[example: The Standard Smooth Structure on the Sphere]
Let $N=(0,\ldots,0,1)$ and $S=(0,\ldots,0,-1)$, and write a point of $\mathbb R^{n+1}$ as $(u,z)$ with $u\in\mathbb R^n$ and $z\in\mathbb R$. Define
\begin{align*}
\varphi_N(u,z)=\frac{u}{1-z}
\end{align*}
on $S^n\setminus\{N\}$ and
\begin{align*}
\varphi_S(u,z)=\frac{u}{1+z}
\end{align*}
on $S^n\setminus\{S\}$. If $a=\varphi_N(u,z)$, then $u=a(1-z)$ and $|u|^2+z^2=1$, so
\begin{align*}
|a|^2(1-z)^2+z^2&=1,\\
|a|^2(1-z)^2-(1-z)(1+z)&=0,\\
(1-z)\bigl(|a|^2(1-z)-(1+z)\bigr)&=0.
\end{align*}
Since $(u,z)\ne N$, we have $1-z\ne0$, hence
\begin{align*}
|a|^2(1-z)&=1+z,\\
|a|^2-1&=(|a|^2+1)z,\\
z&=\frac{|a|^2-1}{|a|^2+1}.
\end{align*}
Therefore
\begin{align*}
u
&=a\left(1-\frac{|a|^2-1}{|a|^2+1}\right)\\
&=a\left(\frac{2}{|a|^2+1}\right)\\
&=\frac{2a}{|a|^2+1},
\end{align*}
so
\begin{align*}
\varphi_N^{-1}(a)=\left(\frac{2a}{|a|^2+1},\frac{|a|^2-1}{|a|^2+1}\right).
\end{align*}
This point lies on $S^n$ because
\begin{align*}
\left|\frac{2a}{|a|^2+1}\right|^2+\left(\frac{|a|^2-1}{|a|^2+1}\right)^2
&=\frac{4|a|^2}{(|a|^2+1)^2}+\frac{(|a|^2-1)^2}{(|a|^2+1)^2}\\
&=\frac{4|a|^2+|a|^4-2|a|^2+1}{(|a|^2+1)^2}\\
&=\frac{|a|^4+2|a|^2+1}{(|a|^2+1)^2}\\
&=1.
\end{align*}
The same calculation with $1+z$ in place of $1-z$ gives
\begin{align*}
\varphi_S^{-1}(b)=\left(\frac{2b}{|b|^2+1},\frac{1-|b|^2}{|b|^2+1}\right).
\end{align*}
On the overlap $S^n\setminus\{N,S\}$, the coordinate $a=\varphi_N(u,z)$ is nonzero. Indeed, if $a=0$, then $u=0$, and $|u|^2+z^2=1$ gives $z=\pm1$, so the point is either $N$ or $S$. For $a\in\mathbb R^n\setminus\{0\}$,
\begin{align*}
(\varphi_S\circ\varphi_N^{-1})(a)
&=\frac{\frac{2a}{|a|^2+1}}{1+\frac{|a|^2-1}{|a|^2+1}}\\
&=\frac{\frac{2a}{|a|^2+1}}{\frac{2|a|^2}{|a|^2+1}}\\
&=\frac{a}{|a|^2}.
\end{align*}
Similarly, for $b\in\mathbb R^n\setminus\{0\}$,
\begin{align*}
(\varphi_N\circ\varphi_S^{-1})(b)
&=\frac{\frac{2b}{|b|^2+1}}{1-\frac{1-|b|^2}{|b|^2+1}}\\
&=\frac{\frac{2b}{|b|^2+1}}{\frac{2|b|^2}{|b|^2+1}}\\
&=\frac{b}{|b|^2}.
\end{align*}
Each component of $a\mapsto a/|a|^2$ is the rational function $a_i/(a_1^2+\cdots+a_n^2)$, whose denominator is nonzero on $\mathbb R^n\setminus\{0\}$. Thus both transition maps are smooth on the overlap. The two stereographic charts cover $S^n$, and their smooth transition maps make them a smooth atlas on the sphere.
[/example]