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Pre-Calc Homework Solutions 120

# Pre-Calc Homework Solutions 120 - 120 dy dx Chapter 3...

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29. } d d y x } 5 } d d x } csc 2 1 (sec x ) 5 1 2 2 } d d x } (sec x ) 5 2 } ) sec x ) ˇ 1 ta w n w 2 w x w } sec x tan x 5 2 } ) s s e e c c x x t t a a n n x x ) } 5 2 5 2 } ) s s i i n n x x ) } 5 2 sign (sin x ), x } p 2 } , p , } 3 2 p } Alternate method: On the domain 0 # x # 2 p , x } p 2 } , x } 3 2 p } , we may rewrite the function as follows: y 5 csc 2 1 (sec x ) 5 } p 2 } 2 sec 2 1 (sec x ) 5 } p 2 } 2 cos 2 1 (cos x ) 5 5 5 5 Therefore, } d d y x } 5 5 Note that the derivative exists at 0 and 2 p only because these are the endpoints of the given domain; the two-sided derivative of y 5 csc 2 1 (sec x ) does not exist at these points. 30. } d d u r } 5 } d d u } 1 } 1 1 2 1 c s o in s u u } 2 2 5 2 1 } 1 1 2 1 c s o in s u u } 21 2 5 2 1 } 1 1 2 1 c s o in s u u } 21 2 5 2 1 } 1 1 2 1 c s o in s u u } 21 2 31. Since y 5 ln x 2 is defined for all x 0 and } d d y x } 5 } x 1 2 } } d d x } ( x 2 ) 5 } 2 x x 2 } 5 } 2 x } , the function is differentiable for all x 0. 32. Since y 5 sin x 2 x cos x is defined for all real x and } d d y x } 5 cos x 2 ( x )( 2 sin x ) 2 (cos x )(1) 5 x sin x , the function is differentiable for all real x .
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