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: 25

Answer

$y'=-\dfrac{2x^2-1}{\sqrt{1-x^2}}$

Work Step by Step

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