<s>[INST] What do I do next when trying to find the derivative of this fraction? I'm trying to find the derivative of this equation: $-\frac{3(x-6)}{2\sqrt{9-x}}$
The quotient rule: $\frac{d}{dx}[\frac{f(x)}{g(x)}]=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}$ where $g(x)$ and $f(x)$ are functions.
So I take out the constants and I'm left with $\frac{-3}{2}\frac{d}{dx}[\frac{(x-6)}{\sqrt{9-x}}]$.
$f(x)$ is $(x-6)$ and $g(x)$ is $(9-x)^{1/2}$. This is what it expands to:
$$\frac{-3}{2} \left[\frac{(9-x)^{1/2}\frac{d}{dx}[x-6]-(x-6)\frac{d}{dx}[(9-x)^{1/2}]}{((9-x)^{1/2})^2}\right]$$
After simplifying, I get:
$$\frac{-3}{2}\left[\frac{(9-x)^{1/2}(1)-(x-6)[(\frac{1}{2})(9-x)^{-1/2}(-1)]}{9-x}\right]$$
$$\frac{-3}{2}\left[\frac{\sqrt{9-x}-(x-6)(-1)}{2(9-x)\sqrt{9-x}}\right]$$
$$\frac{-3}{2}\left[\frac{\sqrt{9-x}+x-6}{2(9-x)\sqrt{9-x}}\right]$$
$$\frac{-3\sqrt{9-x}-3x+18}{4(9-x)\sqrt{9-x}} \text{ ??}$$
What did I do wrong? Wolfram Alpha says the answer is: $\frac{3(x-12)}{4(9-x)^{3/2}}$
 [/INST] Until here: $$\frac{-3}{2}[\frac{(9-x)^{1/2}(1)-(x-6)[(\frac{1}{2})(9-x)^{-1/2}(-1)]}{9-x}]$$ you're good. This simplifies to: $$-\frac{3}{2}\left[\frac{\sqrt{9-x}+ \frac{1}{2}(x-6)\frac{1}{\sqrt{9-x}}}{9-x}\right] = -\frac{3}{2}\left[\frac{\color{red}{(9-x)}+ \frac{1}{2}(x-6)}{(9-x)^{3/2}}\right] = -\frac{3}{4} \frac{18-2x+x-6}{(9-x)^{3/2}},$$which is readily seen to agree with Wolfram's answer. Where I pointed in red, you kept $\color{blue}{\sqrt{9-x}}$.

More detailed:
$$\begin{align}\frac{\sqrt{9-x}+ \frac{1}{2}(x-6)\frac{1}{\sqrt{9-x}}}{9-x} \cdot\frac{\color{red}{\sqrt{9-x}}}{\color{blue}{\sqrt{9-x}}} &= \frac{\left(\sqrt{9-x}+ \frac{1}{2}(x-6)\frac{1}{\sqrt{9-x}}\right)\color{red}{\sqrt{9-x}}}{(9-x)\color{blue}{\sqrt{9-x}}} \\ &= \frac{\sqrt{9-x}\color{red}{\sqrt{9-x}}+\frac{1}{2}(x-6)\require{cancel}\cancel{\frac{1}{\sqrt{9-x}}}\cancel{\color{red}{\sqrt{9-x}}}}{(9-x)^{3/2}} \\ &= \frac{9-x+\frac{1}{2}(x-6)}{(9-x)^{3/2}}\end{align}$$
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