# Differentiability on vector values function

I just had my first lecture of my analysis course, and we were introduced the differentiation on general euclidean space where the derivative is regarded as a linear transformation.

Define $f\colon\mathbb R^n\to\mathbb R$ by $f(x)=|x|$ (the norm of x). Determine the set of points at which f is differentiable and find the derivative there.

I am crazy comparing this to the absolute value function ($f\colon \mathbb R\to\mathbb R$ defined by $f(x)=|x|$), yet quite confused about how to write proofs using the definition of the linear-transformation form derivatives.

I think the answer should be $f(x)$ is anywhere differentiable except $0$ in $\mathbb R^n$, however I cannot write down the proof. I wish I can get some help here.

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Hint: Express $|x|$ as a square root of scalar product of vectors. Then try to make the scalar product look like $f(x)+A(x)+o(x)$ where $A$ is a linear transformation. Then use the fact that $x\mapsto \sqrt{x}$ is differentiable if $x\neq 0$.
The easiest way to tackle this is to write out the norm function in the variables. It is easy to check that all partial derivatives exist and are continuous whenever $x\ne 0$.