Differentiation 2

f^{-1}(x)=g(x), \Rightarrow f(g(x))=x

sof^{!}(g(x))g^{!}(x)=1

g^{!}(x)=\frac{1}{f^{!}(g(x))}

at , g^{!}(\pi )=\frac{1}{f^{!}(g(\pi ))}

now f(g(\pi ))=\pi

notice that f(\frac{\pi }{2})=\pi
hence
g(\pi )=\frac{\pi }{2}
( since f has to be one one for its inverse to be defined )

hence g^{!}(\pi )=\frac{1}{f^{!}(\frac{\pi }{2})}

{f^{!}(\frac{\pi }{2})}=3, hence

g^{!}(\pi )=\frac{d}{dx}(f^{-1}(x))=\frac{1}{f^{!}(\pi /2)}=\frac{1}{3}