Consider the SDE \begin{equation}\tag{1} \mathop{dS_t}=S_t\left(r\mathop{dt}+\sigma(t,S_t)\mathop{dW_t}\right), \end{equation}
Suppose that $\int_0^{t} X_s\mathop{dW_s}=\ln S_t-\ln S_0$ and $\mathop{dX_t}=a_t\mathop{dt}+b_t\mathop{dW_t}$. Then \begin{equation} X_s=X_0+\int_0^{s} a_u\mathop{du}+\int_0^{s} b_u\mathop{dW_u}, \end{equation} so that \begin{equation} \int_0^{t} X_s\mathop{dW_s}=\int_0^{t} X_0 \mathop{dW_s} + \int_0^{t}\int_0^{s} a_u\mathop{du}\mathop{dW_s}+\int_0^{t}\int_0^{s} b_u\mathop{dW_u}\mathop{dW_s}. \end{equation} Applying Itô's lemma to $(1)$ and letting $\sigma_t:=\sigma(t,S_t)$ for convenience, \begin{equation} \ln S_t=\ln S_0+\int_0^{t}\left(r-\frac{\sigma_s^2}{2}\right)\mathop{ds}+\int_0^{t}\sigma_s\mathop{dW_s}. \end{equation} Hence, \begin{equation} rt-\frac{1}{2}\int_0^{t}\sigma_s^2\mathop{ds}+\int_0^{t}\sigma_s\mathop{dW_s}=X_0 W_t + \int_0^{t}\int_0^{s} a_u\mathop{du}\mathop{dW_s}+\int_0^{t}\int_0^{s} b_u\mathop{dW_u}\mathop{dW_s}. \end{equation} Is there any way to solve for $a_t$ and $b_t$ in terms of $\sigma_t$ and $r$? My intention here is to describe the evolution of the derivative of returns $d(\ln S_t)$ with respect to Brownian motion $\mathop{dW_t}$ (since the time derivative is undefined). I would expect that this implicitly involves taking the second derivative of $S_t$ w.r.t. $\mathop{dW_t}$ and I am unsure of whether this makes sense rigorously.