Given $x\in\mathbb{R}^n$ I now that the first derivative of the absolute value function is$$\frac{\mathrm d|x|}{\mathrm dx}=\frac{x}{|x|}.$$
But what is the second derivative of it? I need it in order to compute the second derivative for the function $u:\mathbb{R}^n\rightarrow\mathbb{R}$, $u(x)=|x|^\alpha$ with $\alpha>0$.
Let $f:\Bbb R^n\setminus\{0\}\to\Bbb R,\, x\mapsto|x|=\sqrt{x\cdot x}$, where the dot is the euclidean dot product. Hence
$$\partial f(x)y=\frac{x}{|x|}\cdot y\tag1$$
where $\partial$ stay for Fréchet derivative. Thus
$$\partial^2 f(x)[y,z]=\partial_x[\partial f(x) y] z=\frac{z\cdot y}{|x|}-\frac{(x\cdot y)(x\cdot z)}{|x|^3}\tag2$$
You can check that the above holds using partial derivatives on the identity $x\cdot y=\sum_{k=1}^n x_k y_k$ or using the identity $\partial^2 f(x)[y,z]=H_f(x)y\cdot z$, where $H_f(x)$ is the Hessian of $f$ at $x$.
Of course when $n=1$ then is clear that $\partial^2 f(x)[y,z]=0$ for any chosen pair $y,z\in\Bbb R$.