Differentiate with respect to x:

y = (arcsin 2x)^3

I obtained:

dy/dx = 6(arcsin 2x)^2

although, according to the answer, what I gave was either du/dx or dy/du. Can someone please point out how to get du/dx and dy/du in this case. I dont know how to apply the standard inverse trig function for sin^-1 here.