University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.3 - Hyperbolic Functions - Exercises - Page 417: 6


$\dfrac{ x^2- x^{-2}}{2}$

Work Step by Step

Use formula: $ \sinh x= \dfrac{e^x-e^{-x}}{2}$ Thus, $ \sinh ( 2 \ln x)= \dfrac{e^{2 \ln x}-e^{- 2 \ln x}}{2}=\dfrac{e^{\ln x^2}-e^{\ln x^{-2}}}{2}$ Hence, $ \sinh ( 2 \ln x)=\dfrac{ x^2- x^{-2}}{2}$
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