# 10 dy dx x y y cos2 d d tan so sec2 dt 2 dt x dt x dt

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Unformatted text preview: e ﬁrst says dx dx y x the (black) slope is = − and the second says the (gray) slope is , and these are negative reciprocals x y of each other. 1 dx dy = 15y 2 + 1, = ; dy dx 15y 2 + 1 dy dy 1 dy check: 1 = 15y 2 + , = dx dx dx 15y 2 + 1 y = f −1 (x), x = f (y ) = 5y 3 + y − 7, y = f −1 (x), x = f (y ) = 1/y 2 , check: 1 = −2y −3 55. dx dy = −2y −3 , = −y 3 /2; dy dx dy dy , = −y 3 /2 dx dx 1 dx dy = 10y 4 + 3y 2 , = ; dy dx 10y 4 + 3y 2 1 dy dy dy check: 1 = 10y 4 + 3y 2 , = dx dx dx 10y 4 + 3y 2 y = f −1 (x), x = f (y ) = 2y 5 + y 3 + 1, Exercise Set 4.4 56. 112 dx dy 1 = 5 − 2 cos 2y , = ; dy dx 5 − 2 cos 2y 1 dy dy = check: 1 = (5 − 2 cos 2y ) , dx dx 5 − 2 cos 2y y = f −1 (x), x = f (y ) = 5y − sin 2y , EXERCISE SET 4.4 1. 2. 1 (2) = 1/x 2x 3. 2(ln x) 1 x = 2 ln x x 1 (3x2 ) = 3/x x3 4. 1 (cos x) = cot x sin x 6. 1 √ 2+ x 5. 1 sec2 x (sec2 x) = tan x tan x 7. (1 + x2 )(1) − x(2x) 1 − x2 1 8. = x/(1 + x2 ) (1 + x2 )2 x(1 + x2 ) 1 ln x 9. 3x2 − 14x x3 − 7x2 − 3 x3 11. 13. 1 (ln x)−1/2 2 10. 1 x = 1 √ 2x ln x 1 − sin(ln x) x 12. −2x3 (ln 2)(3 − 2x) 1 x 1 x = 1 √ =√ 2 x(2 + x) 1 x ln x + (3x2 ) ln x = x2 (1 + 3 ln x) 1 2 2(ln x)(1/x) 2 = 1 + ln x 14. 15. 3x2 log2 (3 − 2x) + 1 √ 2x 16. ln x x 1 + ln2 x 2 sin(ln x) cos(ln x) log2 (x2 − 2x) 3 sin(ln x2 ) 1 = x x + 3x log2 (x2 − 2x) 17. 2x(1 + log x) − x/(ln 10) (1 + log x)2 18. 7e7x 20. −10xe−5x 21. x3 ex + 3x2 ex = x2 ex (x + 3) 22. − 23. 2x − 2 (x2 − 2x) ln 2 1/[x(ln 10)(1 + log x)2 ] 19. 2 (ex + e−x )(ex + e−x ) − (ex − e−x )(ex − e−x ) dy = dx (ex + e−x )2 = 2 1 1/x e x2 (e2x + 2 + e−2x ) − (e2x − 2 + e−2x ) = 4/(ex + e−x )2 (ex + e−x )2 24. ex cos(ex ) 25. (x sec2 x + tan x)ex tan x 26. (ln x)ex − ex (1/x) dy ex (x ln x − 1) = = dx (ln x)2 x(ln x)2 27. (1 − 3e3x )e(x−e 28. 15 2 x (1 + 5x3 )−1/2 exp( 1 + 5x3 ) 2 29. (x − 1)e−x x−1 =x −x 1 − xe e −x 30. 1 [− sin(ex )]ex = −ex tan(ex ) cos(ex ) 3x ) 113 Chapter 4 31. dy 1 + dx xy 32. 1 dy = dx x tan y 33. d 3x 1 ln cos x − ln(4 − 3x2 ) = − tan x + dx 2 4 − 3x2 34. d dx 35. ln |y | = ln |x| + 36. ln |y | = x dy +y dx = 0, x sec2 y dy y =− dx x(y + 1) dy dy tan y + tan y , = dx dx x(tan y − sec2 y ) 1 [ln(x − 1) − ln(x + 1)] 2 = 1 2 1 1 − x−1 x+1 2x dy 1 1 3 ln |1 + x2 |, = x 1 + x2 + 3 dx x 3(1 + x2 ) 1 dy 1 [ln |x − 1| − ln |x + 1|], = 5 dx 5 5 x−1 1 1 − x+1 x−1 x+1 1 1 ln |x2 − 8| + ln |x3 + 1| − ln |x6 − 7x + 5| 3 2 √ dy (x2 − 8)1/3 x3 + 1 2x 3x2 6x5 − 7 = + −6 6 − 7x + 5 2 − 8) 3 + 1) dx x 3(x 2(x x − 7x + 5 37. ln |y | = 1 ln |x| 2 sin x cos x tan3 x 1 dy 3 sec2 x √ = − cot x − tan x + dx tan x 2x x 38. ln |y | = ln | sin x| + ln | cos x| + 3 ln | tan x| − 39. f (x) = 2x ln 2; y = 2x , ln y = x ln 2, 40. f (x) = −3−x ln 3; y = 3−x , ln y = −x ln 3, 41. f (x) = π sin x (ln π ) cos x; y = π sin x , ln y = (sin x) ln π , 1 y = ln 2, y = y ln 2 = 2x ln 2 y 1 y = − ln 3, y = −y ln 3 = −3−x ln 3 y 1 y = (ln π ) cos x, y = π sin x (ln π ) cos x y 42. f (x) = π x tan x (ln π )(x sec2 x + tan x); 1 y = π x tan x , ln y = (x tan x) ln π , y = (ln π )(x sec2 x + tan x) y y = π x tan x (ln π )(x sec2 x + tan x) 43. ln y = (ln x) ln(x3 − 2x), dy = (x3 − 2x)ln x dx 1 dy 3x2 − 2 1 =3 ln x + ln(x3 2x), y dx x − 2x x 1 3x2 − 2 ln x + ln(x3 − 2x) x3 − 2x x sin x dy sin x 1 dy = + (cos x) ln x, = xsin x + (cos x) ln x y dx x dx x 44. ln y = (sin x) ln x, 45. ln y = (tan x) ln(ln x), 1 1 dy = tan x + (sec2 x) ln(ln x), y dx x ln x tan x dy = (ln x)tan x + (sec2 x) ln(ln x) dx x ln x Exercise Set 4.4 46. 114 1 dy 2x 1 =2 ln x + ln(x2 + 3), y dx x +3 x 1 2x ln x + ln(x2 + 3) x2 + 3 x ln y = (ln x) ln(x2 + 3), dy = (x2 + 3)ln x dx 47. y = Ae2x + Be−4x , y = 2Ae2x − 4Be−4x , y = 4Ae2x + 16Be−4x so y + 2y − 8y = (4Ae2x + 16Be−4x ) + 2(2Ae2x − 4Be−4x ) − 8(Ae2x + Be−4x ) = 0 48. y = Aekt , dy/dt = kAekt = k (Aekt ) = ky 49. (a) f (x) = kekx , f (x) = k 2 ekx , f (x) = k 3 ekx , . . . , f (n) (x) = k n ekx (b) f (x) = −ke−kx , f (x) = k 2 e−kx , f (x) = −k 3 e−kx , . . . , f (n) (x) = (−1)n k n e−kx 50. dy = e−λt (ωA cos ωt − ωB sin ωt) + (−λ)e−λt (A sin ωt + B cos ωt) dt = e−λt [(ωA − λB ) cos ωt − (ωB + λA) sin ωt] 1 1 exp − 2 2πσ x−µ σ 2 f (x) = √ 1 1 exp − 2 2πσ x−µ σ 2 =√ 51. = −√ 1 d − dx 2 − x−µ σ x−µ σ 1 1 (x − µ) exp − 3 2 2πσ x−µ σ 2 1 σ 2 54. y = −xe−x + e−x = e−x (1 − x), xy = xe−x (1 − x) = y (1 − x) y = −x2 e−x /2 + e−x /2 = e−x /2 (1 − x2 ), xy = xe−x /2 (1 − x2 ) = y (1 − x2 ) (a) logx e = 1 d ln e 1 = , [logx e] = − ln x ln x dx x(ln x)2 (b) 53. (a) (b) 52. logx 2 = ln 2 d ln 2 , [logx 2] = − ln x dx x(ln x)2 2 β = 10 log I − 10 log I0 , (a) (c) 55. 2 dβ dI dβ dI = I =10I0 = I =100I0 2 2 dβ 10 = dI I ln 10 1 db/W/m2 I0 ln 10 1 100I0 ln 10 dk q (T − T0 ) = k0 exp − dT 2T0 T (b) dβ dI I =100I0 db/W/m2 − q 2T 2 =− qk0 q (T − T0 ) exp − 2T 2 2T0 T 56. because xx is not of the form ax where a i...
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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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