Is a derivative a limit of a sequence of continuous functions? Let $f:(a,b) \rightarrow \mathbb R$ be differentiable. Is it the derivative $f'$ equal  the limit of a sequence of continuous functions? 
I know that is true when $f$ is defined on $[a,b]$, because we can  extend $f$ on the whole $\mathbb R$  by putting $f'(b)(x-b)+f(b)$ for $x>b$ and $f'(a)(x-a)+f(a)$ for $x<a$. Then 
$$
f'(x)=\lim_{n \rightarrow \infty} n [f(x+\frac{1}{n})-f(x)].
$$
 A: It is the limit of the continuous functions $$\left\{  \frac{f(x+h)-f(x)}{h} \right\}_{h>0}.$$
Is that what you were asking?
A: What about the following?
Let $f\colon I\to {\mathbb R}$, where $I=(a,b)$, and let $x_0\in (a,b)$ be an interior point of this interval. Then we can define a sequence of continuous functions such that
$$g_n(x)=
  \begin{cases}
    \frac{f(x+1/n)-f(x)}{1/n} & x\in I\setminus(x_0-1/n,x_0+1/n), x<x_0 \\
    \frac{f(x)-f(x-1/n)}{1/n} & x\in I\setminus(x_0-1/n,x_0+1/n), x>x_0\\
    f'(x_0) & x=x_0
  \end{cases}
$$
and for $x\in (x_0-1/n,x_0+1/n)\setminus\{x_0\}$ we define $f_n(x)$ as an arbitrary continuous extension of the above values. (For example, by a piecewise linear function.)
It is easy to see that $g_n$ converges to $f'$ pointwise.
(Note that for large enough $n$ the values $f(x+1/n)$, $f(x-1/n)$ are defined.)
Let me just say that the original question can be formulated as: Does a derivative (of a function defined on an open interval) belong to the first Baire class.
