$\det( J_f^T J_f)$ $f:R^n \rightarrow R^{n+1}, f(x) =\frac{(x_1,\dots,x_n,1) }{(1+x_1^2+\dots+x_n^2)^{\frac{1}{2}}}  $
$J_f$ is the Jacobi matrix of f
I want to calculate $\det( J_f^T J_f)$ but:
$J_f^t J_f =  \left( \begin{array}{cc} |(x,1)|^2 - x_1^2 & -x_1 x_2 & \dots & -x_1 x_n \\ \vdots & & & \vdots \\- x_1 x^n & -x_2 x^n & \dots & |(x,1)|^2 -x_n^2 \end{array} \right) \cdot \frac{1}{|(x,1)|^{4}}$
I don't know how it's possible to find a general solution for this determinant.
It seems impossible to me.
 A: Let me use $n = 3$ to make this easier to write and picture
The trick here is that the matrix
$$\begin{pmatrix} x^2 & xy & xz \\ xy & y^2 & yz \\ xz & yz & z^2 \end{pmatrix} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \begin{pmatrix} x & y & z \end{pmatrix}$$
is rank 1.
So we have
$$J_f^\top J_f = \frac{1}{c^2}\left( cI - u u^\top \right)$$
where $c = (1 + x^2 + y^2 + z^2)$ and $u^\top = \begin{pmatrix} x & y & z \end{pmatrix}$.
And now we use a common formula for the determinant of a rank-1 update:
$$\det(I + uv^\top) = 1 + u^\top v.$$
See e.g. Determinant of rank-one perturbations of (invertible) matrices or https://en.wikipedia.org/wiki/Matrix_determinant_lemma
So we obtain
\begin{align}
\det(J_f^\top J_f) &= \frac{1}{c^{2n}}\det(cI - u u^\top) \\
&= \frac{1}{c^n} \det\left(I - \frac{1}{c} u u^\top \right) \\
&= \frac{1}{c^n} \left( 1 - \frac{1}{c} u^\top u \right) \\
&= \frac{1}{c^{n+1}} \left[ (1 + x_1 + \dots + x_n) - (x_1 + \dots + x_n) \right] \\
&= \frac{1}{c^{n+1}}.
\end{align}
So the final answer is
$$\det(J_f^\top J_f) = \frac{1}{(1 + x_1 + \dots + x_n)^{n+1}}.$$
