Differential of a rotated f(x, y) surface I often hit this problem : 
Consider a surface defined by the equation $z = f(x, y)$,
the differentials of this function are $\frac{\partial f}{\partial x}\mathrm{d}x$ and $\frac{\partial f}{\partial y}\mathrm{d}y$. 
But suppose that the surface is rotated by any space rotation $R$  : 
the points of the surface are now $\left(\matrix{x'\\y'\\z'}\right) = R\left(\matrix{x\\ y\\ f(x, y)}\right)$. 
What can be said about the differentials of the rotated surface?
I would like to write something about $\mathrm{d}x'$, $\mathrm{d}y'$ and $\mathrm{d}z'$ for the points on the rotated surface... 

Edit : 
I am looking for an expression of $\frac{\partial z'}{\partial x'}$ and $\frac{\partial z'}{\partial y'}$, which is the differential of the rotated surface as a function of $x'$ and $y'$.
 A: Your map $F(x,y) = R\left(\matrix{x\\ y\\ f(x, y)}\right)$ is the function composition $R \circ G$ where $R$ is a linear map (the rotation) and $G: (x,y) \mapsto (x,y,f(x,y))$.
Then you probably know what the derivative of a function composition is?
If not, the derivative of $F$ is $F^\prime=R^\prime \circ G^\prime$. As $R$ is linear, it is equal to its derivative. Hence $F^\prime = R \circ G^\prime$.
Can you figure out what is the derivative of $G$ which is a map from $\mathbb{R}^2$ to $\mathbb{R}^3$?
You're right $\frac{\partial G}{\partial x} = \left(\matrix{1 \\ 0 \\ \frac{\partial f}{\partial x}}\right)$ so: $G^\prime(x,y) = \left(\matrix{1 & 0\\ 0 & 1 \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y}}\right)$.
Finally:
$$F^\prime(x,y) = R \circ\left(\matrix{1 & 0\\ 0 & 1 \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y}}\right)$$
Now considering your additional question in the edit part.
You have $(x^\prime, y^\prime, z^\prime)=F(x,y)$. You can rewrite the equality as:
$(x^\prime, y^\prime)=\varphi_1(x,y)$ and $z^\prime=\varphi_2(x,y)$.
Providing $\varphi_1$ is locally invertible at $(x,y)$, this is equivalent (locally) to $(x,y)=\varphi_1^{-1}(x^\prime, y^\prime)$ and $z^\prime=\varphi_2(x,y) = (\varphi_2 \circ \varphi_1^{-1})(x^\prime,y^\prime)$.
Using theorems on composition and inverse of derivatives you can then find $\frac{\partial z'}{\partial x'}$ and $\frac{\partial z'}{\partial y'}$.
A: Your surface $S$ initially has a representation of the form
$$F:\quad {\bf w}=(u,v)\mapsto {\bf x}=\bigl(u,v,f(u,v)\bigr)\ .$$
You then rotate $S$ in space by means of some rotation map $R$, resp. some orthogonal matrix $[R]$:
$${\bf x}\mapsto {\bf x}'=R{\bf x}\ .$$
The rotated surface $S':=R(S)$ then has a representation
$$G:=R\circ F:\quad {\bf w}\mapsto {\bf x}'=R\bigl(F({\bf w})\bigr)\ .$$
The chain rule then says that the derivative of $G$ is given by
$$dG({\bf w})=dR\bigl(F({\bf w})\bigr)\circ dF({\bf w}\bigr)\ .$$
Since $R$ is a linear map one has $dR({\bf x})=R$ for all ${\bf x}$, so that in terms of matrices we simply get
$$\bigl[dG({\bf w})\bigr]= [R]\>\left[\matrix{1&0\cr 0&1\cr f_{.1}(u,v)&f_{.2}(u,v)\cr}\right]\ .$$
