Answer
$y'=\frac{2e^x}{1+e^{2x}}$
Work Step by Step
$y=ln(\frac{1+e^x}{1-e^x})$
$=ln(1+e^x)-ln(1-e^x)$
$y’=\frac{e^x}{1+e^x}-\frac{-e^x}{1-e^x}$
$=\frac{e^x(1-e^x+1+e^x)}{1+e^{2x}}$
$=\frac{2e^x}{1+e^{2x}}$
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: 46
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