Partial Derivative of an Integral

If $f(t)$ is a deterministic function of $t$ and $B_{n}$ is a brownian motion and:

$Z =\int^t_0 f(s)dB(s)$

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?...

$\frac{\partial z}{\partial t} = f(t)$

and

$\frac{\partial z}{\partial B} = f(t)dB(t)$

Looking to apply the Ito formula on a bigger problem but stuck on this. Thanks.

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You're right I think. I was trying to work out a characteristic function for this and if you set dt to zero it computes. Thanks. FYI ${\partial z \over \partial B} = f(t)$ I believe... – Dirk Calloway Nov 27 '12 at 12:32
Do you know if/why my answer is incorrect (if yours is correct)? I know that it's false to say that "you can't write partial derivatives w.r.t. BM". For example apply Ito's lemma to the function $f(t,x)$, where $x$ is $B(t)$ (in fact this is one of the most standard routines in stochastic finance). Applying Ito's lemma you will end up having to evaluate $\frac{\partial f}{\partial x}(t,x)$. – Jase Dec 4 '12 at 7:27