Background
Suppose we have a state $\mathbf{x}(t) \in \mathbb{R}^{d \times 1}$ evolving according to \begin{equation} \mathrm{d}\mathbf{x}(t) = \mathbf{F}(\mathbf{x}(t)) \mathrm{d}t + \mathbf{G}(\mathbf{x}(t)) \mathrm{d}\mathbf{w}(t), \end{equation} where $\mathbf{F}: \mathbb{R}^{d \times 1} \to \mathbb{R}^{d \times 1}$ is a Lipschitz continuous drift coefficient and $\mathbf{G}: \mathbb{R}^{d \times 1} \to \mathbb{R}^{d \times d}$ is a Lipschitz continuous diffusion coefficient. $\mathbf{w}(t)$ is the standard Weiner process.
Now, suppose we have an "energy function" $E(\mathbf{x}): \mathbb{R}^d \to \mathbb{R}_{\geq 0}$ that (according to Ito's Lemma) evolves with the SDE of $\mathbf{x}(t)$ as $$ \mathrm{d}E(\mathbf{x}) = \nabla_\mathbf{x} E(\mathbf{x})^\top \mathbf{F}(\mathbf{x}) \mathrm{d}t + \frac{1}{2} \mathrm{Tr}\left( \mathbf{G}(\mathbf{x}) \mathbf{G}(\mathbf{x})^\top \nabla_\mathbf{x}^2 E(\mathbf{x}) \right) \mathrm{d}t + \nabla_\mathbf{x} E(\mathbf{x})^\top \mathbf{G}(\mathbf{x}) \mathrm{d}\mathbf{w}(t), $$ which is essentially the "velocity" of the energy function.
Question
How can I compute the "acceleration" of the energy function? Would it be an SDE equal to $\mathrm{d}E/\mathrm{d}t$? Or, would it be equal to $\mathrm{d}^2E$? I guess Ito's lemma would need to be applied again.
Could I apply Ito's lemma to the drift and diffusion coefficients of $\mathrm{d}E$, i.e. by letting \begin{align} A(\mathbf{x}) &= \nabla_\mathbf{x} E(\mathbf{x})^\top \mathbf{F}(\mathbf{x}) \mathrm{d}t + \frac{1}{2} \mathrm{Tr}\left( \mathbf{G}(\mathbf{x}) \mathbf{G}(\mathbf{x})^\top \nabla_\mathbf{x}^2 E(\mathbf{x}) \right) \mathrm{d}t \\ B(\mathbf{x}) &= \nabla_\mathbf{x} E(\mathbf{x})^\top \mathbf{G}(\mathbf{x}) \end{align} and computing \begin{align} \mathrm{d}A(\mathbf{x}) =& \nabla_\mathbf{x} A(\mathbf{x})^\top \mathbf{F}(\mathbf{x}) \mathrm{d}t + \frac{1}{2} \mathrm{Tr}\left( \mathbf{G}(\mathbf{x}) \mathbf{G}(\mathbf{x})^\top \nabla_\mathbf{x}^2 A(\mathbf{x}) \right) \mathrm{d}t + \nabla A(\mathbf{x})^\top \mathbf{G}(\mathbf{x}) \mathrm{d}\mathbf{w}(t) \\ \mathrm{d}B(\mathbf{x}) =& \nabla_\mathbf{x} B(\mathbf{x})^\top \mathbf{F}(\mathbf{x}) \mathrm{d}t + \frac{1}{2} \mathrm{Tr}\left( \mathbf{G}(\mathbf{x}) \mathbf{G}(\mathbf{x})^\top \nabla_\mathbf{x}^2 B(\mathbf{x}) \right) \mathrm{d}t + \nabla B(\mathbf{x})^\top \mathbf{G}(\mathbf{x}) \mathrm{d}\mathbf{w}(t), \end{align} then $$ \mathrm{d}^2E(\mathbf{x}) = \mathrm{d}A(\mathbf{x})\mathrm{d}t + \mathrm{d}B(\mathbf{x})\mathrm{d}\mathbf{w}(t)? $$ Would this make sense as the "acceleration" of $E$?
To be clear as you can see from your SDE formula, the process is not-differentiable $E(x)=...\int g(t) dW_{t}$ because Brownian motion isn't. So you don't even have first time-derivative. You only get at best Holder continuity for $Y_{t}:=E(x_{t})$.