# 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.

$$x = \frac{y^2}{4} + 2y .$$

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Try completing the square. Note that you can simply apply implicit differentiation also, to find the derivative. – JavaMan Oct 18 '11 at 3:32
Is your function $x = \frac{y^2}{4} + 2y$, or $x = \frac{y^2}{4+2y}$? The problem with the slash notation is that the absence of parentheses makes it ambiguous. – Arturo Magidin Oct 18 '11 at 3:33
What you have is a quadratic equation for $y$; $(y^2/4)+2y-x=0$. What methods do you know for solving quadratic equations? – Gerry Myerson Oct 18 '11 at 3:49
When you have a nice relationship, differentiate immediately. That usually works out better. Presumably the problem in the post arose when you were trying to solve some other problem. If one knew what that problem was, an efficient approach could be suggested. – André Nicolas Oct 18 '11 at 3:52
The whole point of implicit differentiation is to find $y'$ without having to first "solve for $y$". Don't try: use implicit differentiation! – Arturo Magidin Oct 18 '11 at 4:01

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.

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