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: 14

Answer

$\cosh (2x+1)$

Work Step by Step

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