Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises: 6

Answer

$y'=\dfrac{e^{x}}{(1-e^{x})^{2}}$

Work Step by Step

$y=\dfrac{e^{x}}{1-e^{x}}$ Differentiate using the quotient rule $y'=\dfrac{(1-e^{x})(e^{x})'-(e^{x})(1-e^{x})'}{(1-e^{x})^{2}}=\dfrac{(1-e^{x})(e^{x})-(e^{x})(-e^{x})}{(1-e^{x})^{2}}=...$ Evaluate the products in the numerator $...=\dfrac{e^{x}-e^{2x}+e^{2x}}{(1-e^{x})^{2}}=\dfrac{e^{x}}{(1-e^{x})^{2}}$ $y'=\dfrac{e^{x}}{(1-e^{x})^{2}}$
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