How to find the jacobian of the following? I am stuck with the following problem that says : 

If $u_r=\frac{x_r}{\sqrt{1-x_1^2-x_2^2-x_3^2 \cdot \cdot \cdot-x_n^2}}$ where  $r=1,2,3,\cdot \cdot \cdot ,n$, then prove that the jacobian of $u_1,u_2,\cdot \cdot, u_n$ with respect to $x_1,x_2,\cdot \cdot, x_n$ is $(1-x_1^2-x_2^2-\cdot \cdot \cdot -x_n^2)^{-\frac12}$


My try: Now, I have to calculate the value of 
\begin{vmatrix}
 \frac{\delta u_1}{\delta x_1} &  \frac{\delta u_1}{\delta x_2} &  \frac{\delta u_1}{\delta x_3} & \cdots &  \frac{\delta u_1}{\delta x_n} \\ 
 \frac{\delta u_2}{\delta x_1} &  \frac{\delta u_2}{\delta x_2} &  \frac{\delta u_2}{\delta x_3} & \cdots &  \frac{\delta u_2}{\delta x_n} \\
\vdots & \vdots & \vdots & \ddots &\vdots \\
 \frac{\delta u_n}{\delta x_1} &  \frac{\delta u_n}{\delta x_2} &  \frac{\delta u_n}{\delta x_3} & \cdots &  \frac{\delta u_n}{\delta x_n} \\ 
\end{vmatrix}
Now,the value of $\frac{\delta u_1}{\delta x_1}=\frac{1-2x_1^2}{\{1-x_1^2\}^\frac32}$ , $\frac{\delta u_1}{\delta x_2}=\cdot \cdot =\frac{\delta u_1}{\delta x_n}=0$..
So, things are getting complicated. Can someone show me the right direction?
 A: As the partial derivatives read:
$$
\frac{\partial u_i}{\partial x_j}=
\frac{\delta_{ij}}{\left(1-\sum_{k=1}^nx_k^2\right)^{1/2}}
+\frac{x_ix_j}{\left(1-\sum_{k=1}^nx_k^2\right)^{3/2}},
$$
the jacobian matrix is of a form:
$$
{\cal J}=c(I+ v^Tv),
$$
with $c=\frac1{\left(1-\sum_{k=1}^{n}x_k^2\right)^{1/2}}$ and $v=\frac{(x_1,x_2,\dots, x_n)}{\left(1-\sum_{k=1}^{n}x_k^2\right)^{1/2}}$.
Therefore by the Matrix determinant lemma:
$$\begin{array}{}
\det{\cal J}&=c^n\det(I+ v^Tv)=c^n(1+vv^T)\\
&\displaystyle=\frac{1}{\left(1-\sum_{k=1}^{n}x_k^2\right)^{\frac n2}}
\left[1+\frac{\sum_{k=1}^{n}x_k^2}{1-\sum_{k=1}^{n}x_k^2}\right]\\
&\displaystyle=\frac{1}{\left(1-\sum_{k=1}^{n}x_k^2\right)^{\frac n2+1}}.
\end{array}
$$
As stated already in a comment the correct result deviates from that claimed.
A: The partial derivatives seems easy to find due to symmetry:-
Applying chain rule we get $$ u_{r,i}=\frac{x_ix_r}{\sqrt{1-x_1^2-x_2^2-x_3^2 \cdot \cdot \cdot-x_n^2}}$$ if i
!= r
$$u_{i,i}=\frac{1}{\sqrt{1-x_1^2-x_2^2-x_3^2 \cdot \cdot \cdot-x_n^2}}-\frac {x_i^2}{{\sqrt{1-x_1^2-x_2^2-x_3^2 \cdot \cdot \cdot-x_n^2}}}$$
Then finding the determinant is the task ......
here $u_{i,r} $ denotes partial derivative.
A: Hint:
Since
$$
\eqalign{
  & u_{\,r}  = {{x_{\,r} } \over {\sqrt {1 - x_{\,1} ^2  - x_{\,2} ^2  \cdots  - x_{\,n} ^2 } }}  \cr 
  & u_{\,r}  = {\partial  \over {\partial x_{\,r} }}\sqrt {1 - \left( {x_{\,1} ^2  + x_{\,2} ^2  \cdots  + x_{\,n} ^2 } \right)}   \cr 
  & {\partial  \over {\partial x_{\,k} }}u_{\,r}  = {{\partial ^{\,2} } \over {\partial x_{\,r} \partial x_{\,k} }}\sqrt {1 - \left( {x_{\,1} ^2  + x_{\,2} ^2  \cdots
  + x_{\,n} ^2 } \right)}  = {\partial  \over {\partial x_{\,r} }}u_{\,k}  \cr} 
$$
you get the determinant of a matrix of type
$$
{\bf I} + {\bf v}\,{\bf v}^T 
$$
