If both series $\sum_{n=1}^{\infty} f_{\mathrm{n}}(x)$ and $\sum_{n=1}^{\infty} f_{\mathrm{n}}^{\prime}(x)$ converge uniformly on I and $f_{n}^{\prime}$ is continuous for every $n \in \mathbb{N}$, then the function $f(x) = \sum_{n=1}^{\infty}f_{n}(x)$ is differantiable on I and we have $f^{\prime}(x)=\sum_{n=1}^{\infty}f_{n}^{\prime}(x)$
My Work:
We want to prove that $f^{\prime}(x) =\displaystyle \sum_{n=1}^{\infty}f_{n}^{\prime}(x)$ in other words $\displaystyle\lim_{n \to \infty} f_{n}(x) = \displaystyle\sum_{n=1}^{\infty}f_{n}(x)$ we know that $\displaystyle\sum_{n=1}^{\infty}f_{n}(x)$ converge uniformly. Then $\displaystyle\lim_{n \to \infty}f_{n}(x)=f(x)$ exist. Also we can say that from the information given in question, $f_{n}$ is differantiable because $f_{n}^{\prime}$ is continuos.
And from above we can say that $\displaystyle\lim_{n \to \infty}f_{n}^{\prime}=f^{\prime}$ since $f_{n}^{\prime}$ is continuous. Therefore, that is, $f^{\prime}=\displaystyle\sum_{n=1}^{\infty}f_{n}^{\prime}$.
Does it sufficient to prove? I am not sure...
You said nothing about $I$. I will assume that it is an interval of $\Bbb R$.
For each $N\in\Bbb N$, let $F_N=\sum_{n=1}^Nf_n$. Then$$F_N'=\left(\sum_{n=1}^Nf_n\right)'=\sum_{n=1}^Nf_n'$$and the goal is to prove that $\lim_{N\to\infty}F_N$ is differentiable and that$$\left(\lim_{N\to\infty}F_N\right)'=\lim_{N\to\infty}F_N'.\tag1$$
Since each $F_N'$ is continuous, it is Riemann-integrable on any interval $[a,b]\subset I$. Fix $a\in I$. Then$$F_N(x)=F_N(a)+\int_a^xF_N'(t)\,\mathrm dt.$$Since $(F_N')_{N\in\Bbb N}$ converges uniformly on $I$, it converges uniformly on each subinterval of $I$, and therefore\begin{align}\lim_{N\to\infty}F_N(x)&=\lim_{N\to\infty}\left(F_N(a)+\int_a^xF_N'(t)\,\mathrm dt\right)\\&=\lim_{N\to\infty}F_N(a)+\lim_{N\to\infty}\int_a^xF_N'(t)\,\mathrm dt\\&=\lim_{N\to\infty}F_N(a)+\int_a^x\lim_{N\to\infty}F_N'(t)\,\mathrm dt,\end{align}since the convergence is uniform. Therefore, by the Fundamental Theorem of Calculus, $(1)$ holds.