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I'm trying to find the derivative of the function $$ f(x)= \begin{cases} \frac{1}{n^2},& \text{if }x=\frac{m}{n}\text{with gcd}(m,n)=1\\ 0,&\text{if }x\text{ is irrational.} \end{cases} $$

That it is continuous in $\Bbb R\smallsetminus\Bbb Q$, and that the derivative (supposed it existed) has to be $0$ in $x\in \Bbb R\smallsetminus\Bbb Q$ is clear (since $\lim_{h\to0}\frac{f(x+h)}{h}=0 $ for $x+h \in \Bbb R\smallsetminus\Bbb Q$). But I'm struggling to find the lim for $x+h$ in $\Bbb Q$.

Any help is appreciated

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    [Hurwitz-Borel's theorem](https://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(number_theory)) is your friend. – 2017-02-03
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    Thank you! Do I understand correctly that this means the function has no derivative? Because I can create a sequence $a_i=\frac{m_i}{n_i}$ that converges to x but $$\lim_{n\to inf}\frac{\frac{1}{n_i^2}}{x-a_i}\ge\sqrt 5$$ – 2017-02-03
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    That's right, your sequence (together with another sequence having limit 0) proves there is no limit, hence no derivative. Thank _you_ for the nice counterexample, BTW; I wasn't familiar with it before. – 2017-02-03

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