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I am currently learning about $\mathcal{N}=(2,2)$ supersymmetry and have come up against what is probably a really silly question.

The $\mathcal{}N=(2,2)$ superspace consists of bosonic coordinates $x^{\pm}$ and fermionic/Grassmann coordinates $\theta^{\pm},\bar{\theta}^{\pm}$. A (scalar), superfield is a function defined on this superspace. Due to the anticommuting nature of Grassmann numbers it can be expanded into a finite Taylor series with at most $16$ non-vanishing terms and bosonic functions for the coefficients: $$ \Phi(x^{\mu},\theta^{\alpha},\bar{\theta}^{\beta}) = f_{0}(x^{\mu}) + f_{+}(x^{\mu})\theta^{+}+\ldots $$

We define the covariant derivatives acting on superspace: $$ D_{\pm}=\frac{\partial}{\partial \theta^{\pm}}-i\bar{\theta}^{\pm}\partial_{\pm}\\ \bar{D}_{\pm}=-\frac{\partial}{\partial \bar{\theta}^{\pm}}+i\theta^{\pm}\partial_{\pm} $$

A superfield $\Phi$ which satisfies $\bar{D}_{\pm}\Phi=0$ is called a chiral superfield. I have seen it claimed in many places that the product of chiral superfields is again a chiral superfield. This seems fair enough, but I have been unable to prove it.

The desired result follows easily if the covariant derivatives obey a (possibly graded), product rule, since in this case: $$ \bar{D}_{\pm}(\Phi\Psi)=\bar{D}_{\pm}(\Phi)\Psi\pm\Phi\bar{D}_{\pm}(\Psi)=0 $$ and I have also seen it claimed that such a rule holds, although again I have been unable to prove it. My attempts have proceeded in the obvious way, namely by examining $\frac{\partial}{\partial \bar{\theta}^{\pm}}\left(\Phi(x^{\mu},\theta^{\alpha},\bar{\theta}^{\beta})\Psi(x^{\mu},\theta^{\alpha},\bar{\theta}^{\beta})\right)$, but because some terms involve Grassmann numbers and others don't I have been unable to proceed. Any help would be much appreciated!

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Hint: Each term in the covariant derivatives consists of only 1st-order differential operators, and hence, satisfies a $\mathbb{Z}_2$-graded Leibniz rule.

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