How to prove that $f(x)=\frac{x}{\sqrt{1+|x|^2}}$ is smooth?

Question

I have a function $f(x)=\frac{x}{\sqrt{1+|x|^2}}$ on the open ball $B_1(0)$ in $\mathbb{R}^n$. I have to show that this function is a diffeomorphism between $\mathbb{R}^n$ and $B_1(0)$. I have already proved that the function is bijective with inverse $\frac{x}{\sqrt{1-|x|^2}}$ and it is continuous. How to prove that the function is smooth?

Thanks!

Answer 1

You can apply the theorem that the sum, product, quotient (whenever defined), square root (of a positive smooth function), and composition of smooth functions are smooth. In your examples, all the component functions $x$, $|x|^2$, $1 \pm |x|^2$, and so on are all smooth.

Answer 2

Just use the fact that compositions of smooth functions is smooth and ratio of two smooth functions is smooth as long as the denominator never vanishes.