Let $f:\mathbb R^n\to\mathbb R^n$ be given by the equation $f(\mathbf x)=\|\mathbf x\|^2 \mathbf x$. Show that $f$ is of class $C^\infty$ and that $f$ carries the unit ball $B(\mathbf 0;1)$ onto itself in a one-to-one fashion. Show, however, that the inverse function is not differentiable at $\mathbf 0$.
How does one differentiate a function involving the Euclidean norm? It's simple enough if it was just the norm itself, but multiplied by a vector I'm not sure how to go about it.
On
If $f_1(x) = \|x\|^2$, then $f_1(x+h) = f_1(x)+ 2 x^T h + \|h\|^2$, so $Df_1(x)(h) = 2 x^T h$.
If $f_2(x) = x$, then $f_2(x) = f_2(x)+ h$, so $D f_2(x)(h) = h$.
The product rule gives $D (f_1 \cdot f_2) (x) (h) = Df_1(x)(h) f_2(x)+ f_1(x) D f_2(x)(h)$, so we have $D (f_1 \cdot f_2) (x) (h) = 2 x^T h x + \|x\|^2h = (2 x x^T + \|x\|^2I) h$.
On
Let $f : \mathbb R^n \to \mathbb R^n$ be defined by
$$f (\mathrm x) := \| \mathrm x \|_2^2 \, \mathrm x = (\mathrm x^\top \mathrm x) \, \mathrm x$$
Hence,
$$\begin{array}{rl} \mathrm d f &= (\mathrm d \mathrm x^\top \mathrm x) \, \mathrm x + (\mathrm x^\top \mathrm d \mathrm x) \, \mathrm x + (\mathrm x^\top \mathrm x) \, \mathrm d \mathrm x\\\\ &= (\mathrm x \mathrm x^\top) \, \mathrm d \mathrm x + (\mathrm x \mathrm x^\top) \, \mathrm d \mathrm x + \| \mathrm x \|_2^2 \, \mathrm d \mathrm x\\\\ &= (2 \, \mathrm x \mathrm x^\top + \| \mathrm x \|_2^2 \, \mathrm I_n) \, \mathrm d \mathrm x\end{array}$$
Thus, the Jacobian matrix of $f$ is
$$\nabla f (\mathrm x) = 2 \, \mathrm x \mathrm x^\top + \| \mathrm x \|_2^2 \, \mathrm I_n$$
Hint: To make things easier for you, let's work on $n=2$ as always...
$f(x) = f(x_1,x_2) = \begin{pmatrix} (x_1^2+x_2^2)x_1 \\ (x_1^2+x_2^2)x_2 \end{pmatrix}$