Answer
$y'=-\dfrac{x}{\sqrt[4]{(9-x^2)^3}}$
Work Step by Step
$u=(9-x^2)$; $\dfrac{du}{dx}=-2x$
$y(u)=2u^{\frac{1}{4}};\dfrac{dy}{du}=\dfrac{1}{2\sqrt[4]{u^3}}$
$\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}=(-2x)(\dfrac{1}{2\sqrt[4]{(9-x^2)^3}})=-\dfrac{x}{\sqrt[4]{(9-x^2)^3}}$
