$y'$ at $(0, 0)$ of $x^2 = y^2$

Question

After noticing that implicit differentiation allows you to find an expression for a derivative for relations that aren't even functions (ex. $\frac{dx}{dy}$ where $x$ is defined by $y = x^4 + ax^3 + bx^2+cx + d$), I have been playing around with similar relations, such as $x^2 = y^2$. Implicitly I can find that $$x = y \cdot y' = \pm x \cdot y'$$ According to this, $y' = \{1, -1\} $ when $x \neq 0$ and $y' = \Bbb R $ when $x = 0$. However, looking at the graph, I am not convinced why $y'$ should be $\Bbb R$ instead of $\{1, -1 \}$.

Instead, I would intuitively expect $y' = \{1, -1\}$ for all $y$, or perhaps algebraically $(y')^2 = 1$. (But I couldn't get my derivative into this form without division by $0$). Looking at the graph, I would find it more natural to draw tangent lines at $(0, 0)$ with slope $1$ or $-1$ compared to, say, $\pi$ or anything else which satisfies $0 = 0\cdot y'$ . I would appreciate if someone could explain the discrepancy!

Answer 1

You have to be careful about when implicit differentiation works. It relies on the implicit function theorem:

If $F = F(x,y)$ is a continuously differentiable function of two variables, then the level set $\{(x,y) \mid F(x,y) = 0\}$ implicitly defines $y$ as a function of $x$ in a neighborhood of a point $(a,b)$ provided that $\dfrac{\partial F}{\partial y}(a,b) \not= 0$.

In your example $F(x,y) = x^2 - y^2$ and $\dfrac{\partial F}{\partial y}(0,0) = 0$ so the implicit function theorem doesn't hold.

Answer 2

Talking about $y'(x)$ presupposes that you have a function $y=y(x)$ to begin with. But there is no neighbourhood of the origin where the equation $y^2=x^2$ defines $y$ as a function of $x$ (since for each $x \neq 0$ you have two choices of $y$ which satisfies that equation). So the question is meaningless.

(By the way, if $y(x)$ is a real-valued function, then $y'(x)$ is a real-valued function too, so it's also meaningless to say that the value $y'(0)$ is a set like $\mathbb{R}$ or $\{-1,1\}$. If $y'(0)$ exists, it's a real number.)