Graphical explanation of the difference between $C^1$ and $C^2$ function? We are all aware of the intuitive (graphical) explanation of the concepts of continuous and differentiable function. Whenever these two concepts are formally defined, the following elementary explanations are given:

A continuous function is a function whose graph has no "holes" or "jumps", and a differentiable function is a function whose graph has no "corners".

This is a non continuous function:

This is a non differentiable continuous function: 

And this is a differentiable continuous function:

Is there a "graphical" or intuitive explanation of the difference between a $C^1$ function and a differentiable function with discontinuous derivatives? What about a function that is $C^1$ but not $C^2$ because it does not have second derivatives? Or what about a function that is $C^1$ and has second derivatives but they are not continuous? What about the difference between a $C^1$ function and a $C^{\infty}$ function? 
 A: If $f$  is everywhere differentiable, but $f'$ is not continuous at some point, then $f'$ has to be very discontinuous there, because otherwise $f'$ could not satisfy the intermediate value requirement. As an example consider the function $f(x):=x^2\sin{1\over x}$ $(x\ne0)$, $f(0):=0$.
On the other hand there are beautiful functions which are $C^1$, but not everywhere twice differentiable, e.g., the function $f(x):=0$ for $x\leq0$ and $:=x^2$ for $x\geq0$. Here $f''(0)$ is undefined, and has a jump discontinuity there.
A: In general the differences are very subtle, and I don't know of any good way to visualize them. For instance, here is an example of a function that is $C^2$ but whose second derivative is not differentiable at $x=0$, the function $y=|x|^3$:

(It does not have a third derivative at $x=0$ because the second derivative is $6|x|$.) Can you tell visually that this function does not have a third derivative at $x=0$? I can't, although this could just be a poor example.
Hopefully you're familiar with the famous example of the Weierstrass function, which is $C^0$ but nowhere differentiable:

It's visually clear, I think, that this function is not differentiable anywhere: the surface is too rough.
But it gets harder when you look at the next level. By integrating the Weierstrass function, we obtain a function that is $C^1$ but nowhere second-differentiable:

It looks smooth enough to be differentiable, but I wouldn't know how to tell if the second derivative exists just by looking at the picture.
In summary, I think that our visual intuitions about graphs are too imprecise to fully capture the mathematical notion of smoothness.
A: Indentifying $C^1$ curves is quite straightforward - slope along the curve must change smoothly. Another way to put it is that when there is no pointy spot along a continuous curve, then it's $C^1$. This is because if it were to have such a pointy spot, slope changes abruptly, and this means its first derivative is not continuous (in other words, not $C^1$.) Very informally speaking, let say you put one of your fingers on a curve and run it along the curve. If you notice some sort of sudden change of your arm's muscle tension while doing it, then you can tell that the curve is not $C^1$ but proper $C^0$. However, if it feels smooth like butter, then it's safe to say that the curve is $C^1$.
It's quite tricky to tell if a curve is proper $C^1$, but it's doable in some extent. The best way one could come up with is trying to see how curvature or osculating cirlce changes along the curve. A proper $C^1$ curve has, at some point along the curve, abruptly (or, not continuously) changing osculating circle. This is because the curvature of a curve is directly related to its second derivative.
For a smoother ($n\geq 2$) curve, the curvature changes continuously so you'll find continuously changing osculating circle along the line. However, there is hardly a way to graphically explain/see a subtle difference between curves smoother than that of $C^1$.
To sum up, a $C^0$ curve is continuous, proper $C^0$ has abruptly changing slope, $C^1$ has smoothly changing slope, proper $C^1$ has abruptly changing osculating circle. A smoother curve ($C^2$ or higher) has smoothly changing osculating circle, but hard to distinguish its degree of smoothness properly.
A: The Nash-Kuiper theorem says that there is an embedding of the hyperbolic space, $\mathbb{H}^{2}$ in $\mathbb{R}^{3}$ which is $C^{1}$ but not $C^{2}$, because curvature, a $C^{2}$ characteristic, prohibits this for smoother embeddings.  Using those methods, Borrelli, et al. did some $C^{1}$ embeddings, and did computer graphics that give you an intuition for it, in http://www.pnas.org/content/109/19/7218.full.pdf. Have a look, their method is quite general for producing intuitive graphics of $C^{1}$ but not $C^{2}$ surfaces by convex integration.
