Answer
$f'(z)=-\frac{e^{\frac{z}{(z-1)}}}{(z-1)^{2}}$
Work Step by Step
$f'(z)=\frac{d}{dz}[e^{\frac{z}{(z-1)}}]$
Apply chain rule.
$f'(z)=[e^{\frac{z}{(z-1)}}][-(z-1)^{-2}]$
Hence, $f'(z)=-\frac{e^{\frac{z}{(z-1)}}}{(z-1)^{2}}$
Calculus 8th Edition
by Stewart, James
Published by
Cengage
ISBN 10:
1285740629
ISBN 13:
978-1-28574-062-1
Chapter 6 - Inverse Functions - 6.2 Exponential Functions and Their Derivatives - 6.2 Exercises - Page 418: 44
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