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: 11


$$ y' =-6 \sinh (9-3t) \cosh (9-3t).$$

Work Step by Step

Since $ y=\cosh^2 (9-3t)$, then the derivative $ y'$, using the chain rule, is given by $$ y'= 2 \cosh (9-3t) ( \cosh (9-3t))'=2(-3) \sinh (9-3t) \cosh (9-3t) \\ =-6 \sinh (9-3t) \cosh (9-3t).$$
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