Applying Ito's formula to $D^x(t)=(\partial/\partial x)S(t)$ Let $\gamma:[0,T]\times (0,\infty)\to \mathbb{R}$ be continuous, bounded above, $(\partial/\partial s [\gamma(t,s)])$ is continuous in $(s,t)$. Consider the following process $$dS_{\gamma}^x(t)=S^x_{\gamma}(t)[r(t)dt+\gamma(t,S^x_{\gamma}(t))dW(t)]$$ where $W(t)$ is a one dimensional Brownian motion and $r(t)$ is an adapted process and $S^x_{\gamma}(0)=x$. 
Now define the process $D^x(t)=(\partial/\partial x)S^x_{\gamma}(t)$. And this apparently satisfies $$dD^x(t)=D^x(t)\big[r(t)dt+\frac{\partial}{\partial s}\rho(t,S^x_{\gamma}(t))dW(t)\big]\rightarrow(1)$$ where $\rho(t,s):=s\gamma(t,s). $ I am not quite sure how equation (1) can be derived since $\gamma$ depends on $S^x_{\gamma}$. Maybe the idea is to apply Ito's formula to $D^x(t)$ by defining a function $f(t,s):=(\partial/\partial x)s $ and then the ito rule gives us $df=f_tdt+f_sdS^x_{\gamma}(t)+\frac{1}{2}f_{ss}dS^x_{\gamma}(t)dS^x_{\gamma}(t)$. I am not sure if what I am doing is even slightly correct. Any help is appreciated!
 A: For your Ito process $S_{\gamma}^x(t)$, we focus on its time-differentiation as we write ${\rm d}S_{\gamma}^x(t)$. Since $x$ is an auxiliary parameter independent from $t$, we have that ${\rm d}$ and $\partial/\partial x$ commute with each other, i.e.,
$$
\frac{\partial}{\partial x}{\rm d}S_{\gamma}^x(t)={\rm d}\left(\frac{\partial}{\partial x}S_{\gamma}^x(t)\right)={\rm d}D^x(t).
$$
This is not a magic; it is just as if
$$
\frac{\partial}{\partial x}\left(\frac{\partial}{\partial t}f(x,t)\right)=\frac{\partial}{\partial t}\left(\frac{\partial}{\partial x}f(x,t)\right)
$$
for some differentiable function $f=f(x,t)$.
Therefore, act $\partial/\partial x$ on both sides of
$$
{\rm d}S_{\gamma}^x(t)=S_{\gamma}^x(t)\left(r(t)\,{\rm d}t+\gamma(t,S_{\gamma}^x(t))\,{\rm d}W(t)\right),
$$
and we obtain
\begin{align}
{\rm d}D^x(t)&=\frac{\partial}{\partial x}{\rm d}S_{\gamma}^x(t)\\
&=\frac{\partial}{\partial x}\left(S_{\gamma}^x(t)\left(r(t)\,{\rm d}t+\gamma(t,S_{\gamma}^x(t))\,{\rm d}W(t)\right)\right)\\
&=\frac{\partial}{\partial x}S_{\gamma}^x(t)\left(r(t)\,{\rm d}t+\gamma(t,S_{\gamma}^x(t))\,{\rm d}W(t)\right)\\
&+S_{\gamma}^x(t)\frac{\partial}{\partial x}\left(r(t)\,{\rm d}t+\gamma(t,S_{\gamma}^x(t))\,{\rm d}W(t)\right)\\
&=D^x(t)\left(r(t)\,{\rm d}t+\gamma(t,S_{\gamma}^x(t))\,{\rm d}W(t)\right)\\
&+S_{\gamma}^x(t)\,\frac{\partial}{\partial x}\gamma(t,S_{\gamma}^x(t))\,{\rm d}W(t)\\
&=D^x(t)\left(r(t)\,{\rm d}t+\gamma(t,S_{\gamma}^x(t))\,{\rm d}W(t)\right)\\
&+S_{\gamma}^x(t)\,\frac{\partial\gamma}{\partial s}(t,S_{\gamma}^x(t))\frac{\partial}{\partial x}S_{\gamma}^x(t)\,{\rm d}W(t)\\
&=D^x(t)\left(r(t)\,{\rm d}t+\gamma(t,S_{\gamma}^x(t))\,{\rm d}W(t)\right)\\
&+S_{\gamma}^x(t)\,\frac{\partial\gamma}{\partial s}(t,S_{\gamma}^x(t))\,D^x(t)\,{\rm d}W(t)\\
&=D^x(t)\left(r(t)\,{\rm d}t+\left(\gamma(t,S_{\gamma}^x(t))+S_{\gamma}^x(t)\,\frac{\partial\gamma}{\partial s}(t,S_{\gamma}^x(t))\right){\rm d}W(t)\right).
\end{align}
Note that
\begin{align}
&\gamma(t,S_{\gamma}^x(t))+S_{\gamma}^x(t)\,\frac{\partial\gamma}{\partial s}(t,S_{\gamma}^x(t))\\
&=\left(\gamma(t,s)+s\frac{\partial\gamma}{\partial s}(t,s)\right)\bigg|_{s=S_{\gamma}^x(t)}\\
&=\frac{\partial\rho}{\partial s}(t,s)\bigg|_{s=S_{\gamma}^x(t)}\\
&=\frac{\partial\rho}{\partial s}(t,S_{\gamma}^x(t)).
\end{align}
Hence, we eventually obtain
$$
{\rm d}D^x(t)=D^x(t)\left(r(t)\,{\rm d}t+\frac{\partial\rho}{\partial s}(t,S_{\gamma}^x(t))\,{\rm d}W(t)\right),
$$
as is desired.
In the above derivation, we merely use the commutativity between ${\rm d}$ and $\partial/\partial x$; Ito's formula is not necessary here.
