If $f(t)$ is a deterministic function of $t$ and $B_{n}$ is a brownian motion and:
$Z =\displaystyle\int^t_0 f(s)d\left(B(s)\right)$
How does one take the partial derivatives wrt to $t$ and $B_n$ on an integral like this?
I know $dZ = f(t)dB(t)$
Is this just?...
$\dfrac{\partial z}{\partial t} = f(t)$
and
$\dfrac{\partial z}{\partial B} = f(t)dB(t)$
Looking to apply the Ito formula on a bigger problem but stuck on this. Thanks.