Let $\mathcal{F}$ be a family of analytic functions in the open unit disk $\mathbb{D}$. Let $\mathcal{F}$'= {$f' : f \in \mathcal{F}$}. Suppose $\mathcal{F}'$ is normal and that $\sup \{ |f(0)| : f \in \mathcal{F} \}$ is finite. Show that $\mathcal{F}$ is normal.
I'm not sure how to go about solving this problem.
I know that given a sequence $(f_n)$ in $\mathcal{F}$, there will be a subsequence $(f_{n_k})$ such that $(f_{n_k}')$ converges to an analytic function $g$ on $\mathbb{D}$. Since $\mathbb{D}$ is simply connected, $g$ will have a primitive on $\mathbb{D}$, which gives a candidate for the limit of a convergent subsequence of $(f_n)$ but I'm not sure where to go from here or how to use that $\sup \{ |f(0)| : f \in \mathcal{F} \}$ is finite. I've also tried using Cauchy's Integral Formula, which gives for all $f \in \mathcal{F}$
$$f(0) = \frac{1}{2 \pi i} \int_{|z| = 1} \frac{f(z)}{z} dz, $$ and hence the supremum over $\mathcal{F}$ of the integrals on the right hand side is finite, but again I'm not sure how to use that. Any help would be appreciated!