Implicit differentiation. Confusing assumption.

Question

I've encountered this example of implicit differentiation. The problem asks us to find the derivative of $y^2 + x^2 = 4$

How can they assume that y is a function x when it clearly is not? If we isolate y, we get: $y = \pm \sqrt{4-x^2}$. So then how can we assumt that y is a function of x?

On a high level, I'm still a bit confused. So I guess implicit differentiation is useful when we have relations between y and x that aren't functions. That's why the $\frac{dy}{dx}$ is in term of 2 variables. $\frac{dy}{dx}$ has two values depending on the value of x and the value of y since at x, there are 2 values of y. Is that right?

Answer 1

First, let's double-check the answer they found:

$$ y^2+x^2=4 \rightarrow y=\pm\sqrt{4-x^2} $$ Case 1: $y=\sqrt{4-x^2}$

$$ \frac{dy}{dx}=\frac{1}{2}(4-x^2)^{-0.5}(-2x)=-\frac{x}{\sqrt{4-x^2}}=-\frac{x}{y} $$

Case 2: $y=-\sqrt{4-x^2}$

$$ \frac{dy}{dx}=-\frac{1}{2}(4-x^2)^{-0.5}(-2x)=\frac{x}{\sqrt{4-x^2}}=\frac{x}{-(-\sqrt{4-x^2})}=-\frac{x}{y} $$

Now to your confusion. When you said $y$ is not a function of $x$ because $y=\pm\sqrt{4-x^2}$, you were contradicting yourself because clearly, $y$ is a function of $x$, namely $y=f(x)=\sqrt{4-x^2}$ or $y=f(x)=-\sqrt{4-x^2}$. Which one is the function depends on the conditions on $y$. For example, if the problem says that $y$ is always positive, we will know that $y=\sqrt{4-x^2}$. $$$$The ambiguity in the actual form of $y$, however, does not affect the relation between $\frac{dy}{dx}$ and $x$, as found in the answer by using implicit differentiation. $$$$ Note that we must know whether $y$ is a function of $x$ or $x$ is a function of $y$ when doing implicit differentiation. This is important because $y=f(x)$ does not necessarily mean a function $x=f^{-1}(y)$ exists.

Answer 2

Osama Ghani has pointed out the right idea to have in mind. I'll just add a bit of detail to what he is hinting at.

We know the set of solutions to the equation $$y^2+x^2=4$$ is simply the circle of radius 2. You are certainly correct in pointing out that the circle is not the graph of a differentiable function of $x$. However, let's not forget that the operation of differentiation is a local one -- we take the derivative at each point of a given function.

Then the crucial idea is that for each point on the circle $(x,y)\in \{(x,y)\in \mathbb{R}^2:x^2+y^2=4\}$, if we intersect a small ball (centered at $(x,y)$) of some radius $\delta$ with the circle, we do in fact get the graph of a function -- it looks like an arc on the circle. This function depends on the chosen point and its domain is restricted according to the radius $\delta$. The observation you made is simply that these local functions are not the restrictions of a globally defined function.

Note also that the final answer for $\frac{dy}{dx}$ (when solved implicitly) depends on both $x$ and $y$ and not simply $x$. This again is because $y$ is not globally a function of $x$. You need to specify a pair $(x,y)$ to give a locally defined function $y(x)$, whose graph contains the point $(x,y)$.