Derivative of Function of Brownian motion I have a smooth function that takes as input a Brownian motion $B_t$.  My question is how does one find the time derivative of the expectation?  In other words, how do you calculate 
$\frac{d}{dt} \mathbb{E} f(B_t)$.  
 A: We know that: $B_t \sim N(0,t)$
Thus:
$$\mathbb{E}[f(B_t)] = \dfrac{1}{\sqrt{2\pi t}}\int_{-\infty}^\infty f(x)e^{\frac{-x^2} {2t}}dx$$
To calculate $\dfrac{d}{dt}\mathbb{E}[f(B_t)]$, use the following property: 
$$\dfrac{d}{dt}\left(\int_{a(t)}^{b(t)}g(x,t) dx\right) = \int_{a(t)}^{b(t)}\dfrac{\partial g(x,t)}{\partial t} dx+ b'(t)g(b(t),t)-a'(t)g(a(t),t)$$
Therefore,
$$\begin{align}
\dfrac{d}{dt}\mathbb{E}[f(B_t)] &= \dfrac{d}{dt}\left(\dfrac{1}{\sqrt{2\pi t}}\int_{-\infty}^\infty f(x)e^{\frac{-x^2} {2t}}dx\right)\\
&= -\dfrac{1}{2t\sqrt{2\pi t}}\int_{-\infty}^\infty f(x)e^{\frac{-x^2} {2t}}dx+\dfrac{1}{2t^2\sqrt{2\pi t}}\int_{-\infty}^\infty x^2f(x)e^{\frac{-x^2} {2t}}dx
\end{align}$$
A: $\mathbb{E}[ f(B_t) ] = \frac{1}{ \sqrt{ 2 \pi t}} \int_{\mathbb{R}} f(x) e^{-x^2/{2t}} dx =: g(t)$
Note how all that matters is the pdf at time $t$.
You can now differentiate $g(t)$ using product rule + under the integral sign.
It is definately not the same thing as $\mathbb{E} [ (d/dt) f(B_t)]$. As you point out, this latter expression doesn't make sense.
