Answer
$g'(x)=-\dfrac{3x}{\sqrt{4-3x^2}}$
Work Step by Step
$u=4-3x^2$; $\dfrac{du}{dx}=-6x$
$g(u)=\sqrt{u};\dfrac{d}{du}g(u)=\dfrac{1}{2\sqrt{u}}.$
$\dfrac{d}{dx}g(x)=\dfrac{d}{du}g(u)\times\dfrac{du}{dx}=(-6x)(\dfrac{1}{2\sqrt{4-3x^2}})=-\dfrac{3x}{\sqrt{4-3x^2}}$
