# 4x 68x1 x2 16 2 2 1 x 1 x2 18 5 x sin 2x

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Unformatted text preview: x) + (sin x) = 0. (b) 28. y = A cos x − B sin x, y = −A sin x − B cos x so y + y = (−A sin x − B cos x) + (A sin x + B cos x) = 0. (a) f (x) = cos x = 0 at x = ±π/2, ±3π/2. (b) f (x) = 1 − sin x = 0 at x = −3π/2, π/2. (c) f (x) = sec2 x ≥ 1 always, so no horizontal tangent line. (d) f (x) = sec x tan x = 0 when sin x = 0, x = ±2π, ±π, 0 81 30. Chapter 3 (a) 0.5 0 2p -0.5 (b) y = sin x cos x = (1/2) sin 2x and y = cos 2x. So y = 0 when 2x = (2n + 1)π/2 for n = 0, 1, 2, 3 or x = π/4, 3π/4, 5π/4, 7π/4 31. x = 10 sin θ, dx/dθ = 10 cos θ; if θ = 60◦ , then dx/dθ = 10(1/2) = 5 ft/rad = π/36 ft/deg ≈ 0.087 ft/deg 32. s = 3800 csc θ, ds/dθ = −3800 csc θ cot θ; if θ = 30◦ , then √ √ √ ds/dθ = −3800(2)( 3) = −7600 3 ft/rad = −380 3π/9 ft/deg ≈ −230 ft/deg 33. D = 50 tan θ, dD/dθ = 50 sec2 θ; if θ = 45◦ , then √ dD/dθ = 50( 2)2 = 100 m/rad = 5π/9 m/deg ≈ 1.75 m/deg 34. (a) From the right triangle shown, sin θ = r/(r + h) so r + h = r csc θ, h = r(csc θ − 1). (b) dh/dθ = −r csc θ cot θ; if θ = 30◦ , then √ dh/dθ = −6378(2)( 3) ≈ −22, 094 km/rad ≈ −386 km/deg (a) d4k d87 d3 d4·21 d3 d4 sin x = sin x, so sin x = sin x; sin x = 3 4·21 sin x = 3 sin x = − cos x 4 4k 87 dx dx dx dx dx dx (b) 35. d100 d4k cos x = 4k cos x = cos x dx100 dx d [x sin x] = x cos x + sin x dx d2 [x sin x] = −x sin x + 2 cos x dx2 d3 [x sin x] = −x cos x − 3 sin x dx3 36. d4 [x sin x] = x sin x − 4 cos x dx4 By mathematical induction one can show d4k [x sin x] = x sin x − (4k ) cos x; dx4k d4k+1 [x sin x] = x cos x + (4k + 1) sin x; dx4k+1 d4k+3 d4k+2 [x sin x] = −x sin x + (4k + 2) cos x; [x sin x] = −x cos x − (4k + 3) sin x; dx4k+2 dx4k+3 d17 [x sin x] = x cos x + 17 sin x Since 17 = 4 · 4 + 1, dx17 37. (a) all x (b) all x (c) x = π/2 + nπ , n = 0, ±1, ±2, . . . (d) x = nπ , n = 0, ±1, ±2, . . . (e) x = π/2 + nπ , n = 0, ±1, ±2, . . . (f ) x = nπ , n = 0, ±1, ±2, . . . (g) x = (2n + 1)π , n = 0, ±1, ±2, . . . (h) x = nπ/2, n = 0, ±1, ±2, . . . (i) all x Exercise Set 3.5 38. (a) 82 d cos(x + h) − cos x cos x cos h − sin x sin h − cos x [cos x] = lim = lim h→0 h→0 dx h h = lim cos x h→0 cos h − 1 h − sin x sin h h = (cos x)(0) − (sin x)(1) = − sin x (b) 1 d d cos x(0) − (1)(− sin x) sin x [sec x] = = = = sec x tan x dx dx cos x cos2 x cos2 x (c) sin x(− sin x) − cos x(cos x) d d cos x [cot x] = = dx dx sin x sin2 x − sin2 x − cos2 x −1 = = − csc2 x sin2 x sin2 x 1 cos x d sin x(0) − (1)(cos x) d = − 2 = − csc x cot x [csc x] = = dx dx sin x sin2 x sin x = (d) 39. f (x) = − sin x, f (x) = − cos x, f (x) = sin x, and f (4) (x) = cos x with higher order derivatives repeating this pattern, so f (n) (x) = sin x for n = 3, 7, 11, . . . 40. (a) (b) sin h cos h h sin h h cos h 1 =1 h→0 1 tan x + tan h − tan x tan(x + h) − tan x d [tan x] = lim = lim 1 − tan x tan h h→0 h→0 dx h h tan h = lim lim h→0 h→0 h = lim = tan x + tan h − tan x + tan2 x tan h tan h(1 + tan2 x) = lim h→0 h→0 h(1 − tan x tan h) h(1 − tan x tan h) = lim tan h tan h sec2 x 2 h = sec x lim = lim h→0 h(1 − tan x tan h) h→0 1 − tan x tan h tan h h = sec2 x = sec x lim (1 − tan x tan h) lim h→0 2 h→0 41. d tan(x + y ) − tan y tan(y + h) − tan y = lim = (tan y ) = sec2 y x→0 h→0 x h dy 43. Let t be the radian measure, then h = lim (a) 180 t and cos h = cos t, sin h = sin t. π π cos h − 1 cos t − 1 cos t − 1 = lim = lim =0 lim t→0 180t/π h→0 h 180 t→0 t (b) sin h sin t sin t π π = lim = lim = t→0 180t/π h→0 h 180 t→0 t 180 (c) π d cos h − 1 sin h [sin x] = sin x lim + cos x lim = sin x(0) + cos x(π/180) = cos x h→0 h→0 h dx h 180 lim EXERCISE SET 3.5 d3 (x + 2x) = 37(x3 + 2x)36 (3x2 + 2) dx 1. f (x) = 37(x3 + 2x)36 2. f (x) = 6(3x2 + 2x − 1)5 d (3x2 + 2x − 1) = 6(3x2 + 2x − 1)5 (6x + 2) = 12(3x2 + 2x − 1)5 (3x + 1) dx 83 3. Chapter 3 −3 7 x f (x) = −2 x3 − d dx x3 − 7 x = −2 x3 − 7 x −3 3x2 + 7 x2 4. f (x) = (x5 − x + 1)−9 , f (x) = −9(x5 − x + 1)−10 5. 9(5x4 − 1) d5 (x − x + 1) = −9(x5 − x + 1)−10 (5x4 − 1) = − 5 dx (x − x + 1)10 f (x) = 4(3x2 − 2x + 1)−3 , f (x) = −12(3x2 − 2x + 1)−4 6. d3 1 3x2 − 2 (x − 2x + 5) = √ f (x) = √ 2 x3 − 2x + 5 dx 2 x3 − 2x + 5 7. f (x) = √ d 3 (4 + 3 x) = √ √ dx 2 4+3 x 4 x 4+3 x 1 8. f (x) = 3 sin2 x 9. 10. d 24(1 − 3x) (3x2 − 2x + 1) = −12(3x2 − 2x + 1)−4 (6x − 2) = dx (3x2 − 2x + 1)4 f (x) = cos(x3 ) √ d (sin x) = 3 sin2 x cos x dx d3 (x ) = 3x2 cos(x3 ) dx √ √ √ √d √ √d√ 3 cos(3 x) sin(3 x) √ f (x) = 2 cos(3 x) [cos(3 x)] = −2 cos(3 x) sin(3 x) (3 x) = − dx dx x 11. f (x) = sec2 (4x2 ) d (4x2 ) = 8x sec2 (4x2 ) dx 12. f (x) = 12 cot3 x d (cot x) = 12 cot3 x(− csc2 x) = −12 cot3 x csc2 x dx 13. f (x) = 20 cos4 x d (cos x) = 20 cos4 x(− sin x) = −20 cos4 x sin x dx 14. f (x) = − csc(x3 ) cot(x3 ) 15. f (x)...
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