what is the absolute value of the derivative of sin2xcos2x?

Another way is to notice that sin(2x)cos(2x) is (1/2)sin(4x), the derivative of which is 4*(1/2)*cos(4x) or just 2*cos(4x).

 So the absolute value is 2|cos(4x)|  

(This is equivalent to geno3141's solution).

To find the derivative of  u = sin(2x),  we need to use the chain rule:  u' = cos(2x)·2  --->  u' =  2cos(2x)

Similarly,  v = cos(2x)  --->  v' = -sin(2x)·2  --->  v' = -2sin(2x)

To find the derivative of  y = sin(2x)·cos(2x), we need to use the product rule:  y = u·v  --->  y' = u·v' + v·u'

y' = sin(2x)·-2sin(2x) + cos(2x)·2cos(2x)     --->  y' = -2sin²(2x) +2cos²(2x)

|y'| = |-2sin²(2x) +2cos²(2x)|  =  2|cos²(2x) - sin²(2x)|