Say we have the function $f:(0,1)\cup\{2\}\rightarrow \mathbb R$ defined by $f(x)=5$.

Is $f$ differentiable or not differentiable (or neither) at $2$?

Is $f$ a differentiable function?

  • $\begingroup$ Differentiabilty at an isolated point is not defined. $\endgroup$ – Kavi Rama Murthy Feb 3 at 7:45
  • $\begingroup$ No. One can only take a limit over such $h\to0$ that $f(a+h)$ is defined. Since for an isolated point all such small enough $h$ must be zero the difference quotients $\frac{f(a+h)-f(a)}{h}$ are not defined for small $h$, and neither is the limit. $\endgroup$ – Conifold Feb 3 at 8:23

Differentiability is defined in terms of sequences of differences $h_1,h_2,h_3\dots$ that are all nonzero and yet converge to zero. Such a sequence does not exist in the given domain, as there does not exist a number $k$ in the domain where $|2-k|=0.2$ or $0.1$ or indeed any number smaller than $1$. Hence differentiability is not defined at isolated points.

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  • $\begingroup$ Is $f$ a differentiable function? $\endgroup$ – iqntt1s Feb 3 at 9:24
  • $\begingroup$ @iqntt1s Over $(0,1)$, yes. At $\{2\}$, no. $\endgroup$ – Parcly Taxel Feb 3 at 9:24

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