Partial derivatives for $\|x\|^2 = x_1^2+\dots+ x_n^2$

Question

Let $f: \mathbb{R^n} \to \mathbb{R}$ and $x \mapsto \|x\|^2 = x_1^2+\dots+ x_n^2\:$ for $n>2.$ Find the partial derivatives at $x$ and determine the derivative at $x$.

This seems to be just a problem dealing with the chain rule for partial derivatives. However, I got pretty confused since I would have to apply it $n$ times, how would this go?

For the second part if I just denote $x=x_0$ I would get $$f(x)-f(x_0) = \nabla f(x_0)\cdot (x-x_0)+ \|(x-x_0)\|\varepsilon(x_0-x)$$ for the derivative.

Answer 1

$$f(x_1,\dots,x_n) = x_1^2+\dots+x_n^2 \implies \frac{\partial f}{\partial x_k} = 2x_k$$

All the terms different from $x_k^2$ are fixed so they become $0$ differentiating.

Answer 2

$\nabla f=(2x_1,\dots,2x_n)$. The $i$th partial is $\partial f/\partial x_i=2x_i$.

So, at $x_0$, we get $\nabla f(x_0)=2x_0$.