I know that if $(f_n)$ is a sequence of differentiable functions on $(a,b)$ with pointwise limit $f$ and if $f'_n \rightarrow g$ uniformly the $f$ is differentiable on $(a,b)$ and $f' = g$.
I want to find an example in which $f'_n \rightarrow g$ pointwise only and $f' \ne g$ and $g$ is continuous (differentiable would be even better). The only example I can find has a discontinuous g, this example is $f_n(x) = \frac{x}{1+nx^2}$($g$ is the function that is $1$ at $0$ and $0$ everywhere else).
Consider $f_n :[0,1] \to \mathbb{R}$ where
$$f_n(x) = \int_0^x n^2t(1-t)^n\, dt = n^2\frac{(x-1)(nx+x+1)(1-x)^n + 1}{n^2 + 3n + 2}$$
We have $f'_n(x) = n^2x(1-x)^n \to g(x)\equiv 0$ as $n \to \infty$ such that $g$ is continuously differentiable.
However, as $n \to \infty$ we have
$$f_n(x) \to f(x) =\begin{cases}1,&0 < x \leqslant 1\\0, &x=0 \end{cases}$$
In this case $f'(0)\neq g(0) = 0$ since $f'(0)$ does not exist.