Answer :
[tex]\ [cos(2\theta)]'=(2cos^2\theta-1)'=2(cos^2\theta)'-(1)'\\\\=2\cdot2cos\theta\cdot(-sin\theta)-0=-2\cdot2sin\theta cos\theta=\boxed{-2sin2\theta}\\\\Used:\\\ [f(x)-g(x)]'=f'(x)-g'(x)\\\{f[g(x)]\}'=f'[g(x)]\cdot g'(x)\\(cosx)'=-sinx\\sin2x=2sinxcosx[/tex]
