Let $r$ a random process defined by :
$$dr_t=\theta(t)dt + \sigma dW_t$$
$\theta$ is deterministic in $t$ and $W$ a brownian motion.
I want to compute $\frac{d}{dt}\mathop{\mathbb{E}}\left[ e^{-\int_t^Tr(s)ds}|\mathscr{F}_t \right]$ by directly computing the derivative (I don't have any problem computing $\mathop{\mathbb{E}}\left[ e^{-\int_t^Tr(s)ds}|\mathscr{F}_t \right]$ first and then applying the operator $\frac{d}{dt}$ ).
What I did :
We know that $r_t$ is markovian so we can write $\mathop{\mathbb{E}}\left[ e^{-\int_t^Tr(s)ds}|\mathscr{F}_t \right] = \mathop{\mathbb{E}}\left[ e^{-\int_t^Tr(s)ds}|r_t \right]$
From there we can define $ q(t,r_t) :=\mathop{\mathbb{E}}\left[ e^{-\int_t^Tr(s)ds}|r_t \right]$
Now taking the derivative :
$$ \frac{d}{dt}q(t,r_t) = \frac{dq(t,r_t)}{dt} + \frac{dq(t,r_t)}{dr_t} \frac{dr_t}{dt} = r_tq(t,r_t)-(T-t)q(t,r_t)\frac{dr_t}{dt}$$
Now I'm stuck with the $\frac{dr_t}{dt}$ because when I write down the solution for $r_t$ :
$$r_t=r_0+\int_0^t\theta(s)ds+\sigma \int_0^tdW_s =r_0+\int_0^t\theta(s)ds+\sigma W_t $$
and I take $\frac{dr_t}{t}$ I get $dW_t$ which is infinity ...
Any help?