Let $X_t= X_0 + \int_0^t \mu(s,X_s)ds + \int_0^t \sigma(s,X_s)dW_s$ where $\mu$ and $\sigma$ are $C^1$ functions satisfying the usual growth restriction and $W_t$ is a $d$-dimensional Brownian motion. Let $f\in C^2(\mathbb{R}^d;\mathbb{R})$ and let $D_t$ denote the Malliavin derivative at time $t$; then how would one go about computing $$ D_tf(X_t) =? $$
So Far: I know that if $X_t= \int_0^t \sigma(s)dW_s$ then by definition 2.2 in these lectures by D. Nualart $$ D_tf(X_t) = \sigma_t^T\left[\frac{\partial f}{\partial x} (X_t)\right]... $$ but what about the general case?
