Find critical points of a function: $y = x  \sqrt{4 - x^2}$ I am having trouble finding the critical points of this function, I was wondering if someone could help me out. 
So I have a function $y = x\sqrt{4 - x^2}$. I know that I have to take the derivative then set that equal to 0 to find the critical points.
$$\begin{align*}
    y  &= x  \sqrt{4 - x^2}\\
    y' &= x  \frac{d}{dx}\sqrt{4 - x^2} + \sqrt{4 - x^2}&& \text{(product rule)}\\
       &= x\left(\frac{1}{2}(4 - x^2)^{-1/2}(-2x)\right) + \sqrt{4 - x^2} &&\text{(chain rule on square root)}\\
    &= \frac{\frac{1}{2}x(-2x)}{(4-x^2)^{1/2}} + \sqrt{4-x^2}\\
&\qquad\qquad\text{(moved to denominator changed the exponent sign)}
\end{align*}$$
This is where I am stuck, I cant seem to figure out how to proceed further, any help would be appreciated
Thanks

Put your last expression together as one fraction by getting a common denominator.  The right hand expression will be
$$\frac{4-x^2}{\sqrt{4-x^2}}$$
so the overall fraction is
$$\frac{-x^2 + 4 - x^2}{\sqrt{4-x^2}} = \frac{4 - 2x^2}{\sqrt{4-x^2}} = \frac{-2(x^2 - 2)}{\sqrt{4-x^2}}$$
which is 0 when $x = \pm \sqrt 2$.
