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:
$u=1-x^2$; $\dfrac{du}{dx}=-2x$
$\dfrac{d}{du}v(u)=\dfrac{1}{2\sqrt{u}}$
$\dfrac{d}{dx}v(x)=\dfrac{d}{du}v(u)\times\dfrac{du}{dx}=(\dfrac{1}{2\sqrt{1-x^2}})(-2x)=-\dfrac{x}{\sqrt{1-x^2}}$
$y'=(\sqrt{1-x^2})(1)+(x)(-\dfrac{x}{\sqrt{1-x^2}})=-\dfrac{2x^2-1}{\sqrt{1-x^2}}$
