Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises: 28

Answer

$f'(z)=-\dfrac{1}{(z-1)^{2}}e^{z/(z-1)}$

Work Step by Step

$f(z)=e^{z/(z-1)}$ 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: $...=\dfrac{z-1-z}{(z-1)^{2}}e^{z/(z-1)}=-\dfrac{1}{(z-1)^{2}}e^{z/(z-1)}$
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