Ln y 1 ln x 2 sin x cos x tan3 x 1 dy 3 sec2 x cot

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Unformatted text preview: = −t/2 and lim(1 − 2x)1/x = lim(1 + t)−2/t = lim[(1 + t)1/t ]−2 = e−2 . 58. If t = 3/x, then x = 3/t and lim (1 + 3/x)x = lim (1 + t)3/t = lim [(1 + t)1/t ]3 = e3 . + + x→0 x→+∞ t→0 t→0 t→0 t→0 107 Chapter 4 EXERCISE SET 4.3 1. y = (2x − 5)1/3 ; dy/dx = 2. dy/dx = 1 2 + tan(x2 ) 3 3. dy/dx = 3 x−1 2 x+2 4. dy/dx = 1 x2 + 1 2 x2 − 5 5. dy/dx = x3 − 2 3 1/2 2 (2x − 5)−2/3 3 −2/3 sec2 (x2 )(2x) = 2 x sec2 (x2 ) 2 + tan(x2 ) 3 9 d x−1 x−1 = 2 x+2 dx x + 2 2(x + 2) −1/2 d x2 + 1 1 x2 + 1 = dx x2 − 5 2 x2 − 5 −1/2 −2/3 1/2 −12x 6x x2 + 1 =− 2 (x2 − 5)2 (x − 5)2 x2 − 5 (5x2 + 1)−5/3 (10x) + 3x2 (5x2 + 1)−2/3 = 12 x (5x2 + 1)−5/3 (25x2 + 9) 3 4 x2 (3 − 2x)1/3 (−2) − (3 − 2x)4/3 (2x) 2(3 − 2x)1/3 (2x − 9) = 6. dy/dx = 3 4 x 3x3 5 15[sin(3/x)]3/2 cos(3/x) [sin(3/x)]3/2 [cos(3/x)](−3/x2 ) = − 2 2x2 7. dy/dx = 8. dy/dx = − 9. 10. 1 cos(x3 ) 2 −3/2 − sin(x3 ) (3x2 ) = 32 x sin(x3 ) cos(x3 ) 2 −3/2 dy dy 2 − 3x2 − y + y − 2 = 0, = dx dx x dy 1 + 2x − x3 1 1 (b) y = = + 2 − x2 so = − 2 − 2x x x dx x 2 − 3x2 − y 2 − 3x2 − (1/x + 2 − x2 ) 1 dy = = = −2x − 2 (c) from part (a), dx x x x (a) 3x2 + x 1 −1/2 dy dy √ y − ex = 0 or = 2ex y 2 dx dx dy (b) y = (2 + ex )2 = 2 + 4ex + e2x so = 4ex + 2e2x dx dy √ = 2ex y = 2ex (2 + ex ) = 4ex + 2e2x (c) from part (a), dx (a) dy x dy = 0 so =− dx dx y 11. 2x + 2y 12. 3x2 − 3y 2 13. x2 14. x3 (2y ) dy x2 − 2y dy dy dy = 6(x + y ), −(3y 2 + 6x) = 6y − 3x2 so =2 dx dx dx dx y + 2x dy dy + 2xy + 3x(3y 2 ) + 3y 3 − 1 = 0 dx dx 1 − 2xy − 3y 3 dy dy = 1 − 2xy − 3y 3 so = (x2 + 9xy 2 ) dx dx x2 + 9xy 2 dy dy + 3x2 y 2 − 5x2 − 10xy + 1 = 0 dx dx 10xy − 3x2 y 2 − 1 dy dy (2x3 y − 5x2 ) = 10xy − 3x2 y 2 − 1 so = dx dx 2x3 y − 5x2 −1/2 Exercise Set 4.3 108 1 dy dy 1 y2 − 2 = 0 so =− 2 2 dx y x dx x 15. − 16. 2x = (x − y )(1 + dy/dx) − (x + y )(1 − dy/dx) , (x − y )2 2x(x − y )2 = −2y + 2x 17. 18. 19. cos(x2 y 2 ) x2 (2y ) 21. dy 1 − 2xy 2 cos(x2 y 2 ) dy + 2xy 2 = 1, = dx dx 2x2 y cos(x2 y 2 ) (1 + csc y )(− csc2 y )(dy/dx) − (cot y )(− csc y cot y )(dy/dx) , (1 + csc y )2 dy 2x(1 + csc y )2 = − csc y (csc y + csc2 y − cot2 y ) , dx 2x(1 + csc y ) dy but csc2 y − cot2 y = 1, so =− dx csc y 2x = 3 tan2 (xy 2 + y ) sec2 (xy 2 + y ) 2xy so 20. dy dy + y2 + dx dx =1 dy 1 − 3y 2 tan2 (xy 2 + y ) sec2 (xy 2 + y ) = dx 3(2xy + 1) tan2 (xy 2 + y ) sec2 (xy 2 + y ) dy (1 + sec y )[3xy 2 (dy/dx) + y 3 ] − xy 3 (sec y tan y )(dy/dx) = 4y 3 , 2 (1 + sec y ) dx dy multiply through by (1 + sec y )2 and solve for dx y (1 + sec y ) dy = to get dx 4y (1 + sec y )2 − 3x(1 + sec y ) + xy sec y tan y 3x d2 y (4y )(3) − (3x)(4dy/dx) 12y − 12x(3x/(4y )) 12y 2 − 9x2 −3(3x2 − 4y 2 ) dy = , = = = = , dx 4y dx2 16y 2 16y 2 16y 3 16y 3 but 3x2 − 4y 2 = 7 so 22. dy x(x − y )2 + y dy so = dx dx x d2 y −3(7) 21 = =− 2 3 dx 16y 16y 3 x2 d2 y dy y 2 (2x) − x2 (2ydy/dx) 2xy 2 − 2x2 y (−x2 /y 2 ) 2x(y 3 + x3 ) = − 2, =− =− =− , 2 4 4 dx y dx y y y5 but x3 + y 3 = 1 so d2 y 2x =− 5 dx2 y 23. y d2 y dy x(dy/dx) − y (1) x(−y/x) − y 2y =− , =− =− =2 dx x dx2 x2 x2 x 24. y dy = , dx y−x (y − x)(dy/dx) − y (dy/dx − 1) d2 y = = 2 dx (y − x)2 = 25. (y − x) y y −y −1 y−x y−x (y − x)2 y 2 − 2xy d2 y 3 but y 2 − 2xy = −3, so =− (y − x)3 dx2 (y − x)3 sin y dy d2 y dy = (1 + cos y )−1 , = = −(1 + cos y )−2 (− sin y ) 2 dx dx dx (1 + cos y )3 109 26. Chapter 4 dy cos y = , dx 1 + x sin y (1 + x sin y )(− sin y )(dy/dx) − (cos y )[(x cos y )(dy/dx) + sin y ] d2 y = 2 dx (1 + x sin y )2 =− 2 sin y cos y + (x cos y )(2 sin2 y + cos2 y ) , (1 + x sin y )3 but x cos y = y , 2 sin y cos y = sin 2y , and sin2 y + cos2 y = 1 so sin 2y + y (sin2 y + 1) d2 y =− 2 dx (1 + x sin y )3 27. 28. √ √ √ √ dy dy x = − ; at (1/ 2, 1/ 2), = −1; at (1/ 2, −1/ 2), dx y dx √ −x dy dy 2, = +1. Directly, at the upper point y = 1 − x =√ = −1 and at the lower point dx dx 1 − x2 √ x dy =√ = +1. y = − 1 − x2 , dx 1 − x2 By implicit differentiation, 2x+2y (dy/dx) = 0, √ √ If y 2 − x + 1 = 0, then y = x − 1 goes through the point (10, 3) so dy/dx = 1/(2 x − 1). By √ implicit differentiation dy/dx = 1/(2y ). In both cases, dy/dx|(10,3) = 1/6. Similarly y = − x − 1 √ goes through (10, −3) so dy/dx = −1/(2 x − 1) = −1/6 which yields dy/dx = 1/(2y ) = −1/6. dy x3 1 dy = 0, so = − 3 = − 3/4 ≈ −0.1312. dx dx y 15 29. 4x3 + 4y 3 30. 3y 2 31. 4(x2 + y 2 ) 2x + 2y dy dy dy dy y+1 + x2 + 2xy + 2x − 6y = 0, so = −2x 2 = 0 at x = 0 dx dx dx dx 3y + x2 − 6y dy dx = 25 2x − 2y dy , dx x[25 − 4(x2 + y 2 )] dy dy = ; at (3, 1) = −9/13 dx y [25 + 4(x2 + y 2 )] dx 32. 2 3 34. (a) x−1/3 + y −1/3 dy dx = 0, √ √ dy y 1/3 = − 1/3 = 3 at (−1, 3 3) dx x y 2 0 x 1 2 –2 (b) dy dy = (x − a)(x − b) + x(x − b) + x(x − a) = 3x2 − 2(a + b)x + ab. If = 0 then dx dx 2 3x − 2(a + b)x + ab...
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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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