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Is the function $x^3$ considered three times differntiable? And what would be the $\mathcal{C}$, notation for it. I am not sure whether it is $\mathcal{C}^3$ or $\mathcal{C}^4$. Because once I have differentiated $x^3$ three times I got $6$. Of Course I can differentiate $6$ once again, but with the same Argument I then could say I can differntiate $0$ one more time. This would mean that $x^3$ is infinitely times differentiable.

If that is the case can someone give me a function which is actually only $n$-times differntiable?

$\mathcal{C}^n$ means if I am not wrong that the function is n- times differentiable and the $n$-th Derivation is continious.

I would like to see an example of such a function please.

RM777
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  • The function $x^3$ is differentiable infinitely many times. Indeed. If you can find a function $f$ that is continuous but not differentiable. Then you can take $F$ to be it's anti-derivative. In this case $F$ is differentiable and $F'=f$ is not. So for instance you can take $f(x)=|x|$. – Yanko Feb 09 '19 at 12:05
  • The function $x^3$ is $\mathcal C^n$ for each $n$. There is no rule which says the function can be in $\mathcal C^n$ only for one choice of $n$ or anything like that. – Wojowu Feb 09 '19 at 12:07
  • https://math.stackexchange.com/questions/78825/example-of-a-function-that-is-not-twice-differentiable – Yanko Feb 09 '19 at 12:08
  • https://math.stackexchange.com/questions/1414729/only-once-differentiable – Yanko Feb 09 '19 at 12:08

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Consider $f(x)=|x|x^3$. Then $f'(x)=4|x|x^2$, $f''(x)=12|x|x$, $f'''(x)=24|x|$, and this is no longer differentiable at $0$.

Hagen von Eitzen
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    Can you just explain how you calculated the first derivative? Haven't you apllied the product rule? Which says $f^{'}(x)=|x|'x^{3}+3x^2|x|$? What is the derivative of $|x|$? and how do you proceed to get the result $4|x|x^2$? – RM777 Feb 13 '19 at 14:11