Answer
$\dfrac{d}{dx}(ln(e^x+e^{-x}))=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}$
Work Step by Step
using theorem 3.17: $\dfrac{d}{dx}(ln(e^x+e^{-x}))=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}$
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
Published by
Pearson
ISBN 10:
0321947347
ISBN 13:
978-0-32194-734-5
Chapter 3 - Derivatives - 3.9 Derivatives of Logarithmic and Exponential Functions - 3.9 Exercises - Page 211: 22
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