Continuity of Derivative at a point.

Solution 1:

This is not exactly an answer to your question, but I think the source of your confusion is that you seem to believe that the left/right hand derivatives $$f'_\pm(a)=\lim_{h\to 0^\pm} \frac{f(a+h)-f(a)}{h}$$ are the same things as the left/right hand limits of the derivative $$\lim_{h\to 0^\pm} f'(a+h).$$ They coincide in simple cases, but not in general. For example, if $$f(x)=\begin{cases}1,&x \ge 0 \\ 0,&x < 0\end{cases}$$ then $f'(x)=0$ for all $x\neq 0$, so $\lim_{x\to 0^\pm} f'(x)=0$, but $f'(0)$ doesn't exist (since $f$ is discontinuous at $x=0$). More precisely, the right hand derivative $f'_+(0)$ is zero, but the left hand derivative ${f}'_{-}(0)$ is undefined.

Solution 2:

Yes, consider $f(x) = \begin{cases} x^2 & x \in \mathbb{Q} \\ 0 & x \in \mathbb{R} \setminus \mathbb{Q} \end{cases}$

$f$ is differentiable at $0$ and nowhere else.

Solution 3:

Yes, it's possible! Consider the function

$$f(x) = x^2 W(x)$$

where $W$ is the Weierstrass function. At $x=0$ the derivative is $0$, but if it were differentiable anywhere else then $W$ would be differentiable too.

Solution 4:

You ask about existence of a derivative in a single point, but in title you say continuity.

As for existence, a derivative $f'(a)$ of a real function $f(x)$ at point $x=a$ is defined as a limit $$\lim_{h\to 0} \frac{f(a+h)-f(a)}h$$ The existence (and the value) of the limit determines a derivative at the chosen point, independent on the existence of the limit in any neighborhood of $a$. As others show, there exist functions which are differentiable at a single point only.

However if you ask for continuity, it requires the derivative to be defined (exist) in some neighbourhood of $a$, so that a limit of the derivative exists: $$\lim_{x\to a}f'(x)$$ Then you can ask if a derivative is continuous at $a$. And there are functions (examples given in other answers) with a derivative discontinuous at some point, although existing in a neighborhood of that point.