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.
 A: Just for fun, an alternate approach:
$$\begin{align}
y&= x\sqrt{9-x}\\
y^2&=9x^2-x^3\\
2yy'&=18x-3x^2\\
y'&=\frac{18x-3x^2}{2y}\\
y'&=\frac{18x-3x^2}{2x\sqrt{9-x}}\\
y'&=\frac{18-3x}{2\sqrt{9-x}}
\end{align}$$
A: Set $$f(x)= x, \quad g(z) = \sqrt{z}\quad \text{ and }\quad h(x)= 9-x,$$
then $$y = f(x)g(h(x)).$$
You should be able to compute
$$f'(x)= 1, \quad g'(z) = \frac{1}{2\sqrt{z}}\quad \text{ and }\quad h'(x)=-1.$$
Now using the multiplication rule and the chain rule for derivatives, we know that
$$y' = g(h(x))f'(x)+f(x)(g(h(x)))' =  g(h(x))f'(x)+f(x)(g'(h(x))h'(x)),$$
replacing by the expressions above, we get
$$y' = \sqrt{9-x}\cdot 1+x\left(\frac{1}{2\sqrt{9-x}}\cdot (-1)\right)= \sqrt{9-x}-\frac{x}{2\sqrt{9-x}}$$
A: Also, for fun, a different approach using logarithms. It might seem it makes things harder, but it actually gets you to look at similar problems in a different way. If $f(x) = x \sqrt{9-x}$, then define $Lf(x) = \log f(x)$. You get 
$$
Lf(x) = \log x + \log \sqrt{9-x} = \log x +\frac{1}{2}\log (9-x)\\
\frac{d L f(x)}{dx} = \frac{f'(x)}{f(x)} = \frac{1}{x}-\frac{1}{2(9-x)} = \frac{3(6-x)}{2x(9-x)}
$$
Hence, 
$$
f'(x) = \frac{3f(x)(6-x)}{2x(9-x)} = \frac{3x \sqrt{9-x}(6-x)}{2x(9-x)} = \frac{3 \sqrt{9-x}(6-x)}{2(9-x)}
$$
A: $$y=x\sqrt{9-x}$$
$$y'=x'\sqrt{9-x}+x(\sqrt{9-x})'=\sqrt{9-x}+x\frac{1}{2\sqrt{9-x}}(9-x)'=$$
$$=\sqrt{9-x}+x\frac{1}{2\sqrt{9-x}}(-1)=\sqrt{9-x}+\frac{-x}{2\sqrt{9-x}}=$$
$$=\frac{2(9-x)-x}{2\sqrt{9-x}}=\frac{18-3x}{2\sqrt{9-x}}$$
A: Let $$ y=(-u+9)(u^{\frac{1}{2}})$$
Then,
 $$y=(-u^{3/2}+9u^{1/2})$$
And the derivate is, $$y'= -{\frac{3}{2}}(u^{1/2})+{\frac{9}{2}}(u^{-1/2})$$
As $u= 9-x$
Then,
$$y'= -{\frac{3}{2}}[(9-x)^{\frac{1}{2}}]+{\frac{9}{2}}[(9-x)^{\frac{-1}{2}}]$$ 
A: An inspection of given expression shows x = 0 and x = 9 are roots.
When sqrt appears, it is often advantageous to remove it by squaring.
$ y^2 = 9 x^2-x^3,  2\ y\ y' = 18\,x -3x^2, 2 \,(y \,y'' + y^{'2})= 18-6 x $
@ x = 6 maximum is found as second derivative is < 0.
From the second relation the derivative is
$$ y'= \dfrac{3(6-x)}{2\sqrt{9-x}}$$
