Answer
$y'=\frac{-2(e^x-e^{-x})}{(e^x+e^{-x})^2}$
Work Step by Step
$y=\frac{2}{e^x+e^{-x}}=2(e^x+e^{-x})^{-1}$
$y’=-2(e^x+e^{-x})^{-2}(e^x-e^{-x})$
$=\frac{-2(e^x-e^{-x})}{(e^x+e^{-x})^2}$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.4 Exercises - Page 352: 47
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