if you solve f '(x)=(2sinx)(cosx)=0 ; how do you get that x=0
in case your wondering i am using the first derivative test and this is already the derivative of the original function which was f(x)=sin^2x
if you solve f '(x)=(2sinx)(cosx)=0 ; how do you get that x=0
in case your wondering i am using the first derivative test and this is already the derivative of the original function which was f(x)=sin^2x
not exactly sure what you mean ...
are you asking why $\displaystyle f'(x) = 0$, or are you asking how to solve $\displaystyle 2\sin{x}\cos{x} = 0$ ?