Is $x^{1/3}$ differentiable at $0$? It occurred to me that functions that are "smooth" but have "infinite slope" may not be considered differentiable at the points where their slope is infinite. An easy example of this is $x^{1/3}$, which is "smooth" in a visual sense (i.e. there are no jumps in the slope, as in a function like
$$
f(x) = \begin{cases}
x & x\ge 0 \\
2x & x< 0
\end{cases}
$$
which is continuous at the origin but has different slopes as we approach $0$ in different directions), however at the origin the derivative approaches infinity.
So there's a discrepancy, between the intuitive sense of "differentiable" and the mathematically rigorous definition. Intuitively I would like to say yes, $x^{1/3}$ is differentiable at $0$, but mathematically I would be forced to say no because by definition the limit of the newton quotient of this function approaches $\infty$, and so does not exist.
Another example is the function $(xy)^{1/3}$ on $\mathbb{R}^2$. Is this function differentiable on the $x$ and $y$ axes? (with $(x,y)\neq(0,0)$, where it is actually not differentiable) Is there a notion of differentiability that reconciles this discrepancy?
 A: The function $x^{1/3}$ is not differentiable at $x=0$, but the graph $\{(x,x^{1/3})\colon x\in\mathbb{R}\}\subseteq \mathbb{R}^2$ is a smooth submanifold, something that for example does not happen with $x^{2/3}$. I believe this latter notion is the one that reconciles your discrepancy.
A: I would like to tell you why should we consider such functions non-differentible,
For functions whose slope is infinity from the both sides, for eg.,$$x^{\frac{1}{3}}$$ 
Their graphs of derivative appear like this,
This gives us a discontinuous function, 
That is why we say graph like $x^{\frac{1}{3}}$ are non-differentible at $x=0$

Many of the comments are like ohhh!  differentible is not same as smooth
  But i tend to disagree with this opinion,
  The guy who discovered this differentiable thing must have done it intuitively and not out of the blue,
  We must rather than accepting definitions as laws, learn the intuition, this helps in being a better mathematician.

A: Functions are not differentiable at points where they have infinite slope, and in that respect equating "differentiable" with "smooth" is not perfect.
As far as functions of two or more variables go, a complication is that in some cases they are considered not differentiable at points where not only do the partial derivatives exist, but all directional derivatives exist. For example:
$$
f(x,y) = r\sin(2\theta) \text{ where } r = \sqrt{x^2+y^2} \text{ and } x=r\cos\theta \text{ and } y = r\sin\theta.
$$
For this function you have
$$
\left.\frac{\partial f}{\partial x} \right|_{(x,y)=(0,0)}  = 0 \text{ and } \left.\frac{\partial f}{\partial y} \right|_{(x,y)=(0,0)} = 0 \tag 1
$$
and further more the directional derivative in every direction exists, but nonetheless the function is not differentiable at $(0,0).$ The reason is that line $(1)$ above would require the tangent plane to coincide with the $x,y$-plane, but the slope in a direction halfway between the directions of the $x$- and $y$-axes is not $0,$ but $1.$
