Question

The problem for session 4:

How can I calculate f'(x) of f(x)=sin2x as 2cos2x?

Have I missed anything?

Answer 1

This uses the chain rule.
the derivative of sin x is cos x,
but we do have sin x. We have sin 2x.

Answer 2

Taking the derivative will give us cos 2x times the derivative of 2x.
Since the derivative of 2x is 2, the answer is (cos 2x)*2, or
2*cos 2x