#### Answer

$f'(x) = -\dfrac{1}{2\sqrt{9-x}} \\$
$ Dom f(x) = (-\infty,9] \\$
$ Dom f'(x) = (-\infty,9)$

#### Work Step by Step

Given $f(x) = \sqrt{9-x}$
It's domain is given by the inequality $9-x\geq0 \longrightarrow 9\geq x \longrightarrow x\leq 9 \\ $
Therefore $Dom f(x) = (-\infty,9]$
$f'(x) = \lim\limits_{h \to 0} \dfrac{f(x+h) - f(x)}{h} = \lim\limits_{h \to 0} \dfrac{\sqrt{9-(x+h)} - \sqrt{9-x}}{h} = \lim\limits_{h \to 0} \dfrac{\sqrt{9-(x+h)} - \sqrt{9-x}}{h} \times \dfrac{\sqrt{9-(x+h)} + \sqrt{9-x}}{\sqrt{9-(x+h)} + \sqrt{9-x}} = \lim\limits_{h \to 0} \dfrac{(\sqrt{9-(x+h)} - \sqrt{9-x})(\sqrt{9-(x+h)} + \sqrt{9-x})}{h(\sqrt{9-(x+h)} + \sqrt{9-x})} = \lim\limits_{h \to 0} \dfrac{(\sqrt{9-(x+h)})^2 - (\sqrt{9-x})^2}{h(\sqrt{9-(x+h)} + \sqrt{9-x})} = \lim\limits_{h \to 0} \dfrac{9-(x+h) - (9-x)}{h(\sqrt{9-x-h} + \sqrt{9-x})} = \lim\limits_{h \to 0} \dfrac{9-x-h - 9+x}{h(\sqrt{9-x-h} + \sqrt{9-x})} = \lim\limits_{h \to 0} \dfrac{-h}{h(\sqrt{9-x-h} + \sqrt{9-x})} = \lim\limits_{h \to 0} \dfrac{-1}{(\sqrt{9-x-h} + \sqrt{9-x})} = \dfrac{-1}{\sqrt{9-x-0} + \sqrt{9-x}} = -\dfrac{1}{\sqrt{9-x} + \sqrt{9-x}} = -\dfrac{1}{2\sqrt{9-x}} \longrightarrow f'(x) = -\dfrac{1}{2\sqrt{9-x}} \\ $
$Dom f'(x)$ is the same as $Dom f(x)$ but with the slight difference that $x\ne 9$ because this would make $f'(x)$ undefined. Therefore $Dom f'(x) = (-\infty,9)$