Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.7 - Hyperbolic Functions - Exercises 7.7 - Page 430: 5



Work Step by Step

Use the hyperbolic function formula as follows: $ \cosh x= \dfrac{e^x+e^{-x}}{2}$ Need to solve. $2 \cosh ( \ln x)$ $2 \cosh ( \ln x)=2[ \dfrac{e^{\ln x}+e^{-\ln x}}{2}] \\ =e^{\ln x}+e^{(-\ln x)} \\=e^{\ln x}+e^{(\ln x^{-1})} \\ =x+x^{(-1)} \\=x+\dfrac{1}{x}$
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