I like to know a rule to derivate this:
$$ g(x) = \sum_{t=a(x)}^{b(x)} f(x,t)$$ $$ a(x) < b(x) $$ $$ a(x), b(x) \in \mathbb{Z} $$ I already tried the chain rule (like in the Leibniz integral rule):
$$ h(a(x),b(x),x) = \sum_{t=a(x)}^{b(x)} f(x,t) $$
$$\frac{d}{dx}h = \frac{\partial h}{\partial a} \frac{da}{dx}+\frac{\partial h}{\partial b} \frac{db}{dx}+\sum_{t=a(x)}^{b(x)} \frac{\partial}{\partial x}f(x,t)$$
Thanks for reading!
As $a(x)$ and $b(x)$ are both differentiable (which I think is reasonable to assume in this context), they are continuous. If the domain of $a$ (similarly for $b$ as well) is some connected open set of $\mathbb{R}$, or in particular $\mathbb{R}$ itself, then we get that the image is also connected, and as the image is a subset of integers, it should be constant.
Thus we have that both $a$ and $b$ are constants, and can be treated as such while differentiating.
ie, $$ g'(x) = \sum_{t = a}^b \frac{ \partial }{\partial x} f(x,t)$$