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: 17


$s'=\dfrac{( \sin \theta + \theta \cos \theta)}{ \sqrt{2\theta \sin \theta}}$

Work Step by Step

Given that $r=\sqrt {2 \theta \sin \theta}$ Apply derivative rules of differentiation: $f(x)=p'(x)q(x)+p(x)q'(x)$ $r'=\dfrac{d}{dx}[\sqrt {2 \theta \sin \theta}]$ $=(2 \sqrt{2\theta \sin \theta})^{-1}\dfrac{d}{dx}[ {2 \theta \sin \theta}]$ $=(2 \sqrt{2\theta \sin \theta})^{-1}( 2 \sin \theta +2 \theta \cos \theta)$ Hence, $s'=\dfrac{( \sin \theta + \theta \cos \theta)}{ \sqrt{2\theta \sin \theta}}$
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