I want to prove the product rule for higher order partial derivatives. It is given on Wikipedia under the name "General Leibniz rule":

$$\partial^\alpha(fg)=\sum_{\beta\leq \alpha}\binom\alpha\beta(\partial^\beta f)(\partial^{\alpha-\beta}g),$$

where $\alpha$ and $\beta$ are multiindices.

My first attempt is to make a recursion by computing (here $i$ is just an index)

$$\partial^{\alpha}\big( \partial^i(fg) \big)=\sum_{\beta\leq\alpha}\binom{\alpha}{\beta}\Big( (\partial^{(\alpha,i)-\beta}f)(\partial^{\beta}g)+(\partial^{\beta}f)(\partial^{(\alpha,i)-\beta}g) \Big).$$

I'm not sure how to continue at this point. The index $i$ can be one of the components of the multiindex $\alpha$. I dont'see how to manage the combinatorics...

related on MathOverflow: Multivariable Higher Order Chain Rule


1 Answer 1


The base case $\alpha=0$ is trivial. Suppose the result holds for all $|\alpha|\le k$. We will show the result for any $\alpha = (\alpha_1,\dots,\alpha_n)$ with $|\alpha| = k+1$. Let $j$ be any index where $\alpha_j$ is not zero, and put $\alpha = \alpha_j e_j + \tilde \alpha$. Without loss of generality, $\tilde \alpha\neq 0$, since if $\tilde \alpha=0$ then $\alpha = |\alpha|e_j$ and we can use the 1D result (see e.g. Wikipedia for proof). So both $\alpha_j e_j$ and $\tilde\alpha$ have length at most $k$. Then $\let\del\partial$ \begin{align} \del^\alpha (fg) &= \del^{\alpha_j e_j} \del^\tilde \alpha(fg) \\ &= \del^{\alpha_j e_j} \sum_{\beta \le \tilde \alpha }\binom{\tilde\alpha}{\beta} \del^\beta f \del^{\tilde \alpha-\beta}g\\ &= \sum_{\beta \le \tilde \alpha } \sum_{\gamma=0}^{\alpha_j}\binom{\tilde\alpha}{\beta} \binom{\alpha_j}{\gamma} \del^{\gamma e_j} \del^\beta f \del^{(\alpha_j - \gamma)e_j} \del^{\tilde \alpha-\beta}g \\ &= \sum_{\substack{\beta \le \alpha \\ \beta_j = 0}} \sum_{\gamma=0}^{\alpha_j}\binom{\alpha}{\beta+ \gamma e_j} \del^{\beta+\gamma e_j} f \del^{\alpha - (\beta + \gamma e_j)} g \\ &= \sum_{\gamma=0}^{\alpha_j} \sum_{\substack{\delta: \delta \le \alpha \\ \delta_j = \gamma}}\binom{\alpha}{\delta} \del^{\delta} f \del^{\alpha - \delta } g \\ &= \sum_{\substack{\delta \le \alpha}}\binom{\alpha}{\delta} \del^{\delta} f \del^{\alpha - \delta } g \end{align} which was what we wanted.


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