Condition for pointwise convergence of $({f_n}')_{n\in\mathbb{N}}$ to $f'$ if $({f_n})_{n\in\mathbb{N}}$ converges pointwise to $f$ Problem

Let $(f_n)_{n\in\mathbb{N}}$ be a sequence of differentiable functions such that $f_n:A(\subseteq \mathbb{R})\to \mathbb{R}$ for all $n\in\mathbb{N}$. Let $(f_n)_{n\in\mathbb{N}}$ converges pointwise to a differentiable function $f$. If $({f_n}')_{n\in\mathbb{N}}$ also converges pointwise to $g$ and $g$ satisfies the intermediate value property prove or disprove that $f'=g$.

For the example given here (see page no. 63, Example 5.17) $g$ doesn't satisfy the intermediate value property and hence doesn't meet the condition of the problem. 
Can anyone help me in proving or disproving it?
 A: Hint: The function $h(x) =\int_0^x \sin (1/t)\,dt$ is differentiable everywhere, with $h'(x) = \sin (1/x)$ for $x\ne 0$ by the FTC, and $h'(0)=0$ because of all the oscillation at $0.$ (You have to do some work to see this last bit. ) Now you have mentioned an example of a sequence $f_n$ converging to $0$ uniformly on $\mathbb R,$ such that $f_n'$ converges pointwise to $g = \chi_{\{0\}}.$ For your problem here, consider the functions $f_n + h.$
A: Since $f_0,f_1,f_2,...f_n$ is continuous and differentiable $n$-times, we say it is of multiplicity $n$. 
We say $\alpha$ is a Zero (root) of multiplicity n $\ni f(\alpha)=f'(\alpha)= f^2(\alpha)=... f^{n-1}(\alpha)$
Since the sequence converges to $1$ limit point, $g \rightarrow f$ is a contracting function. 
By the intermediate value theorem, there exists an $\alpha \in [1,n] \ni \frac{f(1)-f(n)}{1-n}=f'(\alpha)$ 
$\frac{f(1)-f(n)}{1-n}$ converges to a limit point $g$.  
Hence $g=f'$. 
(I have used the theory of numerical solutions of non linear equations)
