 16 4 (c) sech2 x0 = 1 − tanh2 x0 = 1 − cosh x0 = √ 5 4 5 2 =1− 9 16 3 = , sech x0 = , 25 25 5 sinh x0 1 5 = , from = tanh x0 we get sinh x0 = sech x0 3 cosh x0 5 3 4 5 = 4 3 Exercise Set 8.8 6. 284 d cosh x d 1 cschx = =− = − coth x csch x for x = 0 dx dx sinh x sinh2 x d 1 sinh x d sech x = =− = − tanh x sech x for all x dx dx cosh x cosh2 x d cosh x sinh2 x − cosh2 x d coth x = = = − csch2 x for x = 0 dx dx sinh x sinh2 x dy dy dx = cosh y ; so dx dy dx 1 1 =√ for all x. 2 1 + x2 1 + sinh y 7. (a) y = sinh−1 x if and only if x = sinh y ; 1 = 1 dy d [sinh−1 x] = = = dx dx cosh y (b) Let x ≥ 1. Then y = cosh−1 x if and only if x = cosh y ; 1 = 1 dy d [cosh−1 x] = = = dx dx sinh y 1 cosh y − 1 2 = x2 dy dy dx = sinh y , so dx dy dx 1 for x ≥ 1. −1 −1 (c) Let −1 < x < 1. Then y = tanh x if and only if x = tanh y ; thus dy 1 dy d dy dy dx = sech2 y = (1 − tanh2 y ) = 1 − x2 , so [tanh−1 x] = = . 1= dx dy dx dx dx dx 1 − x2 9. 4 cosh(4x − 8) 12. 2 15. 1 11. − csch2 (ln x) x 10. 4x3 sinh(x4 ) sech2 2x tanh 2x 13. 1 csch(1/x) coth(1/x) x2 2 + 5 cosh(5x) sinh(5x) 14. −2e2x sech(e2x ) tanh(e2x ) 16. 6 sinh2 (2x) cosh(2x) 2 4x + cosh (5x) √ √ √ 17. x5/2 tanh( x) sech2 ( x) + 3x2 tanh2 ( x) 18. −3 cosh(cos 3x) sin 3x 1 20. 1+ 22. 1/ 1/x2 19. (−1/x2 ) = − 1 √ |x| x2 + 1 √ (sinh−1 x)2 − 1 1 + x2 26. (sech2 x)/ 1 + tanh2 x 27. − 1 sinh7 x + C 7 = 1/ 9 + x2 23. −(tanh−1 x)−2 /(1 − x2 ) 25. 31. 1+ 1 3 x2 /9 √ 21. 1/ (cosh−1 x) x2 − 1 24. 2(coth−1 x)/(1 − x2 ) 28. 10(1 + x csch−1 x)9 − 1 sinh x cosh x − 1 2 = sinh x = | sinh x| 1, x > 0 −1, x < 0 ex √ + ex sech−1 x 2x 1 − x x √ + csch−1 x |x| 1 + x2 32. 1 sinh(2x − 3) + C 2 33. 2 (tanh x)3/2 + C 3 285 Chapter 8 1 34. − coth(3x) + C 3 37. ln 3 1 − sech3 x 3 39. u = 3x, 40. x = √ ln 3 = 37/375 38. ln(cosh x) 0 ln 2 1 3 √ √ 2u, = ln 5 − ln 3 1 1 du = sinh−1 3x + C 2 3 1+u √ 2 du = −2 √ 2u2 1 u2 −1 √ du = cosh−1 (x/ 2) + C 1 du = − sech−1 (ex ) + C u 1 − u2 √ 41. u = ex , √ 42. u = cos θ, − 1 du = − sinh−1 (cos θ) + C 1 + u2 du = − csch−1 |u| + C = − csch−1 |2x| + C u 1 + u2 √ 43. u = 2x, √ 44. x = 5u/3, 1/2 45. tanh−1 x 0 46. sinh−1 t 1 36. − coth3 x + C 3 35. ln(cosh x) + C √ 3 1 5/3 du = 2 − 25 3 25u 1 u2 −1 = tanh−1 (1/2) − tanh−1 (0) = = sinh−1 √ 0 ln 3 sinh 2x dx = 49. A = √ 0 du = 1 cosh−1 (3x/5) + C 3 1 1 1 + 1/2 ln = ln 3 2 1 − 1/2 2 √ 3 − sinh−1 0 = ln( 3 + 2) ln 3 1 cosh 2x 2 = 0 1 [cosh(2 ln 3) − 1], 2 1 1 1 but cosh(2 ln 3) = cosh(ln 9) = (eln 9 + e− ln 9 ) = (9 + 1/9) = 41/9 so A = [41/9 − 1] = 16/9. 2 2 2 ln 2 ln 2 sech2 x dx = π tanh x 50. V = π 0 = π tanh(ln 2) = 3π/5 0 5 5 (cosh2 2x − sinh2 2x)dx = π 51. V = π 0 1 cosh ax dx = 2, 52. 0 dx = 5π 0 1 sinh ax a 1 = 2, 0 1 sinh a = 2, sinh a = 2a; a let f (a) = sinh a − 2a, then an+1 = an − sinh an − 2an , a1 = 2.2, . . . , a4 = a5 = 2.177318985. cosh an − 2 53. y = sinh x, 1 + (y )2 = 1 + sinh2 x = cosh2 x ln 2 ln 2 cosh x dx = sinh x L= 0 = sinh(ln 2) = 0 1 ln 2 1 (e − e− ln 2 ) = 2 2 2− 1 2 = 3 4 Exercise Set 8.8 286 54. y = sinh(x/a), 1 + (y )2 = 1 + sinh2 (x/a) = cosh2 (x/a) x1 x1 L= cosh(x/a)dx = a sinh(x/a) 0 = a sinh(x1 /a) 0 55. sinh(−x) = cosh(−x) = 1 1 −x (e − ex ) = − (ex − e−x ) = − sinh x 2 2 1 1 −x (e + ex ) = (ex + e−x ) = cosh x 2 2 1 1x (e + e−x ) + (ex − e−x ) = ex 2 2 1x 1x (b) cosh x − sinh x = (e + e−x ) − (e − e−x ) = e−x 2 2 1x 1 (c) sinh x cosh y + cosh x sinh y = (e − e−x )(ey + e−y ) + (ex + e−x )(ey − e−y ) 4 4 56. (a) cosh x + sinh x = = 1 x+y [(e − e−x+y + ex−y − e−x−y ) + (ex+y + e−x+y − ex−y − e−x−y )] 4 1 (x+y) [e − e−(x+y) ] = sinh(x + y ) 2 Let y = x in part (c). The proof is similar to part (c), or: treat x as variable and y as constant, and diﬀerentiate the result in part (c) with respect to x. Let y = x in part (e). Use cosh2 x = 1 + sinh2 x together with part (f). Use sinh2 x = cosh2 x − 1 together with part (f). = (d) (e) (f ) (g) (h) 57. (a) Divide cosh2 x − sinh2 x = 1 by cosh2 x. sinh y sinh x + tanh x + tanh y sinh x cosh y + cosh x sinh y cosh x cosh y = = (b) tanh(x + y ) = sinh x sinh y cosh x cosh y + sinh x sinh y 1 + tanh x tanh y 1+ cosh x cosh y (c) Let y = x in part (b). 1 58. (a) Let y = cosh−1 x; then x = cosh y = (ey + e−y ), ey − 2x + e−y = 0, e2y − 2xey + 1 = 0, 2 √ 2x ± 4x2 − 4 y = x ± x2 − 1. To determine which sign to take, note that y ≥ 0 e= 2 √ so e−y ≤ ey , x = (ey + e−y )/2 ≤ (ey + ey )/2 = ey , hence ey ≥ x thus ey = x + x2 − 1, √ y = cosh−1 x = ln(x + x2 − 1). ey − e−y e2y − 1 , xe2y + x = e2y − 1, = 2y ey + e−y e +1 1 1+x 1+x , y = ln . = (1 + x)/(1 − x), 2y = ln 1−x 2 1−x (b) Let y = tanh−1 x;...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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