Prove that $\frac{rx}{\sqrt{r^2-|x|^2}}$ is a diffeomorphism

Question

Let $B_r=\{x\in\mathbb{R}^n: |x|<r\}$ be the open ball of radius $r > 0 $ about the origin in $\mathbb{R}^n$. Show that the mapping: $$s_r:B_r\rightarrow\mathbb{R}^n, s_r(x) = \displaystyle\frac{rx}{\sqrt{r^2-|x|^2}}, x\in{B_r}$$ is a diffeomorphism.

I tried to prove it using the definition of diffeomorphism which says: $s_r$ is a diffeomorphism if and only if $s_r$ is homeomorphic and if $s_r^{-1}$ is differentiable, but I can't reach any conclusion and this is the only thing I know.

Answer 1

Hint in the case $r=1$: show that $t_1: y \in \mathbb{R}^n \mapsto \frac{y}{\sqrt{1+\|y\|^2}}\in B_1$ is smooth and satisfies $s_1\circ t_1 = \mathrm{id}_{B_1}$, $s_1\circ t_1 = \mathrm{id}_{\mathbb{R}^n}$.

Try to adapt this in the general case $r > 0$.

Answer 2

To check it is a smooth function, you could take it as a vector function. let $ s_r(x_1,...,x_n)=(\frac{rx_1}{\sqrt{r^2-x_1^2-...-x_n^2}},...\frac{rx_i}{\sqrt{r^2-x_1^2-...-x_n^2}},...\frac{rx_n}{\sqrt{r^2-x_1^2-...-x_n^2}}) $,then we get: $$\dfrac{\partial s_r}{\partial x_i}=(\frac{rx_{1}x_i}{(r^2-x_1^2-...-x_n^2)^{\frac{3}{2}}},...\frac{r(r^2-x_1^2-...-x_{i-1}^2-x_{i+1}^2-...-x_{n}^2)}{(r^2-x_1^2-...-x_n^2)^{\frac{3}{2}}},...\frac{rx_{n}x_i}{(r^2-x_1^2-...-x_n^2)^{\frac{3}{2}}}) $$ So,just using the knowlege of the vector analysis to check it. The same way for the $ s_r^{-1} $.