About differentiation under the product integral sign It is known that
$$\dfrac{d}{dx}\int_{a(x)}^{b(x)}f(x,t)~dt=\dfrac{db(x)}{dx}f(x,b(x))-\dfrac{da(x)}{dx}f(x,a(x))+\int_{a(x)}^{b(x)}\dfrac{\partial}{\partial x}f(x,t)~dt.$$
How about
$$\dfrac{d}{dx}\prod_{a(x)}^{b(x)}f(x,t)^{dt}?$$

Because we are in a commutative setting, the product integral satisfies
$$\prod_{a(x)}^{b(x)} f(x,t)^{dt}=\exp\left(\int_{a(x)}^{b(x)}\log f(x,t)\,dt\right).$$
Therefore with the chain rule and the rule you state, we derive
$$\frac{d}{dx}\log \left(\prod_{a(x)}^{b(x)} f(x,t)^{dt}\right)  = \frac{d}{dx}\int_{a(x)}^{b(x)}\log f(x,t)dt$$ $$ = b'(x)\log f(x,b(x))-a'(x)\log f(x,a(x))+\int_{a(x)}^{b(x)}\frac{\partial}{\partial x}\log f(x,t)dt$$
hence
$$\frac{d}{dx}\prod_{a(x)}^{b(x)} f(x,t)^{dt}=\left(\prod_{a(x)}^{b(x)} f(x,t)^{dt}\right)\left(\log \frac{f(x,b(x))^{b'(x)}}{f(x,a(x))^{a'(x)}}+\int_{a(x)}^{b(x)}\frac{\frac{\partial}{\partial x}f(x,t)}{f(x,t)}dt\right).$$
