Derivative of $x^{\beta}$

Question

Let $x\in\mathbb{R}^d$, and $\alpha,\beta\in\mathbb{N}^d$ such that $|\alpha|\in\left\{1,2\right\}$ $|\beta|\ge 3$

What is the result of $$\partial_x^{\alpha}x^{\beta}$$?

Thanks

Answer 1

Let $\alpha=(\alpha_1,\dots,\alpha_n)$ and $\beta=(\beta_1,\dots,\beta_n)$. If $\alpha\le\beta$, that is, $\alpha_k\le\beta_k$, $1\le k\le n$,Then \begin{align} \partial_x^\alpha x^\beta&=\prod_{k=1}^n\frac{\partial^{\alpha_k} x_k^{\beta_k}}{\partial x_k^{\alpha_k}}\\ &=\prod_{k=1}^n\beta_k(\beta_k-1)\dots(\beta_k-\alpha_k+1)x_k^{\beta_k-\alpha_k}\\ &=\Bigl(\prod_{k=1}^n\frac{\beta_k!}{(\beta_k-\alpha_k)!}\Bigr)x^{\beta-\alpha}\\ &=\frac{\beta!}{(\beta-\alpha)!}\,x^{\beta-\alpha}. \end{align}