I am new to Ito Process, so I have a following question. Consider a standard Ito Process, $$X_t=X_0+\int_0^t\mu_sds+\int_0^t\sigma_sdW_s$$ where W is the m-dimentional Brownian motion and X is a n-dimentional process. $\mu$ and $\sigma$ are adaptive to {$\Sigma_t$} generated by the Brownian motion. In somewhere in my textbook, it says the drift and the variance can be derived as following, $$\frac{d}{ds} E(X_s|\Sigma_t)|_{s=t}= \mu_t $$ and $$\frac{d}{ds} Var(X_s|\Sigma_t)|_{s=t}= \sigma_t\sigma_t^T$$ I thought this is trivial and attempted to prove it myself, however it took me nearly an hour, and can't even show the first one..
I tried the following way, $$\frac{d}{ds} E(X_s|\Sigma_t)|_{s=t}= \frac{d}{ds} E(X_0+\int_0^s\mu_sds+\int_0^s\sigma_sdW_s|\Sigma_t)|_{s=t}$$$$=\frac{d}{ds} E(X_0|\Sigma_t)+\frac{d}{ds}E(\int_0^s\mu_sds)+\frac{d}{ds}E(\int_0^s\sigma_sdW_s|\Sigma_t))$$
Then I feel that the last term should be zero by the independent increment property, but not 100% sure I can use here, inside the integral there is $\sigma_s$, then I don't know how to proceed from here and get the above two equation.
Any help will be extremely appreciated!!
Since
$$(t,\omega) \mapsto \int_0^t \sigma_r \, dW_r$$
is a martingale, we have for $s \geq t$
$$\begin{align*} \mathbb{E}(X_s \mid \Sigma_t) &= \mathbb{E}(X_0 \mid \Sigma_t) + \mathbb{E} \left( \int_0^s \mu_r \, dr \mid \Sigma_t \right) + \int_0^t \sigma_r \, dW_r \end{align*}$$
Using Fubini's theorem it's not difficult to show that
$$\mathbb{E} \left( \int_0^s \mu_r \, dr \mid \Sigma_t \right) = \int_0^s \mathbb{E}(\mu_r \mid \Sigma_t) \, dr$$
Thus,
$$\frac{d}{ds} \mathbb{E}(X_s \mid \Sigma_t) = \frac{d}{ds} \int_0^s \mathbb{E}(\mu_r \mid \Sigma_t) \, dr$$
Assuming that $\mathbb{E}(\mu_r \mid \Sigma_t)$ is a sufficiently nice function, we obtain by the fundamental theorem of calculus
$$\frac{d}{ds} \mathbb{E}(X_s \mid \Sigma_t) \bigg|_{t=s} = \mathbb{E}(\mu_s \mid \Sigma_t) \bigg|_{t=s} = \mu_t$$
(The last equality sign holds since $\mu$ is $\Sigma_t$-adapted.)
The proof of the second equality is rather similar. As @TheBridge already mentioned, it's based on Itô's isometry.