Let $S^n=\{(x_1,\dots,x_{n+1})∈ \Bbb{R}^{n+1}\mid x_1^2+\cdots+x_{n+1}^2=1\}$. There is natural inclusion from $S^n$ to $ \Bbb{R}^{n +1}$.
I want to prove this natural inclusion is immersion. I want to prove this by calculating rank of Jacobi matrix. But I'm having trouble because I cannot figure out what the Jacobi matrix is. Thank you for your help.
Fix one of the $2(n+1)$ charts of one of the most natural atlants of $S^n$:
$$\phi_i^+\colon U_i^+:=\{x_i>0\} \to B(0,1)\subseteq \mathbb{R}^{n} \text{ by sending } (x_1,\dots, x_{n+1})\mapsto (x_1,\dots \hat{x_i} \dots, x_{n+1})$$
The inverse map sends $$(t_1,\dots ,t_n)\mapsto (t_1,\dots, \sqrt{1-(t_1^2+\dots t_n^2)}, \dots , t_n)$$
Now that you fixed the atlant, then the inclusion function $i\colon S^n\to \mathbb{R}^{n+1}$ will be locally with respect to the charts $\phi_i^+$
$i\circ (\phi_i^+)^{-1}\colon B(0,1)\to \mathbb{R}^{n+1}$ by sending
$$(t_1,\dots ,t_n)\mapsto (t_1,\dots, \sqrt{1-(t_1^2+\dots t_n^2)}, \dots , t_n)$$
The Jacobian of the function is
$$J=\begin{pmatrix} 1 & \dots & 0\\ 0 & 1 & \dots \\ \dots & \dots \\ \frac{-2t_1}{\sqrt{1-(t_1^2+\dots t_n^2)}} & \dots & \dots \frac{-2t_n}{\sqrt{1-(t_1^2+\dots t_n^2)}}\\ \dots & \dots \\ 0& 0 \dots &1\\ \end{pmatrix}$$
Of course $rk(J)=n$, that shows the differential is injective for any point $U_i^+$ of the chart. You can prove this for each of the $2(n+1)$ charts that cover $S^n$. Hence $i$ is an immersion.