#### Answer

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

#### Work Step by Step

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

Published by
Pearson

ISBN 10:
0321947347

ISBN 13:
978-0-32194-734-5

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

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