Prove that $f:\mathbb{R^2} \rightarrow \mathbb{R} $ , $f(x) = \|x\|^2$ is differentiable. The question is the same as the title. All I could churn out was that if the gradient at x was $\nabla f(x)$ then $$\langle\nabla f(x),x\rangle = 2\|x\|^2 .$$ I was also wondering whether this can be generalised from $\mathbb{R^2}$ to $\mathbb{R^n}$
Edit
The norm can be any norm
 A: This is false for general norms. For instance, the $1$-norm of $x=(x_1,x_2)$ is expressed as $$\|x\|_1=|x_1|+|x_2|.$$ From this we obtain $$f(x)=\|x\|_1^2=|x_1|^2+2|x_1||x_2|+|x_2|^2=x_1^2+2|x_1||x_2|+x_2^2.$$ This immediately implies that $f$ is not everywhere differentiable. We can show this by defining $$g(x)=f(x)-x_1^2-x_2^2$$ and noting that $f$ is differentiable if and only if $g$ is, since they differ by a differentiable function. However, we may easily see that $g$ fails to be differentiable at $(0,\frac12)$. To show this, let $$h(x_1)=g(x_1,\tfrac12)=|x_1|,$$ and note that $h$ is just the absolute value function. But the absolute value function obviously fails to be differentiable at $0$, so $$\frac{\partial g}{\partial x_1}(0,\tfrac12)=h'(0)$$ cannot exist.
A: Abstract proof good for any inner product space $E$: The function is composition of of two differentiable functions: the diagonal $E\longrightarrow E\times E$ and the scalar product (bilinear and continuous).
A: I explain using linear algebra. The vector $\mathbf{x} \in \mathbf{R}^n$ and the vector function $f(\mathbf{x})=\|\mathbf{x} \|_{p}^2$ are defined. Then, the gradient of $f(\mathbf{x})$ is shown below.
$$
\begin{align}
\nabla f(\mathbf{x})=& \cfrac{\partial}{\partial x_{i}}\biggl( \sum _{k=1}^{n} x_{k}^p \biggr)^{\frac{2}{p}} \\
=& 2 \biggl( \sum _{k=1}^{n} x_{k}^p \biggr)^{\frac{2-p}{p}} x_{i}^{p-1} \\
=& 2\|\mathbf{x} \|_{p}^{2-p}  \ \mathbf{x}^{p-1}
\end{align}
$$
These inner product can be described using vectors.
$$
\begin{align}
\langle \nabla f(\mathbf{x}), \mathbf{x} \rangle =& 2\|\mathbf{x} \|_{p}^{2-p}  \ (\mathbf{x}^{\text{T}} \mathbf{x}^{p-1}) \\
=& 2 \biggl( \sum _{k=1}^{n} x_{k}^p \biggr)^{\frac{2-p}{p}} \biggl( \sum _{k=1}^{n} x_{k}^p \biggr) \\
=& 2\|\mathbf{x} \|_{p}^{2-p} \|\mathbf{x} \|_{p}^{p}
\end{align}
$$
Particularly, $p=2$ that gives the Euclidean norm. Therefore, $\nabla f(\mathbf{x})=2 \mathbf{x}$ , $\langle \nabla f(\mathbf{x}), \mathbf{x} \rangle =2 \|\mathbf{x} \|_{2}^{2}$ because $\|\mathbf{x} \|_{2}^{0}=1$. Finally, it can be expressed as follows like your description when $n=1$, $x \in \mathbf{R}$ and $p=2$. 
$$\langle \nabla f(x), x \rangle =  2\|x \|^{2}$$
A: You may write, as $||h|| \to 0$,
$$
\begin{align}
f(x+h)-f(x)&=||x+h||^2-||x||^2
\\&=\langle x+h,x+h \rangle-\langle x,x \rangle
\\&=\langle x,h \rangle+\langle h,x \rangle+\langle h,h \rangle
\\&=\langle x,h \rangle+\langle h,x \rangle+||h||^2
\\&=\langle x,h \rangle+\langle h,x \rangle+o(||h||)
\end{align}
$$ or

$$
f(x+\color{blue}{h})-f(x)=\langle 2x,\color{blue}{h} \rangle+o(||\color{blue}{h}||).
$$

