I tried to show the following estimation and am not sure, if it holds:
Given $U\subseteq\mathbb R$ and $f:\mathbb R\times \mathbb R\to\mathbb R$. Its derivative with respect to the second argument exists and is denoted by $f_b(a,b)$:
$\displaystyle\frac{d}{db}\left(\sup_{a\in U}f(a,b)\right)\ge \inf_{a\in U}f_b(a,b)$
My proof:
There exists a sequence $a_n$ in $U$ such that
$\displaystyle\lim_{n\to\infty}f(a_n,b)=\sup_{a\in U}f(a,b)$
$\forall n\in\mathbb N$ the following holds:
$\sup_{a\in U}f(a,b+h)-\sup_{a\in U}f(a,b)\ge f(a_n,b+h)-\sup_{a\in U}f(a,b)$
Using the differentiability of $f$:
$=f(a_n,b)+h f_b(a_n,b)-\sup_{a\in U}f(a,b)+\mathcal O(h^2)$
This is always greater than
$\displaystyle\ge f(a_n,b)+h\;\inf_{a\in U} f_b(a,b)-\sup_{a\in U}f(a,b)+\mathcal O(h^2)$
As this holds for an arbitrary $n$ and thus $f(a_n,n)$ comes arbitrarily close to $\sup_{a\in U}f(a,b)$ it holds:
$\sup_{a\in U}f(a,b+h)-\sup_{a\in U}f(a,b)\ge h\;\inf_{a\in U} f_b(a,b)+\mathcal O(h^2)$
Deviding by $h$ and performing the limit yields the assertion.