Analogue of Leibniz Rule for Stochastic Integrals

I am probably very late in answering this, but maybe the proof is useful for someone like me coming just now to this question.

Using the integral form of $f$: \begin{align*} Y_t = & - \int_t^T \left[ f(0, u) + \int_0^t \alpha(s, u)ds + \int_0^t \sigma(s, u) dw_s \right] du \,, \end{align*} Using Fubini's theorem (twice): \begin{align*} Y_t = & - \int_t^T f(0, u) du - \int_0^t \int_t^T \alpha(s, u) du ds - \int_0^t \int_t^T \sigma(s, u) du dw_s \,, \\ = & - \int_t^T f(0, u) du - \int_0^t \int_s^T \alpha(s, u) du ds - \int_0^t \int_s^T \sigma(s, u) du dw_s \,, \\ & + \int_0^t \int_s^t \alpha(s, u) du ds + \int_0^t \int_s^t \sigma(s, u) du dw_s \,, \\ = & - \int_0^T f(0, u) du - \int_0^t \int_s^T \alpha(s, u) du ds - \int_0^t \int_s^T \sigma(s, u) du dw_s \,, \\ & + \int_0^t f(0, u) du + \int_0^t \int_0^u \alpha(s, u) ds du + \int_0^t \int_0^u \sigma(s, u) dw_s du \,, \end{align*}

Given that $r(u) = f(u, u)$ and using the definition for $Y_0$ we have that: \begin{align*} Y_t = & Y_0 - \int_0^t \int_s^T \alpha(s, u) du\,ds - \int_0^t \int_s^T \sigma(s, u) du\,dw_s + \int_0^t r(u) du \,. \end{align*} Which is the desired result in the integral form.