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January 17, 2017
Posted by **Joe** on Monday, February 11, 2013 at 11:02pm.

k(x)=sqrt(sin(2x))^3

I think I have the answer and was wondering if i had it correct

k'(x)=2sin(x)^3+8sin(x)

- Calculus -
**Reiny**, Monday, February 11, 2013 at 11:16pmI will interpret your question just the way you typed it

k(x) = √ [ (sin(2x) )^3 ] which is

= ( sin(2x) )^(3/2)

k ' (x) = (3/2)( sin(2x))^(1/2) ( cos(2x)) (2)

= 3 cos(2x) √(sin(2x)) - Calculus -
**Joe**, Monday, February 11, 2013 at 11:18pmok thank you