Answer
$y'=\frac{-2e^x}{(e^x-1)^2}$
Work Step by Step
$y=\frac{e^x+1}{e^x-1}$
$u=e^x+1$
$u’=e^x$
$v=e^x-1$
$v’=e^x$
$y’=\frac{(e^x)(e^x-1)-(e^x)(e^x-1)}{(e^x-1)^2}$
$=\frac{(e^x)(-2)}{(e^x-1)^2}$
$=\frac{-2e^x}{(e^x-1)^2}$
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: 49
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