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I'm struggling to understand the term.

"A function is continuously differentiable of order $k$ at $a$ if the $k$-th derivative of $f$ exists and is continuous on some open interval containing $a$."

$(1)$ Does this mean that a function is continuously differentiable if the $k$-th derivative exists AND $f^{'}$ is continuous on some open interval containing $a.$

OR

$ (2)$ Does this mean that a function is continuously differentiable if the $k$-th derivative exists AND $f$ is continuous on some open interval containing $a.$

So if I have to show if a function is continuously differentiable of order $2$ at $a$ I would need to show that $f''$ exists and $f'$ is continuous?

If anyone could provide me an concrete example with a simple one-variable function that would be appreciated.

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  • $\begingroup$ Keep in mind that if a function is differentiable then it's automatically continuous. $\endgroup$ – littleO Mar 31 '14 at 23:29
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It means that the k'th derivative of $f$ exists in an open interval of $a$ and that k'th derivative is a continuous function in that open interval.

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Example: If $k \geq 1$ is an integer, the function $$ f(x) = \begin{cases} 0 & \text{if $x < 0$,} \\ x^{k+1} & \text{if $x \geq 0$,}\end{cases} $$ is $k$ times, but not $(k+1)$ times, continuously differentiable.

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  • $\begingroup$ Sorry but could you elaborate on that to exactly why? Thanks. $\endgroup$ – Bobby Apr 1 '14 at 1:29
  • $\begingroup$ Let's call the function $f_k$ instead of $f$. The derivative $f_k'$ is proportional to $f_{k-1}$; calculating $f_k'(0)$ requires difference quotients (i.e., the definition), but is straightforward. Sketch the graphs of $f_0$ and $f_1$, then do induction on $k$. $\endgroup$ – Andrew D. Hwang Apr 1 '14 at 12:52

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