Answer
$y'= \frac{2}{3}e^x-e^{-x}$
Work Step by Step
$y = \frac{2e^x+3e^{-x}}{3}$
Using Quotient Rule:
$y'=\frac{(2e^x-3e^{-x})(3) - (2e^x+3e^{-x})(0)}{3^2} = \frac{2e^x-3e^{-x}}{3} = \frac{2}{3}e^x-e^{-x}$
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
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 - Page 160: 49
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