Additional hypotheses needed for the converse of $\mathscr F \subset H(G)$ normal $\implies$ $\mathscr F ' = \{f' : f \in \mathscr F\}$ normal? Problem

Show that if $\mathscr F \subset H(G)$ is normal then $\mathscr F' = \{f' : f \in \mathscr F\}$ is also normal. Is the converse true? Can you add something to the hypothesis that $\mathscr F'$ is normal to insure that $\mathscr F$ is normal?

$H(G)$ is the set of holomorphic functions $G \to \mathbb C$, where $G \subset \mathbb C$ is a domain.
Progress
I think I was able to prove the forward implication using this previous result:
Lemma: $f_n \to f$ in $H(G)$ if and only if $f_n(z) \to f(z)$ uniformly for $z \in \gamma$ for every closed rectifiable $\gamma \subset G$.
The definition of normality is that every sequence in $\mathscr F$ has a convergent subsequence in $H(G)$. By the lemma, the convergence of this subsequence $f_n$ can be viewed as uniform convergence on curves $\gamma$, in which setting it is clear that $f_n' \to f'$. Using the other direction of the lemma then gives that $\mathscr F '$ is normal.
Question
For the converse, is the only additional assumption needed an explicit mention that all the elements of $\mathscr F'$ indeed have primitives?
 A: I'm not sure if this is the purest form of the argument, but I'll settle for an equivalence. Clearly, if $\mathcal{F}$ is normal, then for every sequence $(f_n)\subset\mathcal{F}$ there exists $z_0\in G$ and a subsequence $(f_{n_k})$ such that $(f_n(z_0))$ is convergent.
On the other hand, if there exists $z_0\in G$ such that $(f_n(z_0))$ is convergent, and $(f_n^\prime)$ is uniformly convergent on compact sets, define $g_n\in H(G)$ by $g_n = f_n - f_n(z_0)$. Since for all $n$ we have $g_n(z_0)=0$, for every $z\in G$ and curve $\gamma$ connecting $z_0$ and $z$ it holds that
$$g_n(z) = \int_\gamma g_n^\prime(w)dw = \int_\gamma f_n^\prime(w)dw.$$
Uniform convergence of $(f_n^\prime)$ on compact sets therefore implies uniform convergence of the $g_n$'s, and finally the convergence of $(f_n(z_0))$ implies that $(f_n)$ is also uniformly convergent on compact sets.
Hence:

Provided that $\mathcal{F}^\prime$ is normal, then $\mathcal{F}$ is normal if and only if for every sequence $(f_n)\subset\mathcal{F}$ there exists $z_0\in G$ and a subsequence $(f_{n_k})$  such that $(f_{n_k}(z_0))$ converges.

