Posted by Joe on Monday, February 11, 2013 at 11:02pm.
Find the derivatives of the function
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 11:16pm
I 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 11:18pm
ok thank you