y = tanh^-1(sin(2x)) Find the derivative of the function

This is a composite function, and to differentiate it we need the chain rule,
(f(g(x)))' = f'(g(x))*g'(x).
Here  g(x) = sin(2x)  and  f(z) = tanh^-1(z), so we need their derivatives also. They are known, (sin(2x))' = 2cos(2x)  (we use the chain rule for 2x here),  (tanh^-1(z))' = 1/(1 - z^2).
This way the result is  y'(x) = 1/(1 - sin^2(2x))*2cos(2x),
which is equal to (2cos(2x))/(cos^2(2x)) = 2/(cos(2x)) = 2sec(2x).