Let $$dG_t=\mu(t)dt+\sigma(t)dW_t$$ where $W_t$ is standard Brownian motion and define $$f(t,u)=E[e^{-r(u-t)}(\mu(u)-rG_u)|F_t]$$ where $(F_t)_{t\in [0,T]}$ is a filtration, $T\ge u\ge t$ and $r$ is a constant. I want to find $d_tf(t,u)$.
Here is what I tried: \begin{align} d_tf(t,u)&=E[re^{-r(u-t)}(\mu(u)-rG_u)dt+e^{-r(u-t)}(0)|F_t]\\ &=re^{-r(u-t)}\mu(u)dt-r^2e^{-r(u-t)}E[G_u|F_t]dt\\ \end{align}
Now $G_u=G_t+\int_{t}^u\mu(s)ds+\int_{t}^u\sigma(s)dW_s$ Thus, \begin{align} d_tf(t,u)&=re^{-r(u-t)}\mu(u)dt-r^2e^{-r(u-t)}E[G_t+\int_{t}^u\mu(s)ds+\int_{t}^u\sigma(s)dW_s|F_t]dt\\ &=re^{-r(u-t)}\mu(u)dt-r^2e^{-r(u-t)}(G_tdt+E[\int_{t}^u\mu(s)ds+\int_{t}^u\sigma(s)dW_s|F_t]dt) \end{align} Since $E[\int_{t}^u\sigma(s)dW_s|F_t]=0$, $$d_tf(t,u)=re^{-r(u-t)}\mu(u)dt-r^2e^{-r(u-t)}(G_tdt+E[\int_{t}^u\mu(s)ds|F_t]dt)$$ Is this correct? Is it possible to simplify more(remove expectation or filtration)
I would say first compute $f(t,u)=E[e^{-r(u-t)}(\mu(u)-rG_u)|F_t]$ and then compute the derivative. Otherwise you will need justify the derivation under the conditional expectation.
Here we suppose that $\mu$ is in $L^1_{loc}(\mathbb{R})$ \begin{align} f(t,u) &= E[e^{-r(u-t)}(\mu(u)-rG_u)|F_t]\\ &= e^{-r(u-t)}\mu(u)- e^{-r(u-t)}r\left(G_t+\int_{t}^u\mu(s)ds+E[\int_{t}^u\sigma(s)dW_s|F_t]\right)\\ &= e^{-r(u-t)}\mu(u)- e^{-r(u-t)}r\left(G_t+\int_{t}^u\mu(s)ds\right)\\ \end{align} Therefore, $\partial_t f(t,u) = re^{-r(u-t)}\left[(\mu(u)-\mu(t))- \left(r(G_t+\int_t^u\mu(s)ds) +\partial_tG_t\right)\right]$.