# $n$-th derivative of $\displaystyle \left(\sum_{i=1}^Nx_i\right)^n$

I have no idea how to find the $n$-th derivative of $f:\mathbb{R}^N\rightarrow\mathbb{R},x\mapsto$ $(\sum_{i=1}^N x_{i})^n$. I tried to use the multinomial theorem, as well as only the chain rule. I got to $D^2f(x)$, i.e. the Hessian. I have no idea how a a derivative of degree greater than two even looks like. All I know is that it has to be a linear operator in $\mathcal{L}^n(\mathbb{R}^N,\mathbb{R})$. Thank you very much for your help in advance!

• Did you try to find the first derivative? – 5xum Nov 18 '14 at 12:01
• yes.I would say it is $\bigtriangledown f(x)$ with the i-th entry being $\frac{\partial f}{\partial x_{i}}=\sum_{j=1}^N x_{i}$ but $j\neq i$ – Hans Nov 18 '14 at 12:09
• You would be wrong. Take a look at what happens if $n=N=2$, so $f(x,y)=(x+y)^2$. In that case, $\frac{\partial f}{\partial x} = 2(x+y)$ which is not what you have. – 5xum Nov 18 '14 at 12:15
• Yes of course. I know. forgot the n. – Hans Nov 18 '14 at 12:26
• new suggestion:$\frac{\partial f}{\partial x_{i}}=n(\sum_{j=1}^N x_{j})^{(n-1)}$ – Hans Nov 18 '14 at 12:29

This is a highly degenerate functional, everything will be happening along a single direction. You can interpret your function as $$f(\vec{x})=(\vec{x}\cdot\vec{e})^n$$ where $\vec{e}=(1,1,1,1,\ldots)$. Then it just follows $$\nabla f(\vec{x})=n(\vec{x}\cdot\vec{e})^{n-1}\vec{e}$$ which in our case means partial derivatives in all directions are the same. Now assume index notation to go to higher derivatives.
$$\partial_i \partial_j f(\vec{x})=\partial_i(n(\vec{x}\cdot\vec{e})^{n-1}e_j)=$$ $$=\partial_i(n(\vec{x}\cdot\vec{e})^{n-1})e_j=n(n-1)(\vec{x}\cdot\vec{e})^{n-2}e_i e_j$$ Again, for our boring vector $\vec{e}$, all the components are exactly the same, because $e_i e_j=1$ for every pair $i,j$. The same goes for every higher derivative. At the end you arrive at $$\partial_i \partial_j\partial_k\cdots \partial_n f(\vec{x})=n!$$ for any possible combination of partial derivatives.