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First of all, my understanding of the term "continuously differentiable" is that you can continue to take derivatives of your function ad infinitum. Does this sound right?

The function in question is the following:

$f(x) = (2x-1)^{\frac{1}{3}}$

The function $f$ is differentiable on all of $\mathbb{R}$ but its derivative is not, since $f'(x) = \frac{2}{3}(2x-1)^{-\frac{2}{3}}$ is not differentiable at $x = \frac{1}{2}$. So would I conclude that $f$ is not cont. diff. or that $f$ is cont. diff. on $\mathbb{R} \setminus \{\frac{1}{2}\}$?

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  • $\begingroup$ No, "continuously differentiable" means that function is differentiable and derivative is continuous. $\endgroup$ – Zoran Loncarevic Jan 29 '17 at 17:51
  • $\begingroup$ An addendum: the term "smooth" is usually used to refer to infinitely differentiable (symbolically $C^{\infty}$) functions. $\endgroup$ – rubikscube09 Apr 29 '18 at 17:20
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Usually "continuously differentiable" means that the first derivative of the function is differentiable, not that the function is infinitely differentiable. Since the function $f'$ exists everywhere, but is not continuous everywhere, we would say that $f$ is differentiable, but not continuously differentiable (on $\mathbb{R}$). But it is continuously differentiable on $\mathbb{R} \setminus \{\frac{1}{2}\}$, since the derivative is continuous on that set.

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“you can continue to take derivatives of your function ad infinitum. Does this sound right?”

No, this is called “infinitely differentiable.”

The term "continuously differentiable" means that the function is differentiable, and in addition, its derivative is also continuous. But continuity does not imply differentiability, so the first derivative might not be differentiable again.

For example, take the derivative of the continuously differentiable function:

$f(x) = \frac{1}{2}x|x|$

and you'll get

$f’(x) = |x|$

which is continuous everywhere but not differentiable at $x=0$.

“Continuously differentiable” is only mildly stronger than "differentiable" and is not as strong as "twice differentiable."

Meanwhile, the function that you describe:

$f(x) = (2x-1)^{\frac{1}{3}},$

is NOT differentiable for “all real numbers” because there is a problem at $x = \frac{1}{2}$. The derivative $f'(x) = \frac{2}{3}(2x-1)^{-\frac{2}{3}}$ is undefined at $x = \frac{1}{2}$, because you can’t divide by zero.

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