How can we find the derivative of a circle if a circle is not a function? If we consider the equation of a circle, $x^2+y^2=r^2$, then I understand that $dy/dx$ can be computed in the following way via implicit differentiation:
\begin{align}
2x + 2y\frac{dy}{dx} &= 0 \\
\frac{dy}{dx} &= -\frac{2x}{2y} = -\frac{x}{y} \, .
\end{align}
Although I feel comfortable deriving this result, I don't really understand how I should interpret it. On an intuitive level, the formula $dy/dx = -x/y$ seems to suggest that the gradient of the tangent to any given point $(x,y)$ is $-x/y$. However, since the curve $x^2+y^2=r^2$ fails the vertical line test, it doesn't look like it is even a function. Usually, $dy/dx$ can be thought of as a shorthand for
$$
\lim_{h \to 0}\frac{y(x+h)-y(x)}{h} \, .
$$
However, in this case each $x$-value maps to two $y$-values, and so the limit definition doesn't seem to apply here. So what does $dy/dx$ actually represent in this context?
 A: You can still think of it in terms of the slope of a tangent line, and even in terms of a limit.  However, as you've pointed out, $x^2 + y^2 = r^2$ isn't a function because it fails the vertical line test.  However, by the Implicit Function Theorem we can consider $F(x,y) = x^2 + y^2 - r^2$, and for any $(x_{0},y_{0})$ where $\frac{\partial F}{\partial y}\ne 0$ then there exists some neighborhood around the point $(x_{0},y_{0})$ for which we can express $F(x,y) = 0$ as some function $y = f(x)$.  Note that in this case, $$\frac{\partial F}{\partial y} = 2y,$$ which is zero whenever $y = 0$, so at the points $(r,0)$ and $(-r,0)$.  On the circle.  Anywhere else we can define the curve by either $$y = \sqrt{r^{2} - x^{2}}$$ or $$y = -\sqrt{r^{2} - x^{2}}$$ and these functions are differentiable so long as $y\ne 0$.  To verify that the derivative via the limit definition matches that obtained by implicit differentiation, we can compute as follows for the positive semicircle:
\begin{align}
y' &= \lim_{h\to 0}\frac{\sqrt{r^{2} - (x+h)^{2}} - \sqrt{r^2 - x^2}}{h}\\
&=\lim_{h\to 0}\frac{\sqrt{r^{2} - (x+h)^{2}} - \sqrt{r^2 - x^2}}{h} \cdot \frac{\sqrt{r^{2} - (x+h)^{2}} + \sqrt{r^2 - x^2}}{\sqrt{r^{2} - (x+h)^{2}} + \sqrt{r^2 - x^2}}\\
&=\lim_{h\to 0}\frac{r^{2} - (x+h)^{2} - r^{2} - x^{2}}{h(\sqrt{r^{2} - (x+h)^{2}} + \sqrt{r^2 - x^2})}\\
&=\lim_{h\to 0}\frac{-2xh -h^{2}}{h(\sqrt{r^{2} - (x+h)^{2}} + \sqrt{r^2 - x^2})}\\
&=\lim_{h\to 0}\frac{-2x - h}{\sqrt{r^{2} - (x+h)^{2}} + \sqrt{r^2 - x^2}}\\
&=-\frac{2x}{2\sqrt{r^{2} - x^{2}}}\\
&=-\frac{x}{y}.
\end{align}
Where the last equality comes from the equation $y = \sqrt{r^2 - x^2}.$  We could do the same for the negative semicircle.
A: While each $x\in(-r,\,r)$ is compatible with two choices of $y$, continuous motion along the circumference well-defines the choice of $y$ at each point, giving a local $y$-as-a-function-of-$x$ behaviour wherever $dy/dx$ is finite and nonzero (i.e. $x,\,y$ are both nonzero), which happens at all but four of the circumference's points.
This local behaviour is more easily described in terms of the polar angle $\theta$, and since $x=r\cos\theta,\,y=r\sin\theta$, by the chain rule $\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{x}{-y}=-\cot\theta$.
A: The curve is indeed not the graph of a function. At any point $(x,y)$ on the curve, if an open disk about that point is small enough, then that portion of the curve that is within that neighborhood is the graph of a function, and the slope of the tangent line to the graph of that function is $-x/y.$
Derivatives are local, that is the slope of a curve at a point is determined by the behavior of that curve within a small open neighborhood of the point, no matter how small it is.
Also, the concept of a tangent line to a curve is not limited to curves that are the graph of a function, so
