Calculus 10th Edition

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

Chapter 2 - Differentiation - 2.4 Exercises: 50

Answer

$g'(\theta)=\dfrac{\sec^3{\dfrac{\theta}{2}}+\sec{\dfrac{\theta}{2}}\tan^2{\dfrac{\theta}{2}}}{2}$.

Work Step by Step

Using both the Product Rule and the Chain Rule: Product Rule $g'(\theta)=((u(\theta)(v(\theta))’=u’(\theta)v(\theta)+u(\theta)v’(\theta))$ $u(\theta)=\sec{\dfrac{\theta}{2}} ;u’(\theta)=\dfrac{1}{2}\sec{\dfrac{\theta}{2}}\tan{\dfrac{\theta}{2}}$. $v(\theta)=\tan{\dfrac{\theta}{2}} ;v’(\theta)=\dfrac{1}{2}\sec^2{\dfrac{\theta}{2}} $. $g'(\theta)=(\dfrac{1}{2}\sec{\dfrac{\theta}{2}}\tan{\dfrac{\theta}{2}}(\tan{\dfrac{\theta}{2}})+(\dfrac{1}{2}\sec^2{\dfrac{\theta}{2}})(\sec{\dfrac{\theta}{2}})$ $=\dfrac{\sec^3{\dfrac{\theta}{2}}+\sec{\dfrac{\theta}{2}}\tan^2{\dfrac{\theta}{2}}}{2}$.
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