Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.7 Hyperbolic Functions - 6.7 Exercises - Page 490: 33

Answer

$2x\cosh{(x^2)} $

Work Step by Step

Given: $h(x)=\sinh {(x^2)}$ Now, $h'(x)=\frac{d}{dx}(\sinh {(x^2)})$ On applying the chain rule, we have $h'(x)=(\dfrac{d\sinh {(x^2)}}{dx^2})(\dfrac{dx^2}{dx})$ or, $=(\cosh {(x^2)})(2x)$ or, $=2x\cosh{(x^2)} $
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