Extending to a holomorphic function Let $Z\subseteq \mathbb{C}\setminus \overline{\mathbb{D}}$ be countable and discrete (here $\mathbb{D}$ stands for the unit disc).
Consider a function $f\colon \mathbb{D}\cup Z\to \mathbb{C}$ such that
1) $f\upharpoonright \overline{\mathbb{D}}$ is continuous
2) $f\upharpoonright \mathbb{D}$ is holomorphic
3) if $|z_0|=1$ and $z_n\to z_0$, $z_n\in Z$ then $(f(z_n)-f(z_0))/(z_n-z_0)\to f^\prime(z_0)$
Can $f$ be extended to a holomorphic function on some domain containing $Z$?

As has already been mentioned, the answer is no. The identity theorem tells you that there is at most one way to extend $f$ analytically beyond the unit disk, which shows you that the answer must be negative (as per Henning's comment above). 
Another fact to observe in this context is that there are functions that are analytic on the unit disk and extend smoothly to the boundary, but do not extend analytically beyond the boundary anywhere. (Take a Riemann map to a Jordan domain whose boundary is smooth but not analytic anywhere.) 
However, even without any assumptions on derivatives, a continuous function as in your statement can be uniformly approximated arbitrarily closely by entire functions; this follows from Arakelyan's approximation theorem.
