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: 18


$r'=2 \sqrt { \cos \theta}-\dfrac{ \theta \sin \theta}{\sqrt { \cos \theta}}$

Work Step by Step

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