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: 13


$2 \cosh {\dfrac{x}{3}}$

Work Step by Step

Since, $\dfrac{d}{dx} (\sinh x)=\cosh x$ As we are given that $y=6 \sinh {\dfrac{x}{3}}$ Then, on differentiating , we have $\dfrac{dy}{dx}=6 \cosh {\dfrac{x}{3}} (\dfrac{1}{3})=2 \cosh {\dfrac{x}{3}}$
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