Answer
$y=0$
Work Step by Step
$y=ln(\frac{e^x+e^{-x}}{2})$ $(0,0)$
$=ln(e^x+e^{-x})-ln 2$
$y’=\frac{e^x-e^{-x}}{e^x+e^{-x}}$
$y’(0)=M=\frac{e^0-e^{-0}}{e^0+e^{-0}}$
$=0$
Equation of Tangent:
$y-0=0(x-0)$
$y=0$
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: 60
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