If $\abs{\partial_t f(x,t)}\le g(x)$ with $g\in L^1$, then $F(t)=\int f(x,t)\,d\mu$ is differentiable with $F^\prime=\int\partial_t f\,d\mu$.Suggest a slogan
Let \((X,\mcm,\mu)\) be a measure space, let \(I\subseteq\RR\) be an open interval, and let \(f\colon X\times I\to\CC\) be such that \(x\mapsto f(x,t)\) is integrable for each \(t\in I\). Put \(F(t)\coloneqq\int_X f(x,t)\dmu(x)\). Suppose \(\frac{\partial f}{\partial t}(x,t)\) exists for all \(x\in X\) and \(t\in I\), and there is \(g\in L^1(\mu)\) with \(\abs{\frac{\partial f}{\partial t}(x,t)}\leq g(x)\) for all \(x\in X\) and \(t\in I\). Then \(F\) is differentiable on \(I\) and \[ F'(t) = \int_X \frac{\partial f}{\partial t}(x, t) \dmu(x) \] [@Fol13].