Calculus 8th Edition

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: 44

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, $f'(z)=-\frac{e^{\frac{z}{(z-1)}}}{(z-1)^{2}}$
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