Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.9 Hyperbolic Functions - Exercises - Page 384: 13


$$ y' =\frac{ \sinh }{2\sqrt{\cosh x+1}} .$$

Work Step by Step

Since $ y=\sqrt{\cosh x +1}$, then the derivative $ y'$, using the chain rule, is given by $$ y'=\frac{1}{2} (\cosh x +1)^{-1/2}(\cosh x +1)'=\frac{1}{2} (\cosh x +1)^{-1/2} \sinh x\\ =\frac{ \sinh }{2\sqrt{\cosh x+1}} .$$
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