How to prove that two parametrizations of a surface M give the same tangent plane If you have 2 parametrizations of a surface $M$, $x$ and $y$, such that for some point $P \in M$, $x(u_0,v_0)=y(s_0,t_0)=P$. How do you show that the tangent plane of $M$ at $P$ are the same? That is how do I show that $\text{Span}(x_u,x_v)=\text{Span}(y_s,y_t)$.
 A: HINT: Write $x(u,v)=y(s(u,v),t(u,v))$ and use the chain rule to show that $x_u$ and $x_v$ (evaluated at $(u_0,v_0)$) are linear combinations of $y_s$ and $y_t$ (evaluated at $(s_0,t_0)$). This shows $\text{Span}(x_u,x_v)\subset\text{Span}(y_s,y_t)$. Now what?
A: Ted Shifrin's answer can be generalized:
Let $A$ be an affine space with a real and normed translation space $V$ and let $x\colon U\subset\mathbf{R}^n\to A$ and $y\colon U'\subset\mathbf{R}^n\to A$ be injective maps defined on open subsets of $\mathbf{R}^n$ such that the transition maps $t:=y^{-1}\circ x\colon U\cap U'\to U\cap U'$ and $t^{-1}=x^{-1}\circ y\colon U\cap U'\to U\cap U'$ are differentiable in $a\in U\cap U'$ and $b:=y^{-1}(x(a))\in U\cap U'$. Finally, we require that $x$ and $y$ are differentiable in $a$ and $b$, respectively*.
In this case, $Dt(a)\colon\mathbf{R}^n\to\mathbf{R}^n$ and $Dt^{-1}(b)\colon\mathbf{R}^n\to\mathbf{R}^n$ are isomorphisms inverse to each other** and according to the chain rule,
\begin{equation}
Dy(b)=Dx(a)\circ Dt^{-1}(b)
\end{equation}
and
\begin{equation}
Dx(a)=Dy(b)\circ Dt(a).
\end{equation}
Thus, $Dx(a)(\mathbf{R}^n)=Dy(b)(\mathbf{R}^n)=:S\subset V$.
In your notation,
\begin{equation}
x_i=\lim\nolimits_{\delta\to 0}\frac{x(a_1,\ldots,a_{i-1},a_i+\delta,a_{i+1},\ldots,a_n)-x(a_1,\ldots,a_n)}{\delta}=Dx(a)(e_i)
\end{equation}
and $\text{span}\{x_1,\ldots,x_n\}=S$.*** Thus, we just proved the result you asked for.
Remark: $(x_1,\ldots,x_n)$ is a basis of $S$ if and only if $Dx(a)$ is injective and because of the two centered equations, $Dx(a)$ is injective if and only if $Dy(b)$ is injective.
*If $X$ and $Y$ are affine spaces with real and normed translation spaces $V$ and $W$, a map $f\colon D\subset X\to Y$ is called differentiable in $p\in D$, if $p$ is an inner point of $D$, $f$ is continuous in $p$ and a linear map $A\colon V\to W$ with the following property exsits: For each $0<\epsilon$, there is a $0<\delta$ such that
\begin{equation}
\|f(p+v)-f(p)-A(v)\|\leq\epsilon\|v\|
\end{equation}
for all $v\in V$ with $\|v\|<\delta$. If $f$ is differentiable in $p$, there is a unique linear map with this property, called the derivative of $Df(p)$ of $f$ in $p$.
**You can prove this by using the chain rule and the fact that a map $f\colon A\to B$ is injective if and only if there is a map $g\colon B\to A$ such that $g\circ f=\text{id}_A$ and surjective if and only if there is a map $g\colon B\to A$ such that $f\circ g=\text{id}_B$.
*** If $f\colon D\subset\mathbf{R}^n\to A$ is differentiable in $a=(a_1,\ldots,a_n)\in D$, the partial derivatives
\begin{equation}
\partial_i f(a)=\lim\nolimits_{\delta\to 0}\frac{x(a_1,\ldots,a_{i-1},a_i+\delta,a_{i+1},\ldots,a_n)-x(a_1,\ldots,a_n)}{\delta}\in V
\end{equation}
exist for all $i\in\{1,\ldots,n\}$ and
\begin{equation}
Df(a)(x_1,\ldots,x_n)=\sum_{i=1}^n x_i\cdot \partial_i f(a).
\end{equation}
