## Calculus 8th Edition

$f'(z)=-\frac{e^{\frac{z}{(z-1)}}}{(z-1)^{2}}$
$f'(z)=\frac{d}{dz}[e^{\frac{z}{(z-1)}}]$ Apply chain rule. $f'(z)=[e^{\frac{z}{(z-1)}}][-(z-1)^{-2}]$ Hence, $f'(z)=-\frac{e^{\frac{z}{(z-1)}}}{(z-1)^{2}}$