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: 23



Work Step by Step

Use the hyperbolic functions formulae such as: $ \cosh x= \dfrac{e^x+e^{-x}}{2}$ Thus, $\text{sech} x=\dfrac{1}{\cosh x}=\dfrac{2}{e^x+e^{-x}}$ We are given $y=(x^2+1) \text{sech}(\ln x)$ or, $y=(x^2+1)\dfrac{2}{e^{\ln x}+e^{-\ln x}}=(x^2+1)\dfrac{2}{e^{\ln x}+e^{\ln x^{-1}}}=(x^2+1)\dfrac{2x}{x^2+1}=2x$ Now, $\dfrac{dy}{dx}=2$
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