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?

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.
