## Calculus 8th Edition

$(g^{-1})'(8)=\frac{1}{5}$
Since $f(2)=8$, this implies $g^{-1}(8)=2$ and so, by theorem 7, $(g^{-1})'(8)=\frac{1}{g'(g^{-1}(8))}=\frac{1}{g'(2)}=\frac{1}{5}.$