Business Calc Homework w answers_Part_24

Business Calc Homework w answers_Part_24 - 116 Section 3.9...

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Unformatted text preview: 116 Section 3.9 11. dy dx d (x ) dx 12. dy dx d1 (x dx 13. dy dx d x dx 14. dy dx d 1e x dx 15. dy dx dx 8 dx 16. 17. dy dx 26. 2 2x e1 (1 dy dx d ln (2 dx sin x 2 cos x dy dx d ln (ln x) dx 30. 2 d ln (2x 2) dx 1 ,x 1 x1 27. 2)x dy dx 21 e)x 1 (1 21 dy dx 29. 2)x 1 (1 dy dx 28. 2) (1 e e)x 8x ln 8 d 9x dx dy dx 25. 1 x 9 d csc x 3 dx x d dx (ln 9) ( x) 9 x ln 9 d dx 3csc x (ln 3) (csc x) 3csc x (ln 3)( csc x cot x) 18. d cot x 3 dx 3 cot x 31. d (ln 3) (cot x) dx dy dx 3cot x (ln 3)( csc 2 x) dy 32. dx 19. Use logarithmic differentiation. x ln x ln y ln y d (ln y) dx 1 dy y dx dy dx dy dx ln x ln x 33. ln x ln x d (ln x)2 dx 1 (2 ln x) x 2y ln x x 2x ln x ln x x 34. 36. dy dx ln x 1/ln x ln y 1 ln x ln x ln y 1 y dy dx dy dx x 1/ln x ln y dy dx 35. 20. Use logarithmic differentiation. y e dy dx d (e) dx 21. dy dx d ln (x 2) dx 22. dy dx d (ln x)2 dx 23. dy dx d ln (x 1) dx 24. dy dx 10 d ln x dx 37. dy dx 38. 0, x 2 ln x dy dx 39. dy dx 0 1d 2 (x ) x 2 dx 1 (2x) x2 d (ln x) dx d ( ln x) dx d (ln 10 dx ln x) 2 x 1 d2 (x 1 dx x2 1d ln x ln x dx 1 x) (x) 1 x ,x 2 2 2x 2 cos x) 2x 1) 1 x 1 ln x 2 2) 1 d (2 cos x dx 2 1) x d (2x 2 dx 2x x2 1 1 x ln x (ln x)(1) 1 ln x d d ln x 2 (log4 x 2) dx dx ln 4 1 2 2 1 ln 4 x x ln 4 x ln 2 d dx 2 (ln x) ln 4 ln x d2 dx ln 5 1 ,x 2x ln 5 d d ln x 1/2 (log5 x) dx dx ln 5 1 1 d 1 (ln x) 2 ln 5 dx 2 ln 5 x d log2 (3x dx 3 ,x (3x 1) ln 2 1) (3x 1 ,x x 40. 0 1 x 1 ,x x 0 dy dx 0 1 d (3x 1) ln 2 dx 1) 1 3 1d d log10 (x 1)1/2 log10 (x 1) 2 dx dx 1 1 d 1 (x 1) ,x 2 (x 1) ln 10 dx 2(x 1) ln 10 1 d log2 x dx d ( log2 x) dx d 1 1 d (log2 x) dx log2 x (log2 x)2 dx 1 1 1 or (log2 x)2 x ln 2 x(ln 2)(log2 x)2 d (ln 2 log2 x) dx 1 1 (ln 2) ,x x x ln 2 1 1 ,x x ln 2 0 ln 2 x(ln x)2 d dx (ln 2) (log2 x) 0 d 1 d log3 (1 x ln 3) (1 dx (1 x ln 3) ln 3 dx ln 3 1 1 ,x (1 x ln 3) ln 3 1 x ln 3 ln 3 d (log10 e x) dx d (x log10 e) dx log10 e 1 ln 10 2 ln x x 0 1 1 2) 1 3cot x (ln 3)(csc 2x) y d (x 2 dx x cos x) d (x ln x dx ln x 1 2) d ln (x 2 dx 1 3csc x (ln 3)(csc x cot x) dy dx d ln (x dx d ln 10x dx d (x ln 10) dx ln 10 ln e ln 10 x ln 3) Section 3.9 41. The line passes through (a, e a) for some value of a and has e a. Since the line also passes through the origin, slope m y ea , so a a xe x, we have y (x)(e x) (e x)(1) (x has slope m (a 1 (x 1)e a (a includes the point (0, 0), so we have: need to use a y 43. y 0) 46. 0e 0, or y ln y y ln (sin x)(1) 47. dA dt ln (sin x)] ln y (tan x)(ln x) At t d [(tan x)(ln x)] dx 1 (tan x) (ln x)(sec 2 x) x tan x y (ln x)(sec 2 x) x tan x x tan x (ln x)(sec 2 x) x dA dt 1 1 (2x) 2 x2 1 1 x 2d ln (x 3 dx 21 (1) 3x 1 x x2 1) 2 1 3(x xx 11 (x 1)2/3 x 1) x x2 2 1 3(x 1) d 1 t/140 dt 2 d 20 2 t/140 dt (2 ln (x tan x) 1d ln (x 2 2 dx 1) 20 20(2 x tan x 2 ln (x 3 1) 2 20 (2 ln y d ln y dx 1 dy y dx dy dx dy dx y dy dx ln (sin x)] (sin x)x[x cot x 1 x dy dx x(x 2 1)1/2 (x 1)2/3 1 ln (x 2 2 d ln x dx 1 dy y dx d [x ln (sin x)] dx 1 (x) (cos x) sin x y[x cot x ln x d ln y dx x ln (sin x) d ln y dx 1 dy y dx dy dx dy dx ln ln y ln (sin x)x 6 5(2x 5) x(x 2 1)1/2 (x 1)2/3 ln y x. 6 5(2x 5) 3)4(x 2 1) 1/5 (2x 5)3 x x2 1 (x 1)2/3 y t/140 )(ln 2) d dt t/140 )(ln 2) t 140 1 140 t/140 )(ln 2) 7 2 days, we have (2 1/70 )(ln 2) 7 0.098 grams/day. This means that the rate of decay is the positive rate of approximately 0.098 grams/day. 48. (a) (b) d ln (kx) dx 5)] 3 1 (2) 5 2x 5 2x 5(x 2 1) 3) 2x 5(x 2 1) 3) 3 ln (2x 5) 1 1 (2x) 5 x2 1 4 5(x (x 4 5(x (sin x)x ln y 44. dy dx 0. The equation of the normal line is 1 (x (0 1)e 0 41 5x 3 y 1) 3) 3d ln (2x 5 dx 1) dy dx ln (x 2 3) 4d ln (x 5 dx 1 dy y dx ae a. The desired normal line 1 0 (0 a) ae a (a 1)e a a 0 ae a (a 1)e a 1 0a ea (a 1)e a 1 a 0 or ea 0 (a 1)e a 1 The equation e a 0 has no solution, so we (a 1)e a 1 [4 ln (x 5 1d ln (x 2 5 dx 1 and its equation is 1)e a a) 1 (x 3)4(x 2 1) ln 5 (2x 5)3 d (ln y) dx 1)e x, so the normal line through the point (a, ae a) y 3)4(x 2 1) 1/5 (2x 5)3 (x ln ln y 1. Hence, the slope is e and the equation is 3)4(x 2 1) 1/5 (2x 5)3 (x ln y 0 and we have 0 ex. 42. For y 3)4(x 2 1) (2x 5)3 (x 5 y ln y the slope is also given by m ea ea a 45. 117 1d kx kx dx d d ln (kx) (ln k dx dx 1 d 0 ln x x dx k kx ln x) 1 x 1) 118 Chapter 3 Review 2x ln 2, f (0) 49. (a) Since f (x) (b) f (0) lim f (h) h→0 f (0) 20 ln 2 lim 2h 20 h→0 h s Chapter 3 Review Exercises ln 2. lim h→0 h (pp. 172–175) 2h 1 (c) Since quantities in parts (a) and (b) are equal, lim 2h h→0 1 dy dx d5 x dx dy dx d (3 dx 3. dy dx d (2 sin x cos x) dx d 2(sin x) (cos x) dx ln 2. h (d) By following the same procedure as above using 7x, we may see that lim g(x) 1. 2. h 7h h→0 1 ln 7. h y3 3 dy dx 4 dy dx d 2x dx 2x 1, or a dx a dx (d) y2 1. y1 dx ln a, then a will equal a x if and only if dx So if y3 ln a y2 5. ds dt 1, or a 12 x 2 d dx k x and ds dt 2 d cot t dt 2 22 csc t t2 7. dy dx a x if and d dx 8. dy dx 1 . x c) 9. dr d Therefore, at any given value of x, these two curves will 10. dr d 53. (a) Since the line passes through the origin and has slope (b) The graph of y y positive x e. Therefore, ln x x for all e 11. to see that 2d2 t dt t 1 3/2 x 2 e 1)2 x. e . Therefore, e is bigger. 1 d sec (1 d 2x 3) 1 2 t 2t) 2 t2 1/2 x ) ( 2x 1)(1) 1 1 3 ) tan (1 d tan 2 (3 d (2) 2x sec (1 3 ) tan (1 3 )(3) 3) 2 ) 2d ) 2 ) sec 2 (3 2 2 tan (3 d 2 ) )( 2 ) 2 2 ) sec (3 ) d2 (x csc 5x) dx 5x 2 csc 5x cot 5x dy dx d ln dx 13. dy dx d ln (1 dx 14. dy dx d (xe x) dx 12. 2 sin (1 csc 2 1 d (x 2x 1) (x) dx 2 x (2x 1) 3x 1 (x 2)( csc 5x cot 5x)(5) e (d) Exponentiating both sides of ln x e x, we have e e ln x e x, or x e e x for all positive x e. (e) Let x dy dx 2t)( 2) d 1/2 (x dx 1 1 2x3/2 2x 4 tan (3 e. x or ln x 4 (2x x 2 tan (3 ln x lies below the graph of the line (c) Multiplying by e, e ln x 1)(2) 1 1 2 tan (3 x . e x for all positive x e (cos 2x)(2) (2x 1)2 sin (1 csc 2 3 sec (1 have perpendicular tangent lines. 1 , its equation is y e 1)(2) (2x 2t) x 1 1/2 x 2 e. d (ln x dx d sin 2x dx (2x 1 1 2x 52. 2 cos 2 x d cos (1 dt e. a x ln a. This will equal y1 only if ln a d dx 5 We conclude that the graph of y3 is a horizontal line at y ln a. a x if and only if y3 21x 6 2(cos x) (sin x) d (2 sin x cos x) dx 4. 0.693147 1.098613 1.386295 1.609439 dx a dx 21x 2 2 cos 2x ln a 0.693147 1.098612 1.386294 1.609438 (c) 3x 7) 1 4 Alternate solution: 6. 2 1 x 4 5x 4 a. (b) The graph of y3 is always a horizontal line. a 1 x 4 7x 3 2 sin 2 x 50. Recall that a point (a, b) is on the graph of y e x if and only if the point (b, a) is on the graph of y ln x. Since there are points (x, e x) on the graph of y e x with arbitrarily large x-coordinates, there will be points (x, ln x) on the graph of y ln x with arbitrarily large y-coordinates. 51. (a) The graph y4 is a horizontal line at y 12 x 8 x 1 2x csc 5x d x dx e x) (x)(e x x 1 1 x 1 1 (csc 5x)(2x) d (1 e x dx )( 1) (e 2 e x) x 1 ,x 2x x )(1) 0 ex 1 ex xe x e x Chapter 3 Review 15. dy dx d 1 ln x) (e dx dy 16. dx d 1 ln x (e e ) dx d ln (sin x) dx d (ex) dx Alternate solution, using logarithmic differentiation: e (2x)2x y cos x sin x 1d (sin x) sin x dx x2 cot x, for 1 ln (2x) dr 17. dx 1) ), where k is even. d 1 d ln (cos 1 x) cos 1 x dx cos 1 x dx 1 1 1 cos 1 x 1 x2 cos 1x 1 x 2 18. dr d d log2 ( 2) d 19. ds dt d log5 (t dt t ln y 2 7) 1 d2 () ln 2 d (t 2 ln 2 1 d (t 7) ln 5 dt 7) (t 1 , 7) ln 5 7 23. d (8 t ) dt d 8 (ln 8) ( t) dt t 8 t dy dx 24. ds 20. dt x d tan 1x e dx e tan 1 dy du 1 u2 d sin 1 du 1 1 25. d (ln x)2 dx d 2 ln x ln x dx 2y ln x x 2x ln x ln x x dy dt x2 26. x2 dy dt 1 27. (2x)(2x) 1 2 x2 dy dz t 2) cot t 2) 2t 2 4t 2 x 2 1)(2 )(2x ln 2 2) (x 2 1)3/2 (2 2x)[(x 2 x 3 1)(x ln 2 (x 2 1)3/2 2x (2 ) 1) 2 (2 2 )(x ln 2 x x ln 2 (x 2 1)3/2 cos 28. x 2] 1 2 dy dx d (2 dx (2 t 1 1 2t)(2t) 2t z 2) 1 1 (cos 1 z)(1) 2 1 1 z z z2 1 x 1 1 csc x) 1 1) x) x 1 x) 1 2 ( x) x x)2 1 x 1 csc x x 1 1 1 csc 1 1 1 2x 1) 1 x (cot z x ( 1 2t t)(1) 2t] cos z2 (2 csc (2 2x)(x 3 ln 2 x ln 2 (x 2 1)3/2 1 z2 z 1 u2 1 1 2t 2t cot 1 1 x 1 d (z cos 1 z dz 1 1 1 1 (2) (2t)2 1 1 (x (sec (z) (2x) u u 1 d [(1 dt u2 1 ( 2u) sec t2 2 1 1 (2x)(2)] x 1 t 1[(2x)(2x)(ln 2) etan x 1 x2 1 tan dx 1 t d (2x)(2x) dx 1 xd 1 ln t) 2 t2 t x2 2 u2 d (t sec 1 t dt 1 x2 x2 1 (1 d 1 dx [(2x)2x] x2 1 (t) (2x)2x d dx x ln 2 d u 2)2 du 1 u2 2 (ln x)(ln x) d ln y dx 1 dy y dx dy dx dy dx ( 1 ln (x ln x) ln y 1 1 ln (x 2 1) 2 1 d [ln 2 ln x x ln 2 ln (x 2 1)] 2 dx 1 1 1 0 ln 2 (2x) x 2 x2 1 1 x y ln 2 x x2 1 (2x)2x 1 x ln 2 x2 1 x2 1 x ln x ln y dy 22. dx ln x x2 ln ln 8 21. Use logarithmic differentiation. y ln 2 d ln y dx 1 dy y dx dy dx dy dx 2 ln 2 2 ln (2x) ln y values of x in the intervals (k , (k 119 x x 1 2 x z2 ( 2z) 120 29. Chapter 3 Review dy dx 32. Since y d csc 1 (sec x) dx 1 d (sec x) 1 dx 2 sec x sec x 1 tan 2 x sec x dy dx sin x cos x sin x cos x sin x sin x 2 cos 2 1 x ,, 2 ,x 2 3 , we may 2 ,x (sec x) 0 x , d1 d1 x 1, sin cos sin c os (1 1 2 1 sin c os cos 1 1 sin c os cos (1 2 (2x 7) dy dx x xy 2 3 2 2,x x , x 2,x x 2 2 cos )(cos ) (1 sin )(sin ) (1 cos )2 cos2 sin (1 cos )2 all x 0. x 2x 5) 7)(1) (x 7)2 2x x2 3y 5x 4/5 d (5x 4/5) dx 4x 1/5 10y 6/5 5 is defined for all 7 5)(2) 17 , the (2x 7)2 7 . 1 d (1) dx 0 (y y x 2) 2 3 15 d (10y 6/5) dx dy 12y 1/5 dx dy dx d (15) dx 0 4x 1/5 12y 1/5 1 3(xy)1/5 37. Use implicit differentiation. xy d dx 1 [x xy dy dx 0 and 1 xy d (1) dx (y)(1)] 0 dy dx 0 y dy dx y x Alternate method: Since xy Therefore, 1, we have xy dy dx 1 . x2 1, 1. 36. Use implicit differentiation. 2 2 , the function is differentiable for x 2x d d d (xy) (2x) (3y) dx dx dx dy dy x (y)(1) 2 3 dx dx dy (x 3) dx dy dx sin2 sin 1 cos )2 ln x is defined for all x 1d 2 (x ) x 2 dx (x (2x x 2 31. Since y 1 (2x 7 dy and 2 dx 2 1 1 2 34. Since y 2 3 2 Note that the derivative exists at 0 and 2 only because these are the endpoints of the given domain; the two-sided derivative of y csc 1(sec x) does not exist at these points. dr d , which is defined only for x 35. Use implicit differentiation. x 2,x , 1, 0 dy Therefore, dx 30. x 2)3/2 x (1 2 x x, 2 1 function is differentiable for all x x x), x, 2 2x 1 and x)(2x) (cos x) 0 ( 2 x 2)( 1) (1 (1 x 2)2 (1 x x2 21 3 2 x x, 2 1 2 1 x2 (sec x) sec 2 x sin x, the x is defined for all x x2 1 dy dx rewrite the function as follows: 1 (cos x)(1) the function is differentiable for all x On the domain 0 1 1 1 33. Since y Alternate method: csc (x)( sin x) sec x tan x sign (sin x), x y cos x x cos x is defined for all real x and function is differentiable for all real x. sec x tan x sec x tan x 1 cos x 1 cos x sin x 1 and y 1 . x ...
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

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