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Business Calc Homework w answers_Part_24

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

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11. } d d y x } 5 } d d x } ( x p ) 5 p x p 2 1 12. } d d y x } 5 } d d x } ( x 1 1 ˇ 2 w ) 5 (1 1 ˇ 2 w ) x 1 1 ˇ 2 w 2 1 5 (1 1 ˇ 2 w ) x ˇ 2 w 13. } d d y x } 5 } d d x } x 2 ˇ 2 w 5 2 ˇ 2 w x 2 ˇ 2 w 2 1 14. } d d y x } 5 } d d x } x 1 2 e 5 (1 2 e ) x 1 2 e 2 1 5 (1 2 e ) x 2 e 15. } d d y x } 5 } d d x } 8 x 5 8 x ln 8 16. } d d y x } 5 } d d x } 9 2 x 5 9 2 x (ln 9) } d d x } ( 2 x ) 5 2 9 2 x ln 9 17. } d d y x } 5 } d d x } 3 csc x 5 3 csc x (ln 3) } d d x } (csc x ) 5 3 csc x (ln 3)( 2 csc x cot x ) 5 2 3 csc x (ln 3)(csc x cot x ) 18. } d d y x } 5 } d d x } 3 cot x 5 3 cot x (ln 3) } d d x } (cot x) 5 3 cot x (ln 3)( 2 csc 2 x ) 5 2 3 cot x (ln 3)(csc 2 x ) 19. Use logarithmic differentiation. y 5 x ln x ln y 5 ln x ln x ln y 5 ln x ln x } d d x } (ln y ) 5 } d d x } (ln x ) 2 } 1 y } } d d y x } 5 (2 ln x ) 1 } 1 x } 2 } d d y x } 5 } 2 y x ln x } } d d y x } 5 } 2 x ln x x ln x } 20. Use logarithmic differentiation. y 5 x 1/ln x ln y 5 ln x 1/ln x ln y 5 } ln 1 x } ln x ln y 5 1 y 5 e } d d y x } 5 } d d x } ( e ) 5 0, x . 0 21. } d d y x } 5 } d d x } ln ( x 2 ) 5 } x 1 2 } } d d x } ( x 2 ) 5 } x 1 2 } (2 x ) 5 } 2 x } 22. } d d y x } 5 } d d x } (ln x ) 2 5 2 ln x } d d x } (ln x ) 5 } 2 l x n x } 23. } d d y x } 5 } d d x } ln ( x 2 1 ) 5 } d d x } ( 2 ln x ) 5 2 } 1 x } , x . 0 24. } d d y x } 5 } d d x } ln } 1 x 0 } 5 } d d x } (ln 10 2 ln x ) 5 0 2 } 1 x } 5 2 } 1 x } , x . 0 25. } d d y x } 5 } d d x } ln ( x 1 2) 5 } x 1 1 2 } } d d x } ( x 1 2) 5 } x 1 1 2 } , x . 2 2 26. } d d y x } 5 } d d x } ln (2 x 1 2) 5 } 2 x 1 1 2 } } d d x } (2 x 1 2) 5 } 2 x 2 1 2 } 5 } x 1 1 1 } , x . 2 1 27. } d d y x } 5 } d d x } ln (2 2 cos x ) 5 } 2 2 1 cos x } } d d x } (2 2 cos x ) 5 } 2 2 sin co x s x } 28. } d d y x } 5 } d d x } ln ( x 2 1 1) 5 } x 2 1 1 1 } } d d x } ( x 2 1 1) 5 } x 2 2 1 x 1 } 29. } d d x } ln (ln x ) 5 } ln 1 x } } d d x } ln x 5 } ln 1 x } ? } 1 x } 5 } x l 1 n x } 30. } d d y x } 5 } d d x } ( x ln x 2 x ) 5 ( x ) 1 } 1 x } 2 1 (ln x )(1) 2 1 5 1 1 ln x 2 1 5 ln x 31. } d d y x } 5 } d d x } (log 4 x 2 ) 5 } d d x } } l l n n x 4 2 } 5 } d d x } 31 } ln 2 4 } 2 (ln x ) 4 5 } ln 2 4 } ? } 1 x } 5 } x l 2 n 4 } 5 } x l 1 n 2 } 32. } d d y x } 5 } d d x } (log 5 ˇ x w ) 5 } d d x } } ln ln x 5 1/2 } 5 } d d x } 5 } 2 l 1 n 5 } } d d x } (ln x ) 5 } 2 l 1 n 5 } ? } 1 x } 5 } 2 x 1 ln 5 } , x . 0 33. } d d y x } 5 } d d x } log 2 (3 x 1 1) 5 } (3 x 1 1 1) ln 2 } } d d x } (3 x 1 1) 5 } (3 x 1 3 1) ln 2 } , x . 2 } 1 3 } 34. } d d y x } 5 } d d x } log 10 ( x 1 1) 1/2 5 } 1 2 } } d d x } log 10 ( x 1 1) 5 } 1 2 } } ( x 1 1 1 ) ln 10 } } d d x } ( x 1 1) 5 } 2( x 1 1 1 ) ln 10 } , x . 2 1 35. } d d y x } 5 } d d x } log 2 1 } 1 x } 2 5 } d d x } ( 2 log 2 x ) 5 2 } x l 1 n 2 } , x . 0 36. } d d y x } 5 } d d x } } log 1 2 x } 5 2 } (log 1 2 x ) 2 } } d d x } (log 2 x) 5 2 } (log 1 2 x ) 2 } } x l 1 n 2 } 5 2 } x (ln 2)( 1 log 2 x ) 2 } or 2 } x ( l l n n 2 x ) 2 } 37. } d d y x } 5 } d d x } (ln 2 ? log 2 x ) 5 (ln 2) } d d x } (log 2 x ) 5 (ln 2) 1 } x l 1 n 2 } 2 5 } 1 x } , x . 0 38. } d d y x } 5 } d d x } log 3 (1 1 x ln 3) 5 } (1 1 x l 1 n 3) ln 3 } } d d x } (1 1 x ln 3) 5 } (1 1 x ln ln 3 3) ln 3 } 5 } 1 1 1 x ln 3 } , x . 2 } ln 1 3 } 39. } d d y x } 5 } d d x } (log 10 e x ) 5 } d d x } ( x log 10 e ) 5 log 10 e 5 } ln ln 1 e 0 } 5 } ln 1 10 } 40. } d d y x } 5 } d d x } ln 10 x 5 } d d x } ( x ln 10) 5 ln 10 } 1 2 } ln x } ln 5 116 Section 3.9
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41. The line passes through ( a , e a ) for some value of a and has slope m 5 e a . Since the line also passes through the origin, the slope is also given by m 5 } e a a 2 2 0 0 } and we have e a 5 } e a a } , so a 5 1. Hence, the slope is e and the equation is y 5 ex .
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