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:
$u=16-x^2$; $\dfrac{du}{dx}=-2x$
$\dfrac{d}{du}v(u)=\dfrac{1}{2\sqrt{u}}$
$\dfrac{d}{dx}v(x)=\dfrac{d}{du}v(u)\times\dfrac{du}{dx}=-\dfrac{x}{\sqrt{16-x^2}}.$
$y'=(x)(\sqrt{16-x^2})+(\dfrac{x^2}{2})(-\dfrac{x}{\sqrt{16-x^2}})$
$=-\dfrac{x(3x^2-32)}{2\sqrt{16-x^2}}.$
