Question about the differential function on a surface My question is related to this Show that $\text{grad }f=\frac{f_uG-f_vF}{EG-F^2}\textbf{x}_u+\frac{f_vE-f_uF}{EG-F^2}\textbf{x}_v$
In both answers it's assumed that $df_{p}(X_u) = f_{u}$ and $df_{p}(X_v) = f_{v}$ but I don't know why this is true. I'm even confused about what $f_u$ and $f_v$ mean.
I know that $df_{p}(v)$ is such that $df_{p}(v)$ : $T_{p}S$ $\rightarrow$ $\mathbb{R}^3$ and that $\{X_u,X_v\}$ is a basis for $T_{p}S$ but I don't know why mapping basis vectors gives us $f_u$ and $f_v$
 A: It's sort of by definition. Given a parametrization ${\bf x} = {\bf x}(u,v)$ for a surface $S$ and $f:S \to \Bbb R$, one writes $$\frac{\partial f}{\partial u}(q) \doteq \frac{\partial (f\circ {\bf x})}{\partial u}({\bf x}^{-1}(q)) $$for any $q$ in the image of ${\bf x}$. In other words, partial derivatives of a function defined on a surface only make sense once a coordinate system has been chosen. What happens is that people don't have the patience to be precise all the time, so they identify $f$ with $f\circ {\bf x}$ and then you have to struggle to understand what is going on. For a function $F:\Bbb R^n \to \Bbb R$, in general, one has $DF(p)(e_i) = (\partial F/\partial x_i)(p)$, so applying this to the local expression $f \circ {\bf x}$ gives that $${\rm d}f({\bf x}_u) = {\rm d}f({\rm d}{\bf x}(1,0)) = {\rm d}(f \circ {\bf x})(1,0) = \frac{\partial f}{\partial u} = f_u.$$Note that I'm supressing base points above. In any case, ${\rm grad}(f)\circ {\bf x}$ is a combination of ${\bf x}_u$ and ${\bf x}_v$, so write $${\rm grad}(f)\circ {\bf x} = A{\bf x}_u+B{\bf x}_v,$$for functions $A$ and $B$ to be found. Taking products with ${\bf x}_u$ and ${\bf x}_v$ gives $$f_u = EA + FB \quad \mbox{and}\quad f_v = FA + GB.$$So $$\begin{pmatrix}E & F \\ F & G \end{pmatrix} \begin{pmatrix} A \\ B \end{pmatrix} = \begin{pmatrix} f_u \\ f_v \end{pmatrix}\implies \begin{pmatrix} A \\ B \end{pmatrix} = \begin{pmatrix} E & F \\ F & G  \end{pmatrix}^{-1}\begin{pmatrix} f_u \\ f_v \end{pmatrix}.$$This gives the expressions for $A$ and $B$.
