Consider the usual commutation relations of two scalar fields
$$\left[\phi_{m}\left(t,\boldsymbol{x}\right),\pi_{n}\left(t,\boldsymbol{y}\right)\right]=\boldsymbol{i}\delta_{mn}\delta\left(\boldsymbol{x}-\boldsymbol{y}\right),$$
$$\left[\phi_{m}\left(t,\boldsymbol{x}\right),\phi_{n}\left(t,\boldsymbol{y}\right)\right]=\left[\pi_{m}\left(t,\boldsymbol{x}\right),\pi_{n}\left(t,\boldsymbol{y}\right)\right]=0.$$
What's the commutator of $\left[\partial_{i}\phi_{m}\left(t,\boldsymbol{x}\right),\phi_{n}\left(t,\boldsymbol{y}\right)\right]$, where $\partial_{i}\equiv\partial/\partial x^{i}$ is one of the three spatial derivatives?
What about $\left[\partial_{i}\phi_{m}\left(t,\boldsymbol{x}\right),\pi_{n}\left(t,\boldsymbol{y}\right)\right]$ ?
Attempt 1:
$$\begin{array}{cl} \left[\partial_{i}\phi\left(t,\boldsymbol{x}\right),\phi\left(t,\boldsymbol{y}\right)\right] & =\partial_{i}\left[\phi\left(t,\boldsymbol{x}\right),\phi\left(t,\boldsymbol{y}\right)\right]+\left[\partial_{i},\phi\left(t,\boldsymbol{y}\right)\right]\phi\left(t,\boldsymbol{x}\right)\\ & =\left[\partial_{i},\phi\left(t,\boldsymbol{y}\right)\right]\phi\left(t,\boldsymbol{x}\right)\\ & =\left(\partial_{i}\phi\left(t,\boldsymbol{y}\right)\right)\phi\left(t,\boldsymbol{x}\right)-\phi\left(t,\boldsymbol{y}\right)\partial_{i}\phi\left(t,\boldsymbol{x}\right)\\ & =? \end{array}$$