My attempt: here the $f(x)=\lfloor(x\sin(\pi x)\rfloor$. For the function to be differentiable in an interval $(-1,1)$ the right hand derivative of $-1$ must be finite and the left hand derivative of $-2$ must be finite. But i cannot figure out how to find the limit of $(f(-1+h)-f(-1))/h$ as h tends to $0^+$.
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What do "left hand derivative" and "right hand derivative" mean? – YiFan Tey Dec 08 '18 at 05:31
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@YiFan Please refer to this https://math.stackexchange.com/questions/1158510/left-hand-derivative-definition – Onkar Dahale Dec 08 '18 at 05:36
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Isn't $f(x) \equiv 0$ since $f(x) \in [0, 1)$ $\forall x \in (-1, 1)$? Can you prove it? – Alex Vong Dec 08 '18 at 06:05
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@OnkarDahale okay, though I have to say that terminology is quite rare (usually I would just say left hand limit of [...] for example.) – YiFan Tey Dec 08 '18 at 07:14
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Firstly, I want to clarify that since we are talking about the differentiability of $f$ on $(-1, 1)$, not $[-1, 1]$, we don't need to worry about the differentiability of $f$ at the end points.
With this in mind, let's see what we can say about $x \sin(\pi x)$. It is not hard to see that $(-x) \sin(-\pi x) = x \sin(\pi x)$. Therefore, $x \sin(\pi x)$ is an even function on $(-1, 1)$. Furthermore, $0 \le x < 1$ and $0 \le \sin(\pi x) \le 1$ on $[0, 1)$. Hence $0 \le x \sin(\pi x) < 1$ on $[0, 1)$. Finally, using the fact that $x \sin(\pi x)$ is an even function, we conclude $0 \le x \sin(\pi x) < 1$ on $(-1, 1)$.
This means $\lfloor x \sin(\pi x) \rfloor = 0$ on $(-1, 1)$, which is clearly differentiable.
Alex Vong
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