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How I can calculate the derivative of $$f(x) = \left\{ \begin{gathered} {x^2}\quad,\quad{\text{if}}\quad x \in \mathbb{Q} \\ {x^3}\quad,\quad{\text{if}}\quad x \notin \mathbb{Q} \\ \end{gathered} \right.$$ at some $x\in \mathbb{R}$?

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    There are actually _two_ points where this function is continuous: the two points where the graphs of the cubic and quadratic functions intersect. But it's differentiable at only one of them.2011-08-11
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    Exactly. I've seen it.2011-08-12

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HINT:

The derivative exists if $\lim _{y \to x} \dfrac{f(y) - f(x)}{y - x}$ exists. Of course, a limit must be the same along any Cauchy sequence. So at what points does the derivative even exist? (it does exist somewhere)

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    Thanks for the reply, but how can calculate such a limit?. If you only know where the variable estimate limits on the numbers approaching real.2011-08-11
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    @mathsalomon For example, suppose we wanted to calculate it at $x = 2$. Then on irrationals, the limit is 4. On the irrationals, the limit is 12. They're not equal, so it's not differentiable at 2.2011-08-11
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    oh, I see, sounds reasonable. thank you very much.2011-08-11
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The first helpful information to look for is if your function is continuous at any $x$. After all, a function does not have a well-defined derivative where it isn't continuous.

Then, analyze those points where it is continuous. Does it have a derivative there? A hint is that there is always a rational point in between two real numbers (that aren't equal) and that there's always an irrational point in between two real numbers (again, nonequal).

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    I think your very good suggestion, I think I will do so. thanks.2011-08-11
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What makes you think it has a derivative? Doesn't a function have to be continuous to be differentiable?

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    If you plot the function, we could be continued only at $x=0$ and $x=1$.2011-08-11
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    If what you are saying is that the function is continuous only at $x=0$ and $x=1$, then, yes, that's true.2011-08-11
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    Oh yes. That is true.2011-08-12