How does a smooth differential structure hide a discontinuity? A theorem of differential topology states that any $C^1$ manifold admits a compatible $C^{\infty}$ differential structure.  I find this curious on account of such manifolds as the graph of $f(x) = x|x|$, which is $C^1$ but has a discontinuity it it's second derivative.  
Question: How does the compatible smooth structure on the above graph "hide" the discontinuity?
A qualitative answer is fine.
 A: I think a better way to think about it is that a $C^1$ manifold doesn't "know" about any discontinuity in higher derivatives it might have.  You can't define the second derivative of a function on a $C^1$ manifold, so you can't even express the idea that such a derivative would be discontinuous intrinsically.
Let's call your example $X$.  Note that the projection map $p:X\to\mathbb{R}$ sending $(x,x|x|)$ to $x$ is a $C^1$ diffeomorphism.  So actually, $X$ is exactly the same as $\mathbb{R}$ as a $C^1$ manifold.  The $C^1$ manifold structure of $X$ can't tell that it has a "kink" at $(0,0)$: as far as the $C^1$ structure is concerned, $X$ might as well just be $\mathbb{R}$ with no kink.
The only reason you think that your $X$ has a kink in it is because you are imagining it sitting inside $\mathbb{R}^2$, and indeed the map $X\to \mathbb{R}^2$ is not a $C^2$ embedding.  But this embedding is not part of the $C^1$ manifold structure of $X$.  The $C^\infty$ structure you can put on $X$ doesn't "hide" the singularity of $X$ as a subset of $\mathbb{R}^2$; it's just that the map $X\to\mathbb{R}^2$ fails to be a $C^\infty$ embedding.
