multi-index derivative of a power

TheoremIf $i, k$ are multi-indices in $\\mathbb{N}^{n}$ , and $x=(x_{1},\\ldots,x_{n})$ ,then

\\displaystyle\\partial^{i}x^{k}=\\left\\{\\begin{array}[]{ll}\\frac{k!}{(k-i)!}x^{k%-i}&\\mbox{if}\\,\\,i\\leq k,\\\\0&\\mbox{otherwise}.\\end{array}\\right.

Proof. The proof follows from the corresponding rule forthe ordinary derivative; if $i, k$ are in $0,1,2,\\ldots$ , then

\\displaystyle\\frac{d^{i}}{dx^{i}}x^{k}=\\left\\{\\begin{array}[]{ll}\\frac{k!}{(k-%i)!}x^{k-i}&\\mbox{if}\\,\\,i\\leq k,\\\\0&\\mbox{otherwise.}\\end{array}\\right.

(1)

Suppose $i=(i_{1},\\ldots,i_{n})$ , $k=(k_{1},\\ldots,k_{n})$ , and $x=(x_{1},\\ldots,x_{n})$ .Then we have that

	$\\displaystyle\\partial^{i}x^{k}$	$\\displaystyle=$	$\\displaystyle\\frac{\\partial^{\|i\|}}{\\partial x_{1}^{i_{1}}\\cdots\\partial x_{n}^%{i_{n}}}x_{1}^{k_{1}}\\cdots x_{n}^{k_{n}}$
		$\\displaystyle=$	$\\displaystyle\\frac{\\partial^{i_{1}}}{\\partial x_{1}^{i_{1}}}x_{1}^{k_{1}}\\cdot%\\cdots\\cdot\\frac{\\partial^{i_{n}}}{\\partial x_{n}^{i_{n}}}x_{n}^{k_{n}}.$

For each $r=1,\\ldots,n$ , the function $x_{r}^{k_{r}}$ only depends on $x_{r}$ .In the above, eachpartial differentiation $\\partial/\\partial x_{r}$ thereforereduces to the correspondingordinary differentiation $d/dx_{r}$ .Hence, from equation 1, it follows that $\\partial^{i}x^{k}$ vanishesif $i_{r}>k_{r}$ for any $r=1,\\ldots,n$ . If this is not the case, i.e.,if $i\\leq k$ as multi-indices, then for each $r$ ,

\\frac{d^{i_{r}}}{dx_{r}^{i_{r}}}x_{r}^{k_{r}}=\\frac{k_{r}!}{(k_{r}-i_{r})!}x_{%r}^{k_{r}-i_{r}},

and the theorem follows. $\\Box$