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: 20


$r'=\cos (\theta +\sqrt { \theta+1})(1 +\dfrac{1}{2\sqrt { \theta+1}})$

Work Step by Step

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