## Calculus: Early Transcendentals 8th Edition

$f'(z)=-\dfrac{1}{(z-1)^{2}}e^{z/(z-1)}$
$f(z)=e^{z/(z-1)}$ Differentiate using the chain rule: $f'(z)=[e^{z/(z-1)}](\dfrac{z}{z-1})'=...$ Use the quotient rule to find $(\dfrac{z}{z-1})'$: $...=[e^{z/(z-1)}][\dfrac{(z-1)(z)'-(z)(z-1)'}{(z-1)^{2}}]=...$ $...=[e^{z/(z-1)}][\dfrac{(z-1)(1)-(z)(1)}{(z-1)^{2}}]=...$ Simplify: $...=\dfrac{z-1-z}{(z-1)^{2}}e^{z/(z-1)}=-\dfrac{1}{(z-1)^{2}}e^{z/(z-1)}$