Form of Ito's formula for Ito diffusions Suppose we have an Ito diffusion ($Z_{t, z}$) satisfying
$$
Z_{t, z} = z + \int_0^t b(Z_{s,z})ds + \int_0^t \sigma(Z_{s,z})dW_s
$$
where $z\in\mathbb{R}^d, b:\mathbb{R}^d\to\mathbb{R}^d, \sigma:\mathbb{R}^d \to \mathbb{R}^{d\times d}$ and $W_s$ is a $d$-dimensional Brownian motion.
According to this thesis (Eqn. 4.7), Ito's formula looks like this:
$$
f(t, Z_{t, z}) - f(s, Z_{s, z}) = \int_s^t \nabla_x f(r, Z_{r,z})dW_r + \int_s^t \left( \frac{1}{2}\Delta_x f(r, Z_{r,z}) 
 + \frac{\partial}{\partial t}f(r, Z_{r,z}) \right) dr
$$
where $\Delta_x$ is the Euclidean Laplace operator w.r.t. the second coordinate $x$.
I've never seen it in this form, so I wanted to write out the r.h.s. more explicity. How are $\nabla$ and $\Delta$ interpreted correctly here? I'm thinking
$$\nabla_xf(r, Z_{r, z}) = f'(r, Z_{r, z}) \cdot Z_{r, z}'$$
with $f'(r, Z_{r, z}) \in \mathbb{R}^{1\times d}, Z_{r, z} \in \mathbb{R}^{d\times d}$? But is $Z_{r, z}$ even differentiable? And how does it look like? From the same thesis, it seems that
$$
\nabla_xf(r, Z_{r, z}) = f'(r, Z_{r, z}) \cdot \sigma(Z_{r,x})?
$$
I'm having the same problems understanding $\Delta_x$. Any help is appreciated!
 A: The gradient and Laplacian only act on $f$.  The process $Z$ is composed with these functions.  In other words, $\nabla_{x} f(r,Z_{r,z})$ might be clearer if we instead wrote $(\nabla_{x} f)(r,Z_{r,z})$.  Similarly, $\Delta_{x} f(r,Z_{r,z})$ is $(\Delta_{x} f)(r,Z_{r,z})$.  Thus, there is no chain rule involved and no need for any derivatives of $Z$.
(It's like if I wrote $f'(t^{2})$ in single variable calculus.  $t^{2}$ is the argument of $f'$.  The notation is not meant to mean $\frac{d}{dt}\{f(t^{2})\}$.)
$\nabla_{x}f$ is the gradient of $f$ with respect to the $x$ variable.  Its values are vectors in $\mathbb{R}^{1 \times d}$.  $\Delta_{x} f$ is the Laplacian of $f$.  It is a real-valued function.
In (4.7) in the text, it looks like the author is restricting attention to a Brownian motion.  (Notice in the last two integrals on the right-hand side there is a $B_{r}$ in $f$ instead of $X_{r}$.  I don't exactly know what $X_{r}$ is supposed to be, but I suspect it should be $B_{r}$ throughout (4.7).)  In general, if $Z_{r,z}$ is the solution of the SDE
\begin{equation*}
dZ_{t,z} = b(Z_{t,z}) dt + \sigma(Z_{t,z}) dB_{t}, \quad Z_{0,z} = z,
\end{equation*}
then Ito's formula for the composition $t \mapsto f(t,Z_{t,z})$ gives, in this case,
\begin{align*}
df(t,Z_{t,z}) &= \nabla_{x}f(t,Z_{t,z}) d Z_{t,z} + \frac{1}{2} D^{2} f(t,Z_{t,z}) \cdot \langle Z_{\cdot,z} \rangle_{t} + \frac{\partial f}{\partial t}(t,Z_{t,z}) dt \\
&= b(Z_{t,z}) \cdot \nabla_{x} f(t,Z_{t,z}) dt + \sigma(Z_{t,z})^{*} \nabla_{x}f(t,Z_{t,z}) \cdot dB_{t} + \left( \frac{1}{2} \text{tr} (\sigma(Z_{t,z}) \sigma(Z_{t,z})^{*} D^{2}f(t,Z_{t,z})) + \frac{\partial f}{\partial t}(t,Z_{t,z}) \right) dt
\end{align*}
In integral form, this reads:
\begin{equation*}
f(t,Z_{t,z}) = f(0,z) + \int_{0}^{t} \left( b(t,Z_{t,z}) \cdot \nabla_{x}f(t,Z_{t,z}) + \frac{1}{2} \text{tr} (\sigma(t,Z_{t,z}) \sigma(t,Z_{t,z})^{*} D^{2}_{x}f(t,Z_{t,z})) + \frac{\partial f}{\partial t}(t,Z_{t,z}) \right) \, dt + \int_{0}^{t} \sigma(t,Z_{t,z})^{*} \nabla_{x}f(t,Z_{t,z}) \, dB_{t}.
\end{equation*}
