I have been trying to understand SDE notation and what it means but I cant seem to figure it out. Im sorry to post this question here, since this is probably elementary.
It is known to me that
$$ dX_t = f(X_t)dt + g(X_t)dW_t$$ basically means $\int_0 ^t d X_s = X_t - X_0 = \int_0 ^t f(X_t)dt + \int_0 ^t g(X_t)dW_t$. So basically one just "leaves off" the integrals. I've now come across a problem where the situation is the following:
If $T$ is some finite stopping time, let $W^T _t := W_{T \wedge t}$ be a stopped Brownian Motion. For some time $T < t'$ consider the process $f(W^T _t)$, where $f$ is smooth. Show that $df(W^T _{t'}) = 0$. Intuitively this makes sense to me, since "the differential" (or whatever this quantity is called) should measure small changes and this process is constant for big enough $t'$. But I have no idea how to rigorously proof this, since I don't know how to manipulate $df(W^T _t)$ based on the "definition" from above. The only thing that comes to mind would be Itôs Lemma, but then I get a term like $\int_0 ^{t'} df(W^T _{s}) = W_T - W_0$, which, again, I dont know how to recover $df(W)$ from.
In more generality, I also came across something like
$$dX_t = f(X_t)dt + g(X_t)dW_t + dh(X_T).$$
How can I make sense if this? Again, I would assume that, because the third term does not depend on $t$ (but rather on $T$) that it just vanishes. But this is obviously not a rigorous argument and therefor Im really confused. Any help is greatly appreciated!