Suppose we have smooth real functions $f,g$ such that the Riemann–Stieltjes integral $\int_0^t f(s) dg(s)$ is defined for all $t>0$ and is smooth as a function of $t$.
Is their an analytic formula for $\frac{d}{dt}\int_0^t f(s) dg(s)$ in terms of $f$ and $g$ ?
Bonus: If so, is it also the case for the Lebesgue-Stieltjes integral and the stochastic integral ?
If $g$ is smooth (indeed continuously differentiable) then the following identity holds $$ \int_0^t f(s)\mathrm dg(s)=\int_0^tf(s)g'(s)\mathrm ds $$ where the integral on the RHS is a Riemann integral. Then apply the fundamental theorem of calculus (the integrand is certainly continuous) $$ \frac{d}{dt}\int_0^t f(s)\mathrm dg(s)=\frac{d}{dt}\int_0^tf(s)g'(s)\mathrm ds=f(t)g'(t) $$
Edit: The idea is as follows; if $0<x_0<\dots<x_n<t$ is a partition $$ \int_0^t f(s)\mathrm d g(s)\approx\sum_{i=1}^n f(x_*)(g(x_{i})-g(x_{i-1})) $$ for $x_{i-1}\leq x_*\leq x_{i}$but by the Mean value theorem $$ g(x_{i})-g(x_{i-1})=(x_i-x_{i-1})g'(c_i) $$ for $x_{i-1}<c<x_{i}$. In the limit as $\sup_{i}|x_{i}-x_{i-1}|\to 0$, we have $c_i=x_*$ and the usual definition of the Riemann integral.