I was reading this paper on derivatives of the hypergeometric function $F(a,b;c;x)$ w.r.t. the parameters $a$, $b$, and $c$. In the paper the authors simply state without justification $$ \tag{1} \frac{\mathrm d^n}{\mathrm dc^n}F(a,b;c;x)=\sum_{k=1}^\infty\frac{\mathrm d^n}{\mathrm dc^n}\frac{(a)_k(b)_k}{(c)_k}\frac{x^k}{k!}. $$ Now I am willing to accept that this is just a formal representation of the derivatives; however, from a rigorous point of view this certain does not hold in general. For example, we know the series expansion for the hypergeometric function diverges outside the unit disk (with the exception of when $a$ or $b$ are nonpositive integers). So for $|x|>1$, $(1)$ certainly does not hold. After a little digging I found that we can differentiate functional series so long as we can justify a few criteria. In particular one needs to justify uniform convergence of the differentiated series.
So my question is this: Let $x\in[0,1]$, $a,b\in\Bbb R$, and $c>0$ with $c-a-b>0$ so that the hypergeometric series converges absolutely for all $x\in[0,1]$. How can we rigorously justify the differentiation of the hypergeometric series: $$ \frac{\mathrm d}{\mathrm dc}F(a,b;c;x)=\sum_{k=1}^\infty\frac{\mathrm d}{\mathrm dc}\frac{(a)_k(b)_k}{(c)_k}\frac{x^k}{k!}? $$
Note that $$ \sum_{k=1}^\infty\frac{\mathrm d}{\mathrm dc}\frac{(a)_k(b)_k}{(c)_k}\frac{x^k}{k!}=-\sum_{k=1}^\infty\underbrace{\frac{(a)_k(b)_k}{(c)_k}(\psi(c+k)-\psi(c))\frac{x^k}{k!}}_{f^\prime(c)}, $$ where $\psi(z)$ is the digamma function. So we need to show that $\sum_{k=1}^\infty f^\prime(c)$ converges uniformly for all $c>0$.