Isolate a variable in a polynomial function How would I go about isolating $y$ in this function? I'm going crazy right now because I can't figure this out.
The purpose of this is to allow me to derive $f(x)$ afterwards.
$$ x = \frac{y^2}{4} + 2y .$$
 A: Point the Zeroth: ignore points the first and points the second (in the sense that they aren't really the 'right' way of proceeding; they are presented so you can see that they are not the right path to take). 
Point the First: $y$ is not a function of $x$; if you plot this equation on the plane, you'll have a parabola,
$$x = \frac{y^2}{4} + 2y = \left(\frac{y}{2}\right)^2 + 2\left(\frac{y}{2}\right)(2) + 2^2 - 2^2 = \left(\frac{y}{2} + 2\right)^2 - 4.$$
This parabola opens right, so it is not the graph of a function of $x$. 
Point the Second: You can break up the graph into two functions by using the quadratic formula:
$$ \frac{y^2}{4} + 2y - x = 0$$
gives
$$y^2 + 8y - 4x = 0,$$
so
$$y = \frac{-8+\sqrt{64+16x}}{2},\quad\text{or}\quad y = \frac{-8-\sqrt{64+16x}}{2}.$$
We would then need to find the derivatives of each of these two separately, and for any given value of $x$ and $y$, determine which of the two formulas to use. They are not hard, but they are somewhat annoying.
If $y = -4 + \frac{1}{2}\sqrt{64+16x}$, then
$$\frac{dy}{dx} = \frac{1}{4}(64+16x)^{-1/2}(16) = \frac{4}{\sqrt{64+16x}}.$$
Similarly, if $y=-4-\frac{1}{2}\sqrt{64+16x}$, then
$$\frac{dy}{dx} = -\frac{4}{\sqrt{64+16x}}.$$
Point the Third: What you really want to do here is implicit differentiation, which is a way of handling all of these difficulties without having to solve for $y$ first, and without having to worry about "which formula" to use later. Explicitly, from
$$x = \frac{y^2}{4} + 2y,$$
take derivatives on both sides, using the Chain Rule and remembering that $y$ is an (implicit) function of $x$, so that $y'$ needs to be left indicated (we don't know what it is right now):
$$\begin{align*}
x & = \frac{y^2}{4} + 2y\\
\frac{d}{dx}x &= \frac{d}{dx}\left( \frac{y^2}{4} + 2y\right)\\
1 &= \frac{2y}{4}y' + 2y'\\
1&= \frac{y}{2}y' + 2y'\\
1 &= y'\left(\frac{y}{2} + 2\right).\end{align*}$$
Solving for $y'$ gives an implicit definition for $\frac{dy}{dx}$ in terms of $y$ and $x$ (though in this case, $x$ plays no role):
$$y' = \frac{1}{\frac{y}{2}+2} = \frac{2}{y+4}.$$
Point the Fourth: Alternatively, since you have $x$ explicitly as a function of $y$, use the Inverse Function Theorem: taking derivatives with respect to $y$, we have:
$$\frac{dx}{dy} = \frac{1}{2}y + 2,$$
so
$$\frac{dy}{dx} = \frac{1}{\quad\frac{dx}{dy}\quad} = \frac{1}{\frac{1}{2}y + 2} = \frac{2}{y+4}.$$
Point the Fifth: So, do these "implicit formulas" give the same answer as the "explicit ones" we got in Point the Second? Yes! 
If $y = \frac{-8+\sqrt{64+16x}}{2}$, then $y+4 = \frac{\sqrt{64+16x}}{2}$, so
$$\frac{2}{y+4} = \frac{2}{\frac{1}{2}\sqrt{64+16x}} = \frac{4}{\sqrt{64+16x}},$$
same answer as in Point the Second; and if $y=\frac{-8-\sqrt{64+16x}}{2}$ then $y+4 = -\frac{1}{2}\sqrt{64+16x}$, so
$$\frac{2}{y+4} = \frac{2}{-\frac{1}{2}\sqrt{64+16x}} = -\frac{4}{\sqrt{64+16x}},$$
again, same answer as in Point the Second. 
But using implicit differentiation (or in cases like this, when $x$ is an explicit function of $y$, the inverse function theorem) is much easier than first solving for $y$, possibly requiring breaking up the original implicit function into several different explicit functions, and then differentiating. If you were trying to work with the Folium of Descartes ($x^3+y^3=3xy$), you would have to consider three different formulas, each involving a sum of cubic roots that has square roots inside the radicals; if you were trying to work with a function like $y = \sin(x+y)$, you would have a hard time solving for $y$, but using implicit differentiation is pretty easy.
