If we take the implicit derivative of $x^3+x^2-y^2=0$, we find that $\frac{dy}{dx}=\frac{3x^2+2x}{2y}.$ So, the slope of the tangent line should be undefined at any point where y is 0.

To me, the tangent line to the graph of the equation at $x=0$ should not have an undefined slope. It appears that the function is not differentiable at the point $x = 0$.

How do we know if the implicit derivative is undefined due to an actual vertical slope or due to the equation being non-differentiable at that point?

  • $\begingroup$ Consider $\lim_{x\to x_0}f’(x)$. Like $\lim_{ x\to 0} (\sqrt{|x|})’=\infty$ and $\lim_{x\to 0}(e^{-1/x^2})’=0$. Both functions are non-differentiable at zero. The first one can be understood as ‘non-differentiable due to vertical slope’. For the second one, although the limit exists, it is still non-differentiable at zero; in fact, there is an essential singularity. $\endgroup$ – Szeto Jun 13 '18 at 23:30

It looks like you have to be mores specific about what you mean about when $y=0$

plot of $x^3+x^2-y^2=0$

At $(-1,0)$ you have a vertical tangent line $x=-1$ with undefined slope.

At $(0,0)$ you are correct there is not a uniquely defined tangent line. However you can make a uniquely defined tangent line of you parameterize the curve. That is both $x$ and $y$ would be functions of some parameter for example $t$ then the curve will be drawn $(x(t),y(t))$.

In this way of looking at the curve, the self intersection will be two unique $t$-values and you can draw a distinct tangent line for each one and it looks like the slope will actually be defined in that case, but you are adding extra information with the parameterization.

Edit Also I might add that not just any parameterization will work you'd need to pick a differentiable one, that is $\frac{dx}{dt}$ and $\frac{dy}{dt}$ both exist and are not simultaneously zero. This forces no sharp turns along the path you follow. So at the offending point $(0,0)$ you'd need to make sure it follows to the opposite quadrant as $t$ gets larger.

Just to complete what I was saying above I found a differentiable parameterization of the curve

$$ \begin{split} x&=\tan^2 t -1\\ y&=\tan^3 t- \tan t \end{split} $$

Which has derivatives

$$ \begin{split} \frac{dx}{dt}&=2\tan t\sec^2 t \\ \frac{dy}{dt}&=3\tan^2 t \sec^2 t- \sec^2 t \end{split} $$

On the interval $-\frac{\pi}{2} < t < \frac{\pi}{2}$

And the $(0,0)$ point on the graph occurs at the $t$-values $t=-\frac{\pi}{4}$ and $t=\frac{\pi}{4}$

$$ \begin{split} \frac{dx}{dt}|_{t=\frac{\pi}{4}}&=4 \\ \frac{dy}{dt}|_{t=\frac{\pi}{4}}&=4\\ \frac{dy}{dx}|_{t=\frac{\pi}{4}}&=\frac{dy/dy|_{t=\frac{\pi}{4}}}{dx/dt|_{t=\frac{\pi}{4}}}=1 \end{split} $$

$$ \begin{split} \frac{dx}{dt}|_{t=-\frac{\pi}{4}}&=-4 \\ \frac{dy}{dt}|_{t=-\frac{\pi}{4}}&=4\\ \frac{dy}{dx}|_{t=-\frac{\pi}{4}}&=\frac{dy/dy|_{t=-\frac{\pi}{4}}}{dx/dt|_{t=-\frac{\pi}{4}}}=-1 \end{split} $$

I'm fairly sure this curve is well-studied, but don't know the name of it and just felt like deriving something like this myself :)

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