derivative of trig function (chain and product rule)

[mavfan]

find the derivative of:

y=2sin(2x)cos(2x)

this is what i did but im not getting the right answer:

2[sin(2x)(-sin(2x))(2)+(cos(2x)cos(2x))(2)]
4[-sin^2(2x)+cos^2(2x)]

the right answer is supposed to be 4cos(4x)

is there a trig identity ur supposed to be using with that or am i doing something wrong?

EDIT: nevermind i figured it out. you use the cos^2(x)-sin^2(x)=cos2(x)
so 4[cos2(2x)]
=4cos(4x)

thanks anyway...

 

[tkhunny]

Talk about the hard way!

Why not just use sin(2x) = 2sin(x)cos(x)? It's a one-stepper.