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

First, notice that the domain of your function is $[-2,2]$. We will only be working there.
Now, when you take the derivative, there are other critical points besides the stationary points: you must also exclude any point in the domain where the derivative is not defined. In this case, notice that if $x=-2$ or $x=2$, then the derivative is not defined, so $x=-2$ and $x=2$ are critical points.
To move further, make the expression into a single fraction:
$$\begin{align*}
\frac{\frac{1}{2}x(-2x)}{\sqrt{4-x^2}} + \sqrt{4-x^2} &= \frac{-x^2}{\sqrt{4-x^2}} + \frac{\sqrt{4-x^2}\sqrt{4-x^2}}{\sqrt{4-x^2}}\\
 &= \frac{-x^2 + (4-x^2)}{\sqrt{4-x^2}}\\
&= \frac{4-2x^2}{\sqrt{4-x^2}}.
\end{align*}$$
Now, determine the points in the domain where the numerator is $0$ (we already excluded the points where the denominator is $0$). 
