Business Calc Homework w answers_Part_21

Business Calc Homework w answers_Part_21 - 101 Section 3.7...

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Unformatted text preview: 101 Section 3.7 70. For h 1: 72. dG dx d dx 1 2 2x uv 2 uv uv d dx x(x d (x 2 dx cx A G cx) d dx c) 2x 2x x2 c 2 cx x cx (x c) 2 x(x c) [ 2, 3] by [ 5, 5] For h 0.7: s Section 3.7 Implicit Differentiation (pp. 149–157) Exploration 1 1. 2x dy dx [ 2, 3] by [ 5, 5] For h 2y An Unexpected Derivative 2xy 2yy 1 (provided y 0. Solving for y , we find that x). 2. With a constant derivative of 1, the graph would seem to be a line with slope 1. 0.3: 3. Letting x 0 in the original equation, we find that y 2. This would seem to indicate that this equation defines two lines implicitly, both with slope 1. The two lines are y x 2 and y x 2. [ 2, 3] by [ 5, 5] As h → 0, the second curve (the difference quotient) approaches the first (y 2x sin (x 2)). This is because 2x sin (x 2) is the derivative of cos (x 2), and the second curve is the difference quotient used to define the derivative of cos (x 2). As h → 0, the difference quotient expression should be approaching the derivative. 71. (a) Let f (x) 4. Factoring the original equation, we have [(x y) 2][(x y) 2] 0 x y 2 0 or x y 2 0 y x 2 or y x 2. The graph is shown below. x. Then d u dx d f (u) dx f (u) du dx d u du du dx u u. u The derivative of the absolute value function is positive values, at 0. So f (u) 1 for negative values, and undefined 1, 1, u u 0 0. g (x) u evaluates. u (2x)(x 2 9) d x2 9 (x 2 9) x2 9 dx x2 9 d ( x sin x) dx d d x (sin x) (sin x) x dx dx x sin x x cos x x Note: The expression for g (x) above is undefined at x 0, but actually g (0) lim g(0 h→0 h) h g(0) lim h→0 h sin h h Therefore, we may express the derivative as x cos x g (x) 0, x sin x , x 5. At each point (x, y) on either line, x 0 x 0. 0. dy dx 1. The condition y x is true because both lines are parallel to the line y x. The derivative is surprising because it does not depend on x or y, but there are no inconsistencies. Quick Review 3.7 But this is exactly how the expression (b) f (x) 1 for [ 4.7, 4.7] by [ 3.1, 3.1] 1. x y2 x x y1 0 y2 y x, y2 [ 6, 6] by [ 4, 4] x 102 Section 3.7 2. 4x2 9y 2 36 9y 2 36 7. y sin x 4x 2 36 y2 4x 2 3 2 3 y1 4 (9 9 y xy y (sin x x)y y x cos x y x cos x x 2) x2 9 y 2 3 2 9 xy y sin x 2 9 y x cos x x , y2 9 x cos x sin x x y) 8. x(y 2 y (x 2 y) xy 2 y (x 2 y 9. [ 4.7, 4.7] by [ 3.1, 3.1] x2 4y 2 2y) 0 10. y x ,y 22 3 x2 x3 x x 2 y1 x+ x 1/2 x x 2/3 x x x 2/3 x 3/2 x 3/2 x 2 y x 1/2(x x 1/3) x 1/2x x 1/2x 1/3 x 3/2 x 5/6 x) 0 2y)(x (x x2 3 x(x x) xy2 y 3. x 2 3/2 5/6 x Section 3.7 Exercises 1. [ 6, 6] by [ 4, 4] 4. x 2 2 y y2 y y1 9 9 2. x2 9 x2 9 x2, y2 3. 9 y2 4. 5. [ 4.7, 4.7] by [ 3.1, 3.1] 5. x 2 2 y y2 y y1 2x 2x 3 3 x2 2x 3 x2 2x 3 x2, y2 6. 2x 3 dy dx dy dx dy dx dy dx dy dx dy dx x2 [ 4.7, 4.7] by [ 3.1, 3.1] 6. x 2y 2xy x 2y y 4x y 4x y 4x y x2 2xy dy dx 9 5/4 x 4 3 ( 3/5) x 5 d x 3/5 dx 3 8/5 x 5 1 d3 x dx d 1/3 x dx 1 (1/3) x 3 1 1 2/3 x 3 d4 x dx d 1/4 x dx 1 (1/4) x 4 1 1 3/4 x 4 1 d d (2x 5) 1/2 (2x 5)( 1/2) 1 (2x 2 dx dx 1 (2x 5) 3/2(2) (2x 5) 3/2 2 d (1 dx 2 (1 3 2 (1 3 4(1 7. 9 (9/4) 1 x 4 d 9/4 x dx 6x)2/3 1d 6x)(2/3) 6x) 1/3 6x) dx (1 ( 6) 1/3 d (x x 2 1) dx d x x2 1 x2 dx d x (x 2 1)1/2 (x 2 dx 12 x (x 1) 1/2(2x) 2 x 2(x 2 1) 6x) 1/2 (x 2 1 d (x) dx 1)1/2 (x 2 1)1/2 1)1/2 2xy Note: This answer is equivalent to 2x 2 1 x2 1 . 5) Section 3.7 dy x d dx dx x2 (x 2 8. 1)1/2 (x d dx 1)1/2 x x (x 2 x2 1 12 (x 2 x x2 x2 d dx (x 2 1) 1/2 1)1/2 x x2 12. d2 (x ) dx 1 y xy dx y dx x y (2x) (x 1 x 1 (x 2 (x 2x x 2y 9. d2 (x y) dx x 2 dy y(2x) dx x(2y) x xy 2 6 d (xy 2) dx dy y 2(1) dx 2 dy 2xy dx dy x d3 (x ) dx 2 d3 (y ) dx 2 dy 2 dy dx dy 3x 3y y 3 dx (3y 2 3y 18x 18x) dx dy dx dy d2 y dx dy 2y 2y x x d dx (x dx dy dx dy dx y 2) (2xy 2xy 18x 18y x(x x y)2 x y y) x y x3 x 2y x y x2 2 1) 3x 2 y2 13. dy dx 6x 14. (x 1)2 y (x x d (x) dx y)2 dy dx 3x 2 1 dx dy dy dx 1 3x d 3(2x 1/2 dx (2x x2 1/2 (2x 2 x dy dx 2xy 1 1/2 1) 1) 1) 1/3 4/3 d dx 4/3 (2x (x 3/2 (2x 1/2 1) 1/2 3/2 (x 15. 2xy 2 d (1 x 1/2)1/2 dx 1 d (1 x 1/2) 1/2 (1 x 1/2) 2 dx 1 1 1/2 (1 x 1/2) 1/2 x 2 2 1 (1 x 1/2) 1/2x 1/2 4 1)(1) 1)2 1 y(x 1)2 d (y) dx dy 1 dx 3x 2 18y y)2 y(2x) dx 2 18y dx d2 (x y) dx dy (x 3x y)2 y x x2 3y 2 18x 6y x 2 1 1 1)(1) (x y x 2(x 18y (1) dx x dx x(x dx dy y2 2xy x dx y)2 dy y dy y 2) 1 1 x x dy Alternate solution: 3x 2 dx y2 (2xy dx dy x dx 18xy dx dy 11. y y d3 (x ) dx d (18xy) dx dy 2x (x y)2 x 0 dx 10. y dx x2 x(x d (6) dx dx dy x 2) dx dy (2xy 3 dy (x 2y 3/2 1) x dy y) 1 y)2 2x 1)3/2 2 (x dx (x 1 x2 1)(x 2 1)1/2 2 dy y) 1 2x 4/3 d 3(csc x)3/2 dx 9 d (csc x)1/2 (csc x) 2 dx 9 (csc x)1/2( csc x cot x) 2 9 (csc x)3/2 cot x 2 ) 1) dy dx y y dy dx 103 104 16. Section 3.7 dy dx 17. d [sin (x dx 5 [sin (x 4 5 [sin (x 4 x (c) Differentiating both sides of the given equation 5)]5/4 5)]1/4 d sin (x dx 5)]1/4 cos (x f (x) 5) dy dx x 22. (a) If g (t) sin y d (x) dx dy dx 4 4 t 19. d (4t 1/4 dx 4) 3/4 t x 1 t 3/4 g (t) tan xy t 3/4 dy dx dy dx dy dx (c) If g (t) 0 g (t) 0 4 1 (sec2 xy)(y) 1 d (x) dx 1 sin y y sec2 xy (cos y) (cos y x) x2 x 1/3 d2 (x ) dx 2x 1 y 1 x x y2 y cos y d2 (y ) dx 2yy 0 2x x y y x y d dx 1 4/3 x 3 y x 1/3 x x y y2 . y2 y3 7, then 3 2/3 x and f (x) 2 (x)(y ) y2 x2 f (x) 1 3/4 t , which 16 d (1) dx (y)(1) which contradicts the given equation f (x) (b) If f (x) 4t 1/4 and 1 3, then and f (x) 9 5/3 x 10 1 , which matches the given 1 1/4 t , then g (t) 4 2yy y cos y t 3/4 y 1 3 2/3 x 2 21. (a) If f (x) f (x) dy dx dy dx 3/4 16 5/4 t , then g (t) 5 1 Conclusion: (a) and (c) could be true. y x d (xy) dx dy x (y)(1) dx dx 7 . t contradicts the given equation. xy d (sin y) dx dy t (d) If g (t) 23. x 4 equation. 0 x sec2 xy 1 cos2 xy x t 3 7/4 t , which is not 4 gives g (t) 0 d (tan xy) dx d 1 sec2 (xy) (xy) dx dy (sec2 xy)[x (y)(1)] dx 20. 1 , which matches the t 3/4 (b) Differentiating both sides of the given equation d (x) dx (sec2 xy)(x) , which matches 4, then consistent with g (t) 1 1/3 given equation. d (sin y) dx dy cos y dx 1 sec y cos y 1 x Conclusion: (b), (c), and (d) could be true. g (t) 18. 6, then f (x) the given equation. d (tan y) dx dy sec2 y dx 1 cos2 y sec2 y 1 1 4/3 x . 3 3 2/3 x 2 (d) If f (x) 1 4/3 x , so it must be true 3 gives f (x) that f (x) 5) tan y d (x) dx 1/3 x x 1/3 , which matches the given equation. Since our original equation was x 2 substitute 1 for x 2 y 2, giving y y2 1 . y3 1, we may Section 3.7 x2/3 24. d 2/3 (x ) dx y2/3 1 d 2/3 (y ) dx 2 1/3 x 3 0 1 /3 y 1/3 1 /3 x y y x y 1/3 d dx y x 1y 3x 2/3 d 1y 3x 2/3 xy x2 y 1/3 x Slope at (3, 3x 4/3y 1/3 substitute 1 for x y2 25. x2 d2 (y ) dy y 2/3 , giving y 3x4/3y1/3 . 2x y y (x 2)(2y) (x 1) x 1 y (y 2)(2x) 0 2 y2 d2 (y ) dx d (2x) dx y (2y y y 2y 2x 2y) 2)y 2 Slope at ( 2, 1): (a) Tangent: y (b) Normal: y (y 1) 2 (y 1) 2 1 1)3 y 1 y 1 1) 3 or y 1) 4y 1 d (1) dx dy 24 0 dx dy (2y 4) dx dy dx dy dx 1 1 1(x 6 1 x 3 8 3 0 d (0) dx 0 2 1 y 2 1 2 (x 3x 3 or y d (4y) dx 1) 1 y1 d1 dx y 1 (y 2y y x 3 2x 1 d (2x dx 2xy 2 2x 2y 1 (x 3 30. 2x, we may 2x 1) 1, 2xy 2 3(x (b) Normal: y 1)2 (x y3 dy dx 4 x 3 ( 4) or y d (9) dx (a) Tangent: y Since our original equation was y x write y 2 (x 1)2 (x 2 2x) (x 2 1 which gives y 3. d2 (y dx 25 4 9 dy dx dy dx 3 Slope at ( 1, 3): 1 2 26. y 3 x 4 3) 2x 2y y2 2 3) x 2y 2 2 y2 ( 4) or y 3 4 3 4 d 22 (x y ) dx 2x 2 x1 2y y dx 1 dx y (y)(1) (x 1)y y2 y 29 7 x y 4 (x 3 29. 4 x 7 0 (b) Normal: y d (2x) dy 1 2 3 or y 2) 3 (x 4 (a) Tangent: y 7 x 4 3 or y y x d (25) dx 4): 1, we may 1 2x d2 (x ) dy 2yy y 2/3 2x 2y 25 d2 (y ) dx dy 2x 2y dx dy dx y 2/3 Since our original equation was x 2/3 y2 d2 (x ) dx 2/3 1/3 y 1x y 3 x4/3y2/3 2/3 x2 28. y xy x 2/3 1 2) 4 (x 7 (b) Normal: y (y)(1) 4/3 2/3 3 7 (x 4 (a) Tangent: y y dx x (x) 1 y2 xy d2 d d2 d (x ) (xy) (y ) (1) dx dx dx dx dy dy 2x x (y)(1) 2y 0 dx dx dy (x 2y) 2x y dx dy 2x y dx x 2y 2(2) 3 7 Slope at (2, 3): 2(3) 2 4 d (1) dx 2 1/3 y y 3 x2 27. 2) 2) 1 or y 1 or y x x 1 3 105 ...
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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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