The criterion to obtain covariant spinor derivatives in superspace It can be simply said that covariant derivatives in 4d superspace are given by $\mathcal{D}_\mu $, $\mathcal{D}_{\alpha}$ and $\mathcal{D}_{\dot{\alpha}}$, so that they commute with  the representation of the supercharge generator $\epsilon\mathcal{Q}+\bar{\epsilon}\bar{\mathcal{Q}}$, which means that the covariant derivatives of any superfield are again superfields.
But while following the detailed argument to derive the result (Quevedo {arxiv:1011.1491}), I found I don't understand the logic at the very critical point. According to him, the criterion for a field to be a superfield is
\begin{equation}
i[S, \epsilon Q+\bar{\epsilon}\bar{Q}]=i(\epsilon\mathcal{Q}+\bar{\epsilon}\bar{\mathcal{Q}})S=\delta S,
\end{equation}
And $\partial_{\alpha}S$ is not a superfield because
$\delta(\partial_{\alpha}S)=i[\partial_{\alpha}S, \epsilon Q+\bar{\epsilon}\bar{Q}] =i\partial_{\alpha}[S, \epsilon Q+\bar{\epsilon}\bar{Q}] = i\partial_\alpha(\epsilon\mathcal{Q}+\bar{\epsilon}\bar{\mathcal{Q}})S \neq i(\epsilon\mathcal{Q}+\bar{\epsilon}\bar{\mathcal{Q}})(\partial_\alpha S)$.
What I don't get is the second equality. It seems not trivial at all for me that $\partial_{\alpha}$ can be simply extracted outside (and this happens again when $\mathcal{D}$ is substituted for $\partial$), and I feel that we can jump directly to the equality 
$ i[\partial_{\alpha}S, \epsilon Q+\bar{\epsilon}\bar{Q}]  =i(\epsilon\mathcal{Q}+\bar{\epsilon}\bar{\mathcal{Q}})(\partial_\alpha S) $
from the superfield transformation rule. Can anybody explain what the matter is with my understanding? 
 A: For a superfield $S$, an infinitesimal supersymmetric transformation  gives : 
$\delta_\epsilon S = -i(\epsilon\mathcal{Q}+\bar{\epsilon}\bar{\mathcal{Q}})S$
Here $\mathcal{Q}_a$ and ${ \mathcal{\bar Q}}^{\dot a}$ are differential operators applying on the superfield: 
$\mathcal{Q}_a = i \partial_a - (\sigma^\mu \bar \theta)_a \partial_\mu$, 
${ \mathcal{\bar Q}}^{\dot a} = i \partial^{\dot a} - (\bar \sigma^\mu  \theta)^{\dot a} \partial_\mu$
(with notations $\partial_a = \frac{\partial}{\partial \theta_a}, \partial^{\dot a} = \frac{\partial}{\partial \theta^{\dot a}}$)  
So : 
$\delta_\epsilon S = (\epsilon.\partial+ \bar\epsilon.\bar \partial + i (\epsilon \sigma^\mu \bar \theta + \bar  \epsilon \sigma^\mu \theta ) \partial_\mu)S $
So, you have : 
$\partial_a \delta_\epsilon S = \delta_\epsilon \partial_aS-i ( \bar \epsilon \sigma^\mu)_a \partial_\mu S$
Now, looking for "covariant" derivatives means that we are looking for a definition of derivatives which are consistent with the supersymmetry transformations, so we want that the derivative operation and the (infinitesimal) supersymmetry operation, applying on a superfield, commute. 
This is clearly not the case with the derivatives $\partial_a$ (or $\partial^{\dot a}$)
