having a question on the symbol $dN_p$ when writing down its correspondence matrix My question is about the differential of the gauss map $dN_{p}$, to start the convenient expression of the symbol and formula, I have to construct some functions(maps).
First, there is a one dimension parameters line and I use the parameter as $t$, and the function $\overline{α}(t)=(u(t),v(t))$ maps $t$ to $uv$ plane, and which is a plane curve on $uv$ plane.
Second, there is a function $\mathbf{x}$ that takes $\overline{α}(t)$ to a curve on a regular surface, $α(t)=\mathbf{x}(\overline{α}(t))=(x(u(t),v(t)),y(u(t),v(t)),z(u(t),v(t)))$.
Final, there is a map $\mathbf{N}$ represents the unit normal vector on each point of the surface (x,y,z), $N(x,y,z)=(X(x,y,z),Y(x,y,z),Z(x,y,z))$.
If $p$ is a point on the surface(i.e.,$p=α(0)$),then, the linear map $dN_p$ corresponds a 3x3 matrix:
\begin{bmatrix}X_x&X_y&X_z\\Y_x&Y_y&Y_z\\Z_x&Z_y&Z_z\end{bmatrix}
The matrix derive from the chain rule in calculus:
$\frac{dN(α(t))}{dt}=(X_{x}*x'(t)+X_{y}*y'(t)+X_{z}*z'(t), Y_{x}*x'(t)+Y_{y}*y'(t)+Y_{z}*z'(t), Z_{x}*x'(t)+Z_{y}*y'(t)+Z_{z}*z'(t))=\begin{bmatrix}X_x&X_y&X_z\\Y_x&Y_y&Y_z\\Z_x&Z_y&Z_z\end{bmatrix}\pmatrix{x'(t)\\y'(t)\\z'(t)}=(dN)_{3x3}(α'(t))$, here, $x'(t)=\frac{dx(u(t),v(t))}{dt},y'(t)=\frac{dy(u(t),v(t))}{dt},z'(t)=\frac{dz(u(t),v(t))}{dt},$
Now ,I have the question about the partial derivative $\pmatrix{X_x\\Y_x\\Z_x}$,  the definition of it is 
$$\lim_{\Delta t\to 0} \frac{(X(x+Δt,y,z),Y(x+Δt,y,z),Z(x+Δt,y,z))-(X(x,y,z),Y(x,y,z),Z(x,y,z))}{\Delta t},$$ 
but even $(X(x+Δt,y,z),Y(x+Δt,y,z),Z(x+Δt,y,z))$ may not defined because the domain of the map $\mathbf{N}$ is only defined on the surface, not a solid domain, so if we move a little $Δt$, it may go out of the domain(the domain is only on the surface).So, I'm not really sure how we interpret the correspondence matrix of the linear map $dN_p$, maybe there are some concepts need to be clarify, thanks for help.

Thanks Andrew and janmarqz, I think that we couldn't express the correspondence of the linear map $dN_p$ in this way(using (x,y,z) domain when we want to differentiate it at a point on the surface) because as I mention above, the partial derivative column vectors may not exist.
  Instead, we can explained the $dN_p$ is in the form: $dN_{\{u,v\}} \circ \mathbf{τ_{β}}$ where $β=\{x_u, x_v\}$ is the basis of $T_{p}(S)$, and $τ_{β}$ is a linear transformation that express the vector in $T_{p}(S)$ in the basis $\{x_u, x_v\}$ as its coordinate.
$dN_{\{u,v\}}\circ τ_{β}(x_{u}*u'+x_{v}*v')=dN_{\{u,v\}}\left(\begin{array}{c} u'\\ v'\end{array}\right)=N_{u}*u'+N_{v}*v'$, so in this way, the problem is solved and $N_u$ and $Nv$ both exist because $N_u=\frac{\partial\frac{x_{u} \times x_{v}}{\|x_{u} \times x_{v}\|}}{\partial u}$, and I think that it exist. 
  Do others agree with my explaination?

 A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Vec}[1]{\mathbf{#1}}$For definiteness, let $M \subset \Reals^{3}$ be an oriented regular surface, $S^{2} \subset \Reals^{3}$ the unit sphere, and $N:M \to S^{2}$ the Gauss map, whose value at each point $p$ of $M$ is the unit normal vector $N(p)$ associated to the orientation of $M$ (interpreted as a point of the unit sphere).
The differential $dN_{p}$ is a linear transformation from the tangent plane $T_{p}M$ of $M$ at $p$ to the tangent plane $T_{N(p)}S^{2}$ of the sphere at $N(p)$.
If you're trying to represent $N$ in coordinates so you can do calculus, the usual method is to pick a parametrization of $M$ in some neighborhood of $p$. That is, find a non-empty (connected) open set $U \subset \Reals^{2}$ and an injective regular mapping $\Vec{x}:U \to M$ whose image contains $p$. Up to an overall sign,
$$
N \circ \Vec{x}
  = \frac{\Vec{x}_{u} \times\Vec{x}_{v}}{\|\Vec{x}_{u} \times\Vec{x}_{v}\|}.
$$
If $\Vec{x}(u_{0}, v_{0}) = p$, then $dN_{p}$ is (represented by) the differential of this mapping evaluated at $(u_{0}, v_{0})$.
A: By the chain's rule on a composition  $\beta:I\to\Omega$, 
$\mathbf x:\Omega\to\Bbb R^3$ and $N:\Bbb R^3\to\Bbb R^3$, where $I$ is an interval and $\Omega$ is a domain in $\Bbb R^2$ one can control how $N$ varies along tangent directions or along curves in the surface.
So, $N\circ\mathbf x\circ\beta$ gives you the normal vector to a curve $\mathbf x\circ\beta$ in the surface parameterizated by $\mathbf x$. 
But if $C=\mathbf x\circ\beta$ then $d(N\circ C)_t=dN_p\cdot C'(t)$.
So it happens that $d(N\circ\mathbf x)=dN\cdot d\mathbf x$ serves to
connect 
$\Bbb R^2\stackrel{d(N\circ\mathbf x)_p}\longrightarrow\Bbb R^3$ 
with 
$\Bbb R^2\stackrel{d\mathbf x_a}\longrightarrow\Bbb R^3$
and 
$\Bbb R^3\stackrel{dN_p}\longrightarrow\Bbb R^3$ in composition.
Now if $\beta$ goes over some one of the $v$ or $w$ directions, we get
$d(N\circ C)_t=dN_p\cdot \partial_v$ or $dN_p\cdot \partial_w$, being $\partial_v$ and $\partial_w$ the tangent space basis.
Since the tangent space $T_p\Sigma$ is the ${\rm im}\ d\mathbf x_a$ in $\Bbb R^3$ and is affine to $p$, one can consider $dN_p$ as a transformation $T_p\Sigma$ into itself since, as a little calculus shows $dN_p$ is perpendicular to the tangent space. That is, we have
$N\cdot N=1$ then $d(N\cdot N)=dN\cdot N+N\cdot dN=2dN\cdot N$ hence $dN\cdot N=0$. 
