Let ${\phi}_{\boldsymbol{r}}(r=1,2, \ldots, n)$ be a set of $\mathrm{n}$ boson fields with Lagrangian density
$$
\mathscr{L}=\sum_{r=1}^{n} \frac{1}{2} \dot{\phi}_{r}^{2}+G\left(\phi_{r}\right)
$$
and let $\phi_{s}^{\prime}=\phi^{\prime}{ }_{s}\left(\phi_{r}\right)$ be a redefinition of the field $\phi_r$
Suppose we have $\left[\dot{\phi}_{r}, \phi_{s}\right]=\Delta \delta_{r s}$
Now in this article Change of variables and equivalence theorems in quantum field theories they have
$$
\begin{aligned}
\left[\frac{\partial \mathscr{L}}{\partial \phi^{\prime}_s}, \phi^{\prime}_{r}\right] &=\frac{1}{2} \sum_{a}\left\{\left[\dot{\phi}_{a}, \phi_{r}^{\prime}\right] \frac{\partial \phi_{a}}{\partial \phi^{\prime}{ }_{s}}+\frac{\partial \phi_{a}}{\partial \phi_{s}^{\prime}}\left[\dot{\phi}_{a}, \phi^{\prime}_r\right]\right\} \\
&=\frac{1}{2} \sum_{a}\left\{\Delta \frac{\partial \phi_{r}^{\prime}}{\partial \phi_{a}} \frac{\partial \phi_{a}}{\partial \phi_{s}^{\prime}}+\Delta \frac{\partial \phi_{a}}{\partial \phi_{s}^{\prime}} \frac{\partial \phi_{r}^{\prime}}{\partial \phi_{a}}\right\}=\Delta \frac{\partial \phi_{r}^{\prime}}{\partial \phi^{\prime}}=\Delta \delta_{r s} .
\end{aligned} \tag 1
$$
Why we have
$\left[\dot{\phi}_{a}, \phi_{r}^{\prime}\right]=\Delta \frac{\partial\phi_{r}^{\prime}}{\partial \phi_a}$
Best Answer
There is a property of commutators:
$$[A,[A,B]]=0~~\Rightarrow~~[F(A),B]=[A,B]\frac{\partial F}{\partial A}.$$
So, assuming $[\phi_s,\Delta]=0$, you have
$$[\dot{\phi}_a,\phi_r'(\phi)]=-[\phi_r'(\phi),\dot{\phi}_a]=-[\phi_s,\dot{\phi_a}]\frac{\partial\phi'_r}{\partial\phi_s}=\Delta\delta_{as}\frac{\partial\phi'_r}{\partial\phi_s}=\Delta\frac{\partial\phi'_r}{\partial\phi_a}$$