For what $x\in[0,1]$ is $y = \sum\limits_{k = 1}^\infty\frac{\sin( k!^2x )}{k!}$ differentiable? 
For what $x\in[0,1]$ will the function $y = \sum\limits_{k = 1}^\infty\frac{\sin( k!^2x )}{k!}$ be differentiable? How do you know?

Here is the equation expressed more clearly on Wolfram Alpha. The only difference is that 10 should be Infinity (Wolfram apparently can't handle that yet).
I'm trying to understand for what $x\in[0,1]$ this function is differentiable. I've used a computer to plot the graph of $y'$ (the derivative of the function) with the upper limit (top number of sigma) as 10, and then with the upper limit as 11, 12... it looks like these "zig-zags" continue to exist as you "go deeper" into the function.
...so I'm thinking the values of $x\in[0,1]$ that make the function differentiable are all of them... But is my line of thinking correct? How can I validate?

Thanks to great comments from people like Henning Makholm, I have an answer, and I decided to summarize it here. (Please edit this if I have any inaccuracies.)
First of all, I was mistaken in my supposed graphing of the derivative. The derivative of y equates to the sum of the derivatives, namely $y' = \sum\limits_{k = 1}^\infty\ k!cos(k!^2x)$
This sum diverges, meaning it should be impossible to graph the derivative. We know the sum diverges because cos oscillates between -1 and 1.
As Henning Makholm says, "[The graph of y] is nowhere differentiable because for every $x_0$, the difference quotient $\frac{\ f(x)−f(x_0)}{x−x_0}$ can be made arbitrarily large for $x$ arbitrarily close to $x_0$." This means that we have points of non-differentiability at every point in the graph. Also, if we examine the graph of y with increasing magnification, we see that the "zig zags" continue to appear, and this further confirms the non-differentiability of all points.
Therefore, for no $x\in[0,1]$ is the function y differentiable.
