6 1 a dy f xdx 2xdx 41 4 and y x x2 x2

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Unformatted text preview: = cos(1/x2 ) 16. f (x) = 4 tan3 (x3 ) 17. f (x) = 4 sec(x7 ) 18. f (x) = 3 cos2 =− d3 (x ) = −3x2 csc(x3 ) cot(x3 ) dx d 2 (1/x2 ) = − 3 cos(1/x2 ) dx x d d [tan(x3 )] = 4 tan3 (x3 ) sec2 (x3 ) (x3 ) = 12x2 tan3 (x3 ) sec2 (x3 ) dx dx d d [sec(x7 )] = 4 sec(x7 ) sec(x7 ) tan(x7 ) (x7 ) = 28x6 sec2 (x7 ) tan(x7 ) dx dx x x+1 3 cos2 (x + 1)2 d cos dx x x+1 x x+1 sin = 3 cos2 x x+1 − sin x x+1 x x+1 19. f (x) = 5 sin(5x) d [cos(5x)] = − 2 cos(5x) dx 2 cos(5x) 20. f (x) = 3 − 8 sin(4x) cos(4x) d [3x − sin2 (4x)] = dx 2 3x − sin (4x) 2 3x − sin2 (4x) 1 1 2 (x + 1)(1) − x(1) (x + 1)2 Exercise Set 3.5 84 d x + csc(x3 + 3) dx f (x) = −3 x + csc(x3 + 3) −4 = −3 x + csc(x3 + 3) −4 1 − csc(x3 + 3) cot(x3 + 3) = −3 x + csc(x3 + 3) 21. −4 1 − 3x2 csc(x3 + 3) cot(x3 + 3) f (x) = −4 x4 − sec(4x2 − 2) −5 = −4 x4 − sec(4x2 − 2) 22. d x4 − sec(4x2 − 2) dx −5 = −16x x4 − sec(4x2 − 2) d3 (x + 3) dx 4x3 − sec(4x2 − 2) tan(4x2 − 2) −5 d (4x2 − 2) dx x2 − 2 sec(4x2 − 2) tan(4x2 − 2) −2x x(10 − 3x2 ) 23. f (x) = x2 · √ + 2x 5 − x2 = √ 2 5 − x2 5 − x2 √ √ 1 1 − x2 (1) − x(−x/ 1 − x2 ) = 1 − x2 (1 − x2 )3/2 24. f (x) = 25. dy d = x3 (2 sin 5x) (sin 5x) + 3x2 sin2 5x = 10x3 sin 5x cos 5x + 3x2 sin2 5x dx dx 26. √ √ √ √ √ √ √ 1 3 1 dy 1 = x 3 tan2 ( x) sec2 ( x) √ + √ tan3 ( x) = tan2 ( x) sec2 ( x) + √ tan3 ( x) dx 2 2x 2x 2x 27. dy = x5 sec dx = −x3 sec 1 x tan 1 x tan 1 x d dx 1 x 1 x + sec + 5x4 sec 1 x (5x4 ) = x5 sec 1 x tan 1 x − 1 x2 + 5x4 sec 1 x 1 x 28. sec(3x + 1) cos x − 3 sin x sec(3x + 1) tan(3x + 1) cos x − 3 sin x tan(3x + 1) dy = = 2 (3x + 1) dx sec sec(3x + 1) 29. d dy = − sin(cos x) (cos x) = − sin(cos x)(− sin x) = sin(cos x) sin x dx dx 30. d dy = cos(tan 3x) (tan 3x) = 3 sec2 3x cos(tan 3x) dx dx 31. d d dy = 3 cos2 (sin 2x) [cos(sin 2x)] = 3 cos2 (sin 2x)[− sin(sin 2x)] (sin 2x) dx dx dx = −6 cos2 (sin 2x) sin(sin 2x) cos 2x 32. (1 − cot x2 )(−2x csc x2 cot x2 ) − (1 + csc x2 )(2x csc2 x2 ) 1 + cot x2 csc x2 dy = = −2x csc x2 2 )2 dx (1 − cot x (1 − cot x2 )2 33. d d dy = (5x + 8)13 12(x3 + 7x)11 (x3 + 7x) + (x3 + 7x)12 13(5x + 8)12 (5x + 8) dx dx dx = 12(5x + 8)13 (x3 + 7x)11 (3x2 + 7) + 65(x3 + 7x)12 (5x + 8)12 34. dy = (2x − 5)2 3(x2 + 4)2 (2x) + (x2 + 4)3 2(2x − 5)(2) = 6x(2x − 5)2 (x2 + 4)2 + 4(2x − 5)(x2 + 4)3 dx = 2(2x − 5)(x2 + 4)2 (8x2 − 15x + 8) 85 Chapter 3 2 35. dy x−5 =3 dx 2x + 1 dy = 17 dx 1 + x2 1 − x2 16 36. 1 + x2 1 − x2 16 = 17 37. 38. x5 d x−5 =3 dx 2x + 1 2x + 1 1 + x2 1 − x2 d dx 41. 1 + x2 1 − x2 11 33(x − 5)2 = 2 (2x + 1) (2x + 1)4 16 (1 − x2 )(2x) − (1 + x2 )(−2x) (1 − x2 )2 dy (4x2 − 1)8 (3)(2x + 3)2 (2) − (2x + 3)3 (8)(4x2 − 1)7 (8x) = dx (4x2 − 1)16 2 2 7 2(2x + 3) (4x − 1) [3(4x2 − 1) − 32x(2x + 3)] 2(2x + 3)2 (52x2 + 96x + 3) = =− 2 − 1)16 (4x (4x2 − 1)9 dy d = 12[1 + sin3 (x5 )]11 [1 + sin3 (x5 )] dx dx d = 12[1 + sin3 (x5 )]11 3 sin2 (x5 ) sin(x5 ) = 180x4 [1 + sin3 (x5 )]11 sin2 (x5 ) cos(x5 ) dx dy = 5 x sin 2x + tan4 (x7 ) dx 4 4 = 5 x sin 2x + tan4 (x7 ) 40. · 4x 68x(1 + x2 )16 = 2 )2 (1 − x (1 − x2 )18 = 5 x sin 2x + tan4 (x7 ) 39. = 17 2 4 d x sin 2x tan4 (x7 ) dx d d x cos 2x (2x) + sin 2x + 4 tan3 (x7 ) tan(x7 ) dx dx 2x cos 2x + sin 2x + 28x6 tan3 (x7 ) sec2 (x7 ) d dy = cos(3x2 ) (3x2 ) = 6x cos(3x2 ), dx dx d d2 y = 6x(− sin(3x2 )) (3x2 ) + 6 cos(3x2 ) = −36x2 sin(3x2 ) + 6 cos(3x2 ) 2 dx dx d d dy = x(− sin(5x)) (5x) + cos(5x) − 2 sin x (sin x) dx dx dx = −5x sin(5x) + cos(5x) − 2 sin x cos x = −5x sin(5x) + cos(5x) − sin(2x), d d d d2 y = −5x cos(5x) (5x) − 5 sin(5x) − sin(5x) (5x) − cos(2x) (2x) dx2 dx dx dx = −25x cos(5x) − 10 sin(5x) − 2 cos(2x) 42. dy = x sec2 dx 2 d2 y = − sec dx2 x d dx 1 x 1 x 1 x + tan d sec dx 1 x 1 x + 1 = − sec2 x 1 sec2 x2 1 x 1 x + sec2 + tan 1 x d dx 1 , x 1 x = 2 sec2 x3 1 x tan 1 x 43. (1 − x) + (1 + x) 2 d2 y dy = = = 2(1 − x)−2 and = −2(2)(−1)(1 − x)−3 = 4(1 − x)−3 dx (1 − x)2 (1 − x)2 dx2 45. dy dy = −3x sin 3x + cos 3x; if x = π then y = −π and = −1 so y + π = −(x − π ), y = −x dx dx 46. dy dy = 3x2 cos(1 + x3 ); if x = −3 then y = sin(−26) = − sin 26 and = 27 cos 26 dx dx so y + sin 26 = 27(cos 26)(x + 3) Exercise Set 3.5 47. 48. 86 dy dy = −3 sec3 (π/2 − x) tan(π/2 − x); if x = −π/2 then y = −1 and =0 dx dx so y + 1 = (0)(x + π/2), y = −1 dy dy = 3(x − 1/x)2 (1 + 1/x2 ); if x = 2 then y = 27/8 and = 135/16 dx dx 135 135 27 27 = (x − 2), y = x− so y − 8 16 16 2 49. y = cot3 (π − θ) = − cot3 θ so dy/dx = 3 cot2 θ csc2 θ 50. 6 51. au + b cu + d 5 ad − bc (cu + d)2 d [a cos2 πω + b sin2 πω ] = −2πa cos πω sin πω + 2πb sin πω cos πω dω = π (b − a)(2 sin πω cos πω ) = π (b − a) sin 2πω 52. 2 csc2 (π/3 − y ) cot(π/3 − y ) 53. (a) (c) 2 f (x) = x √ −x + 4 − x2 4 − 2x2 4 − x2 = √ 4 −...
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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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