Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.3 Exercises: 54

Answer

$f'(\theta)=\tan(\theta)+\sec(\theta)(\theta(5\tan(\theta)+\sec(\theta))+5).$

Work Step by Step

$f(\theta)=g(\theta)+h(\theta)\rightarrow g(\theta)=5\theta(\sec(\theta))$; $h(\theta)=\theta(\tan(\theta))$ Product Rule $g'(\theta)=((u(\theta)(v(\theta))’=u’(\theta)v(\theta)+u(\theta)v’(\theta))$. $u(\theta)=5\theta ;u’(\theta)=5 $ $v(\theta)=\sec(\theta) ;v’(\theta)=\sec(\theta)\tan(\theta) $ $g'(\theta)=5(\theta\sec(\theta)\tan(\theta)+\sec(\theta))$. Product Rule $h'(\theta)=((u(\theta)(v(\theta))’=u’(\theta)v(\theta)+u(\theta)v’(\theta))$. $u(\theta)=\theta ;u’(\theta)=1 $ $v(\theta)=\tan(\theta) ;v’(\theta)=\sec^2(\theta) $ $h'(\theta)=(\tan(\theta)+\theta(\sec^2(\theta)))$. $f'(\theta)=g'(\theta)+h'(\theta)$ $=\tan(\theta)+\sec(\theta)(\theta(5\tan(\theta)+\sec(\theta))+5).$
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