If $\displaystyle f(x-y)=f(x)g(y)-g(x)f(y)$ and $\displaystyle g(x-y)=g(x)g(y)+f(x)f(y)$ for all $\displaystyle x,y \in \mathbb{R}$ and the right hand derivative of $\displaystyle f(x)$ at zero exists. Find $\displaystyle g'(x)$ at $\displaystyle x=0$