[proofplan] We first prove the two identities when $\beta$ has a single bidegree $(r,s)$. In that case, the ordinary graded Leibniz rule for the [exterior derivative](/theorems/1525) gives an identity for $d(\alpha \wedge \beta)$, and projecting that identity to the two relevant bidegrees isolates the $\partial$ and $\bar\partial$ components. The general case follows by decomposing $\beta$ into its finitely many bidegree components and using linearity of $\partial$, $\bar\partial$, and wedge product. [/proofplan]

[step:Reduce to a homogeneous bidegree component of $\beta$] Assume first that $\beta \in A^{r,s}(X)$ for some integers $0 \leq r,s \leq n$. Since $\alpha \in A^{p,q}(X)$, the total degree of $\alpha$ is $p+q$. The graded Leibniz rule for the exterior derivative gives \begin{align*} d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^{p+q}\alpha \wedge d\beta. \end{align*} Here we are using the standard graded Leibniz rule for $d$ on smooth differential forms, with $\alpha$ homogeneous of total degree $p+q$ (citing a result not yet in the wiki: graded Leibniz rule for the exterior derivative). [/step]

[step:Project the exterior derivative identity to the $\partial$ bidegree]Let $\pi_{a,b}: A^\bullet(X) \to A^{a,b}(X)$ denote the projection onto the component of bidegree $(a,b)$. Since $\alpha \wedge \beta \in A^{p+r,q+s}(X)$, the component of $d(\alpha \wedge \beta)$ in bidegree $(p+r+1,q+s)$ is $\partial(\alpha \wedge \beta)$. Thus \begin{align*} \pi_{p+r+1,q+s}\bigl(d(\alpha \wedge \beta)\bigr) = \partial(\alpha \wedge \beta). \end{align*} Also, \begin{align*} d\alpha = \partial\alpha + \bar\partial\alpha, \end{align*} where $\partial\alpha \in A^{p+1,q}(X)$ and $\bar\partial\alpha \in A^{p,q+1}(X)$. Therefore $\partial\alpha \wedge \beta$ has bidegree $(p+r+1,q+s)$, while $\bar\partial\alpha \wedge \beta$ has bidegree $(p+r,q+s+1)$. Hence \begin{align*} \pi_{p+r+1,q+s}(d\alpha \wedge \beta) = \partial\alpha \wedge \beta. \end{align*} Similarly, \begin{align*} d\beta = \partial\beta + \bar\partial\beta, \end{align*} where $\partial\beta \in A^{r+1,s}(X)$ and $\bar\partial\beta \in A^{r,s+1}(X)$. Thus $\alpha \wedge \partial\beta$ has bidegree $(p+r+1,q+s)$, while $\alpha \wedge \bar\partial\beta$ has bidegree $(p+r,q+s+1)$. Hence \begin{align*} \pi_{p+r+1,q+s}(\alpha \wedge d\beta) = \alpha \wedge \partial\beta. \end{align*} Applying $\pi_{p+r+1,q+s}$ to the exterior derivative Leibniz identity gives \begin{align*} \partial(\alpha \wedge \beta) = \partial\alpha \wedge \beta + (-1)^{p+q}\alpha \wedge \partial\beta. \end{align*}[/step]

[guided]The purpose of this step is to extract only the part of the exterior derivative identity whose bidegree corresponds to $\partial$. Define \begin{align*} \pi_{a,b}: A^\bullet(X) \to A^{a,b}(X) \end{align*} to be the projection onto the smooth forms of type $(a,b)$. Since $\alpha \in A^{p,q}(X)$ and $\beta \in A^{r,s}(X)$, the wedge product $\alpha \wedge \beta$ has bidegree $(p+r,q+s)$. Applying $d = \partial + \bar\partial$ to such a form produces two possible bidegrees: $\partial(\alpha \wedge \beta)$ has bidegree $(p+r+1,q+s)$, and $\bar\partial(\alpha \wedge \beta)$ has bidegree $(p+r,q+s+1)$. Therefore \begin{align*} \pi_{p+r+1,q+s}\bigl(d(\alpha \wedge \beta)\bigr) = \partial(\alpha \wedge \beta). \end{align*} Now examine the right-hand side of the exterior derivative Leibniz identity. Since \begin{align*} d\alpha = \partial\alpha + \bar\partial\alpha, \end{align*} the term $d\alpha \wedge \beta$ splits into two bidegree components. The form $\partial\alpha$ has bidegree $(p+1,q)$, so $\partial\alpha \wedge \beta$ has bidegree $(p+r+1,q+s)$. The form $\bar\partial\alpha$ has bidegree $(p,q+1)$, so $\bar\partial\alpha \wedge \beta$ has bidegree $(p+r,q+s+1)$. Consequently the projection onto bidegree $(p+r+1,q+s)$ keeps exactly the $\partial\alpha$ term: \begin{align*} \pi_{p+r+1,q+s}(d\alpha \wedge \beta) = \partial\alpha \wedge \beta. \end{align*} The same bookkeeping applies to $d\beta$. Since \begin{align*} d\beta = \partial\beta + \bar\partial\beta, \end{align*} with $\partial\beta \in A^{r+1,s}(X)$ and $\bar\partial\beta \in A^{r,s+1}(X)$, the wedge product $\alpha \wedge \partial\beta$ has bidegree $(p+r+1,q+s)$, while $\alpha \wedge \bar\partial\beta$ has bidegree $(p+r,q+s+1)$. Therefore \begin{align*} \pi_{p+r+1,q+s}(\alpha \wedge d\beta) = \alpha \wedge \partial\beta. \end{align*} Applying the projection $\pi_{p+r+1,q+s}$ to \begin{align*} d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^{p+q}\alpha \wedge d\beta \end{align*} therefore gives \begin{align*} \partial(\alpha \wedge \beta) = \partial\alpha \wedge \beta + (-1)^{p+q}\alpha \wedge \partial\beta. \end{align*} The sign is unchanged by projection because it came from the ordinary exterior derivative Leibniz rule, where the sign depends on the total degree $p+q$ of $\alpha$.[/guided]

[step:Project the same identity to the $\bar\partial$ bidegree] Using the same projection map $\pi_{a,b}: A^\bullet(X) \to A^{a,b}(X)$, the component of $d(\alpha \wedge \beta)$ in bidegree $(p+r,q+s+1)$ is $\bar\partial(\alpha \wedge \beta)$. Thus \begin{align*} \pi_{p+r,q+s+1}\bigl(d(\alpha \wedge \beta)\bigr) = \bar\partial(\alpha \wedge \beta). \end{align*} From the bidegree computations above, \begin{align*} \pi_{p+r,q+s+1}(d\alpha \wedge \beta) = \bar\partial\alpha \wedge \beta \end{align*} and \begin{align*} \pi_{p+r,q+s+1}(\alpha \wedge d\beta) = \alpha \wedge \bar\partial\beta. \end{align*} Applying $\pi_{p+r,q+s+1}$ to the exterior derivative Leibniz identity gives \begin{align*} \bar\partial(\alpha \wedge \beta) = \bar\partial\alpha \wedge \beta + (-1)^{p+q}\alpha \wedge \bar\partial\beta. \end{align*} Thus both identities hold when $\beta$ is homogeneous of bidegree $(r,s)$. [/step]

[step:Extend the identities to an arbitrary smooth form $\beta$] Let $\beta \in A^\bullet(X)$ be arbitrary. By the Dolbeault decomposition of smooth complex-valued forms, there are uniquely determined components $\beta_{r,s} \in A^{r,s}(X)$ such that \begin{align*} \beta = \sum_{r=0}^{n}\sum_{s=0}^{n} \beta_{r,s}. \end{align*} The sum is finite because $X$ has complex dimension $n$. Since $\partial$, $\bar\partial$, and wedge product with the fixed form $\alpha$ are linear, we have \begin{align*} \partial(\alpha \wedge \beta) = \sum_{r=0}^{n}\sum_{s=0}^{n} \partial(\alpha \wedge \beta_{r,s}). \end{align*} Applying the homogeneous case to each $\beta_{r,s}$ gives \begin{align*} \partial(\alpha \wedge \beta) = \sum_{r=0}^{n}\sum_{s=0}^{n} \left(\partial\alpha \wedge \beta_{r,s} + (-1)^{p+q}\alpha \wedge \partial\beta_{r,s}\right). \end{align*} By finite additivity and linearity, \begin{align*} \partial(\alpha \wedge \beta) = \partial\alpha \wedge \beta + (-1)^{p+q}\alpha \wedge \partial\beta. \end{align*} The identical argument using the homogeneous $\bar\partial$ identity gives \begin{align*} \bar\partial(\alpha \wedge \beta) = \bar\partial\alpha \wedge \beta + (-1)^{p+q}\alpha \wedge \bar\partial\beta. \end{align*} These are the two asserted graded Leibniz rules for the Dolbeault operators. [/step]