# multi-index derivative of a power

Theorem If $i,k$ 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 for the 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, each partial differentiation $\partial/\partial x_{r}$ therefore reduces to the corresponding ordinary differentiation $d/dx_{r}$. Hence, from equation 1, it follows that $\partial^{i}x^{k}$ vanishes if $i_{r}>k_{r}$ 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. $\Box$

Title multi-index derivative of a power MultiindexDerivativeOfAPower 2013-03-22 13:42:01 2013-03-22 13:42:01 matte (1858) matte (1858) 9 matte (1858) Theorem msc 05-00