Answer
$-2x$
Work Step by Step
Since $f(x)=\sqrt{3-x}$, then $f'(x)=-\frac{1}{2\sqrt{3-x}}$. Hence by Theorem 2 and assuming that $g(x)=f^{-1}(x)$, we have
$$y=\sqrt{3-x}\Longrightarrow y^2=3-x\Longrightarrow x=3-y^2,$$ Therefore,
$$g(x)=3-x^2, \quad g'(x)=\frac{1}{f'(g(x))}\\=-\frac{1}{ (1/(2\sqrt{3-(3-x^2))}) }= -2x.$$
