Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.4 The Product and Quotient Rules - 3.4 Exercises: 49

Answer

$y'= \frac{2}{3}e^x-e^{-x}$

Work Step by Step

$y = \frac{2e^x+3e^{-x}}{3}$ Using Quotient Rule: $y'=\frac{(2e^x-3e^{-x})(3) - (2e^x+3e^{-x})(0)}{3^2} = \frac{2e^x-3e^{-x}}{3} = \frac{2}{3}e^x-e^{-x}$
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