#### Answer

$$g'(x)=\frac{-1}{2\sqrt{9-x}}$$
The domain of $g(x)$ is $(-\infty,9]$
The domain of $g'(x)$ is $(-\infty,9)$

#### Work Step by Step

$$g(x)=\sqrt{9-x}$$
According to definition, $$g'(x)=\lim\limits_{h\to0}\frac{g(x+h)-g(x)}{h}$$
$$g'(x)=\lim\limits_{h\to0}\frac{\sqrt{9-x-h}-\sqrt{9-x}}{h}$$
Multiply both numerator and denominator by $(\sqrt{9-x-h}+\sqrt{9-x})$, the numerator would become $$(\sqrt{9-x-h}-\sqrt{9-x})(\sqrt{9-x-h}+\sqrt{9-x})$$
$$=(9-x-h)-(9-x)$$
$$=9-x-h-9+x$$
$$=-h$$
Therefore, $$g'(x)=\lim\limits_{h\to0}\frac{-h}{h(\sqrt{9-x-h}+\sqrt{9-x})}$$
$$g'(x)=\lim\limits_{h\to0}\frac{-1}{(\sqrt{9-x-h}+\sqrt{9-x})}$$
$$g'(x)=\frac{-1}{\sqrt{9-x}+\sqrt{9-x}}$$
$$g'(x)=\frac{-1}{2\sqrt{9-x}}$$
*Domain of $g(x)$:
$(9-x)$ must be $\ge0$. So, $x\le9$
Therefore, the domain of $g(x)$ is $(-\infty,9]$
*Domain of $g'(x)$:
As before, $(9-x)$ must be $\ge0$. So, $x\le9$
Also, $2\sqrt{9-x}\ne0$. Therefore, $9-x\ne0$, so $x\ne9$
After all, the domain of $g'(x)$ is $(-\infty,9)$