University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Practice Exercises - Page 202: 22

Answer

$y'=\dfrac{sin \sqrt x}{\sqrt x}+\cos\sqrt x$

Work Step by Step

Given that $y=2\sqrt x \sin\sqrt x $ Apply derivative rules of differentiation: $f(x)=p'(x)q(x)+p(x)q'(x)$ $y'=\dfrac{d}{dx}[2\sqrt x \sin\sqrt x]$ $=2[\dfrac{d}{dx}(\sqrt x) \sin\sqrt x+(\sqrt x (\cos\sqrt x)]$ $=2[\dfrac{sin \sqrt x}{2\sqrt x}+\dfrac{\cos\sqrt x}{2}]$ $=2(\dfrac{sin \sqrt x}{2\sqrt x})+2(\dfrac{\cos\sqrt x}{2})$ Hence, $y'=\dfrac{sin \sqrt x}{\sqrt x}+\cos\sqrt x$
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