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I learned that the derivative of a continuous function $f$ (if it exists) is $$ f'(x):=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}, $$ or any other "equivalent" definition. Since $\mathbb{Q}$ is dense, if $f'(x)$ is defined in $\mathbb{R}$, can the derivative be defined over $\mathbb{Q}$ as well?

I feel like I can see $f'$ defined for $f:\mathbb{Q}\to\mathbb{Q}$, where $f(x)=x^2$, and $f'$ failing to be defined for $g:\mathbb{Q}^+\to\mathbb{R}$, where $g(x)=\ln(x)$, or $h:\mathbb{Q}^+\to\mathbb{R}$, where $h(x)=\sqrt{x}$. For this last example $h$, could it be "pointwise differentiable" at square numbers?

Are there any additional rules for whether a function is differentiable over $\mathbb{Q}$, if that's ever possible at all?

Samuel Han
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    Yes, the definition of derivative makes sense in any ordered field. Of course, there are times when (in the rational case) the derivative does not exist but (in the real case) the derivative is irrational. – GEdgar Jun 23 '23 at 11:04
  • Yes you can with any topological field, but the properties are not equivalent: $\mathbb{Q}$-differentiable is weaker than $\mathbb{R}$-differentiable (the usual) which is weaker than $\mathbb{C}$-differentiable (holomorphic). – julio_es_sui_glace Jun 23 '23 at 11:54
  • I would say: it does not matter that the $\mathbb{K}$-derivative is not in $\mathbb{K}$, so each of the function you mentioned are $\mathbb{Q}$-differentiable everywhere. – julio_es_sui_glace Jun 23 '23 at 11:56
  • @JulesBesson In that case you'd need to modify the limit definition to something like "for every sequence $h_n$ that converges to $0$, $[f(x + h_n) -f(x)]/h_n$ is Cauchy". – eyeballfrog Jun 23 '23 at 13:17
  • Yes indeed but working with derivative only in $\mathbb{Q}$ seems very restrictive.. – julio_es_sui_glace Jun 23 '23 at 13:23

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