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I'm hoping someone can clarify this because I can't seem to find a solid answer.

I know functions can be continuously differentiable, but I just read in a textbook "this applies to once continuously differentiable..." but they don't give an example and google seems unhelpful.

Am I right in assuming that once continuously differentiable is equivalent to continuously differentiable?

If not can you please provide an example that clearly demonstrates the difference.

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    $\begingroup$ Yes, what this is emphasising is that one needs not assume the function is twice continuously differentiable, that is there is no need to assume its derivative also has a continuous derivative. $\endgroup$ – Angina Seng Mar 6 '18 at 10:56
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Another simple example

$$x\mapsto \cases{0 & if $x < 0$\\x^2 & otherwise}$$

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  • $\begingroup$ I think the discontinuity in the second derivative is clearer in this much simpler example. Clearly shows it's once continuously differentiable. Thank you =) $\endgroup$ – Kendall Mar 6 '18 at 11:26
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    $\begingroup$ Yes this is simpler to see and assuming $x^n$ we can find examples for $n-1$ times continuously differentiable. You are welcome! Bye $\endgroup$ – user Mar 6 '18 at 11:28
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Once continuously differentiable is indeed equivalent to continuously differentiable, but it emphasis the point that the function may not be more than once continuously differentiable. For example : $$x\mapsto \cases{0 & if $x = 0$\\x^3\sin\left(\frac{1}{x}\right) & otherwise} $$ is exactly one time continuously differentiable.

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  • $\begingroup$ This is due to the second derivative correct? It's undefined at x=0 and thus by having it 0 at x=0 you get the discontinuity on the second derivative? $\endgroup$ – Kendall Mar 6 '18 at 11:07
  • $\begingroup$ @KennyB : if you compute the derivative of it over $\mathbb{R}\setminus \{0\}$ you will have $2x\sin(1/x)-x \cos(1/x)$ which you can show it is the derivative over $\mathbb{R}$ by taking $0$ at $x=0$, but if you derivate again, you will have terms like $\sin(1/x)$ alone and it can't be continuous at $0$. $\endgroup$ – Netchaiev Mar 6 '18 at 12:43

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