<s>[INST] Find the derivative of $y=x\sqrt{9-x}$ "Find the derivative of $y=x\sqrt{9-x}$."
So this is what I have and now I'm stuck.
\begin{align}
y' &= x \frac{d}{dx}\left[(9-x)^{1/2}\right] + (9-x)^{1/2} \frac{d}{dx}(x)\\
   &= x \left[\frac{1}{2}(9-x)^{-1/2}\right] + (9-x)^{1/2} (1)
\end{align}
So I now that I need to multiply and simplify but I don't know where to start. Help!
This problem is actually part of a homework question where I have to analyze a graph and find critical points and min and max.
 [/INST] $$y=x\sqrt{9-x}$$
$$y'=x'\sqrt{9-x}+x(\sqrt{9-x})'=\sqrt{9-x}+x\frac{1}{2\sqrt{9-x}}(9-x)'=$$
$$=\sqrt{9-x}+x\frac{1}{2\sqrt{9-x}}(-1)=\sqrt{9-x}+\frac{-x}{2\sqrt{9-x}}=$$
$$=\frac{2(9-x)-x}{2\sqrt{9-x}}=\frac{18-3x}{2\sqrt{9-x}}$$
</s>