How does one interpret $\dfrac{dx}{dy}$ for a function which isn't invertible?

Question

I was just going through the proof of derivative of inverse functions.

The statement reads:

If $y= f(x)$ is a differentiable function of $x$ such that it's inverse $x=f^{-1}(y)$ exists, then $x$ is a differentiable function of $y$ and it's derivative is $\dfrac{dx}{dy} = \dfrac{1}{\frac{dy}{dx}}, \dfrac{dy}{dx}≠0 $

Which naturally arises few questions, what if the inverse $y=f(x)$ doesn't exist?

My Questions :

1. Does $\dfrac{dx}{dy}$ still have a meaning?
2. If so then what would it mean geometrically?
3. Would $\dfrac{dx}{dy} = \dfrac{1}{\frac{dy}{dx}}$ still hold? And if it does, then why do we even need the invertible condition in the statement?

Answer 1

Notationwise, you should interpret $$\frac {dx}{dy}= \frac{ df^{-1}(y)}{dy}$$

Where we have defined $y = f(x)$ so $x = f^{-1}(y)$, now if the function $f$ is not invertible, we know that the inverse relation $f^{-1}$ is not a proper function even to begin with, so differentiation does not make an awful lot of sense.

Take the parabola $y=x^2$, it does not have one unique inverse over $\mathbb R$, it has two possible 'inverses' $\pm \sqrt x$, this has two slopes for every $x$, opposite and equal and that's why we cannot define a unique derivative. We usually interpret $\frac{dx}{dy}$ as the slope of the inverse function, but for non-invertible functions we run into problems. We simply do not understand what you mean by "the slope" - of what?

Answer 2

Your statement isn't actually true. For a counterexample, take $f(x)=x^3$. The inverse $f^{-1}(y)=y^{\frac13}$ exists, but is not differentiable at $y=0$.