How to prove $f(x,y) = (\frac{x}{\sqrt{1+x^2+y^2}},\frac{y}{\sqrt{1+x^2+y^2}})$ is smooth? When we consider the diffeomorphism between the open disk in $\mathbb{R}^2$ to $\mathbb{R}^2$ itself we have to prove that the following function
$f(x,y) = (\frac{x}{\sqrt{1+x^2+y^2}},\frac{y}{\sqrt{1+x^2+y^2}})$
is smooth.
So, how can I prove that, may I just use the definition of $C^{\infty}$?
 A: The coordinate functions $g_1(x,y)=x$ and $g_2(x,y)=y$, the polynomial $p(x)=1+x^2+y^2$ and the square root function (except for $x=0$) are differentiable. Then so is $\sqrt{1+x^2+y^2}$, since the argument of the root will never be zero. And since the root will also never be zero, both quotients $\frac x  {\sqrt{1+x^2+y^2}}$ and $\frac y  {\sqrt{1+x^2+y^2}}$ give differentiable functions.
Now, having both component functions differentiable, $f$ itself is differentiable.
A: It is a theorem that for $F, G \in \mathcal C^\infty(\mathbb R^p, \mathbb R)$ real functions, with $G$ never vanishing, $H = \frac{F}{G}$ is also $\mathcal C^\infty(\mathbb R^p, \mathbb R)$.
For the proof, you can prove by induction on $n$ that:


*

*If $G \in \mathcal C^n(\mathbb R^p, \mathbb R)$ is never vanishing, $\frac{1}{G}$ is also in $\mathcal C^n(\mathbb R^p, \mathbb R)$

*For $F, G_1 \in \mathcal C^n(\mathbb R^p, \mathbb R)$, $F \cdot G_1$ also belongs to $\mathcal C^n(\mathbb R^p, \mathbb R)$.


Then $\frac{F}{G} = F \cdot G_1$ where $G_1 = \frac{1}{G}$
