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: 26

Answer

$y'=-\dfrac{x(3x^2-32)}{2\sqrt{16-x^2}}.$

Work Step by Step

Product Rule $(y'=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$ $u(x)=\dfrac{x^2}{2} ;u’(x)=x $ $v(x)=\sqrt{16-x^2}$ $v'(x)$ is found using the Chain Rule: $u=16-x^2$; $\dfrac{du}{dx}=-2x$ $\dfrac{d}{du}v(u)=\dfrac{1}{2\sqrt{u}}$ $\dfrac{d}{dx}v(x)=\dfrac{d}{du}v(u)\times\dfrac{du}{dx}=-\dfrac{x}{\sqrt{16-x^2}}.$ $y'=(x)(\sqrt{16-x^2})+(\dfrac{x^2}{2})(-\dfrac{x}{\sqrt{16-x^2}})$ $=-\dfrac{x(3x^2-32)}{2\sqrt{16-x^2}}.$
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