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Pre-Calc Homework Solutions 104

Pre-Calc Homework Solutions 104 - 104 dy dx Section 3.7...

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16. } d d y x } 5 } d d x } [sin ( x 1 5)] 5/4 5 } 5 4 } [sin ( x 1 5)] 1/4 } d d x } sin ( x 1 5) 5 } 5 4 } [sin ( x 1 5)] 1/4 cos ( x 1 5) 17. x 5 tan y } d d x } ( x ) 5 } d d x } (tan y ) 1 5 sec 2 y } d d y x } } d d y x } 5 } sec 1 2 y } 5 cos 2 y 18. x 5 sin y } d d x } ( x ) 5 } d d x } (sin y ) 1 5 cos y } d d y x } } d d y x } 5 } co 1 s y } 5 sec y 19. x 1 tan xy 5 0 } d d x } ( x ) 1 } d d x } (tan xy ) 5 0 1 1 sec 2 ( xy ) } d d x } ( xy ) 5 0 1 1 (sec 2 xy )[ x } d d y x } 1 ( y )(1)] 5 0 (sec 2 xy )( x ) } d d y x } 5 2 1 2 (sec 2 xy )( y ) } d d y x } 5 } 2 1 x 2 se y c 2 se x c y 2 xy } } d d y x } 5 2 } 1 x } cos 2 xy 2 } y x } 20. x 1 sin y 5 xy } d d x } ( x ) 1 } d d x } (sin y ) 5 } d d x } ( xy ) 1 1 (cos y ) } d d y x } 5 x } d d y x } 1 ( y )(1) (cos y 2 x ) } d d y x } 5 2 1 1 y } d d y x } 5 } c 2 os 1 y 1 2 y x } 5 } x 2 1 2 co y s y } 21. (a) If f ( x ) 5 } 3 2 } x 2/3 2 3, then f 9 ( x ) 5 x 2 1/3 and f 0 ( x ) 5 2 } 1 3 } x 2 4/3 which contradicts the given equation f 0 ( x ) 5 x 2 1/3 . (b) If f ( x ) 5 } 1 9 0 } x 5/3 2 7, then f 9 ( x ) 5 } 3 2 } x 2/3 and f 0 ( x ) 5 x 2 1/3 , which matches the given equation. (c) Differentiating both sides of the given equation f 0 ( x ) 5 x 2 1/3 gives f - ( x ) 5 2 } 1 3 } x 2 4/3 , so it must be true that f - ( x ) 5 2 } 1 3 } x 2 4/3 . (d) If f 9 ( x ) 5 } 3 2 } x 2/3 1 6, then f 0 ( x ) 5 x 2 1/3 , which matches the given equation. Conclusion: (b), (c), and (d) could be true. 22. (a) If g 9 ( t ) 5 4 ˇ 4 t w 2 4, then g 0 ( t ) 5 } d d x } (4 t 1/4 2 4) 5 t 2 3/4 5 } t 3 1 /4 } , which matches the given equation. (b) Differentiating both sides of the given equation g 0 ( t ) 5 } t 3 1 /4 } 5 t 2 3/4 gives g - ( t ) 5 2 } 3 4 } t 2 7/4 , which is not consistent with g - ( t ) 5 2 } ˇ 4 4 t w } . (c) If g ( t ) 5 t 2 7 1 } 1 5 6 } t 5/4 , then g 9
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