Partial derivatives + Taylor's Formula in several variables

Question

Given a function $f(x) = (x_1+...+x_n)^k$, how do we show that $$D_1^{j_1}\cdots D_n^{j_n}f(x) = k!$$ if $j_1+...+j_n = k$?

Answer 1

Simplify the problem: let $f_k(x) = x^k, g(x_1,\dots x_n) = \sum_{l=1}^n x_l$.

$\left(\sum_{l=1}^n x_l\right)^k = f\circ g(x_1,\dots,x_n)$.

Now $$ D_j f_k\circ g(x_1,\dots,x_n) = f_k'\circ g(x_1,\dots,x_n) \times 1 =k f_{k-1}\circ g(x_1,\dots,x_n) \\= \frac{k!}{(k-1)!} f_{k-1}\circ g(x_1,\dots,x_n) \\ D_{1}^{j_1} \dots D_{n}^{j_n} f_k\circ g(x_1,\dots,x_n) =\frac{k!}{(k-\sum_{l=1}^n j_l)!} f_{k-\sum_{l=1}^n j_l}\circ g(x_1,\dots,x_n) $$

with an easy induction.

Now if $\sum_{l=1}^n j_l=k$ this becomes, as $f_0 = 1$: $$ D_{1}^{j_1} \dots D_{n}^{j_n} (f_k\circ g)= k! $$