## Calculus 8th Edition

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