When given a function of multiple variables, is it possible to differentiate all the variables simultaneously? I understand that when given a function of two variables, for example $f(x,y)=x^2y+xy^2$, you can perform partial differentiation on it to get $(2xy+y^2,y+2xy)$. However, to my understanding, this is essentially seeing how the function changes when moving either along the $x$-axis or the $y$-axis. One or the other, not both simultaneously. Just as when differentiating a function of one variable we can get a tangential line, is it possible to simultaneously differentiate both variables (instead of differentiating one and then the other) of a two-variable function to get a tangential surface?
I apologize if my understanding of this concept is wrong, any help and/or corrections would be greatly appreciated. If my question seems confusing, please let me know. Thanks.
 A: If $f\colon\mathbb{R}^2\to\mathbb{R}$ is differentiable at a point $(x_0,y_0)$, then
$$
f(x_0+h,y_0+k)-f(x_0,y_0)=\frac{\partial f}{\partial x}(x_0,y_0)\,h+\frac{\partial f}{\partial y}(x_0,y_0)\,k+\epsilon(h,k),
$$
where
$$
\lim_{(h,k)\to(0,0)}\frac{\epsilon(h,k)}{\sqrt{h^2+k^2}}=0.
$$
The partial derivatives give you the ghange in the value of the function when the two variables change independently. Of course this generalizes to any number of variables.
A: Yes, this can be done and has been done. The basic idea is the following: 

Definition. Let $U \subseteq \mathbf R^d$ be open and $f \colon U \to \mathbf R$. $f$ is called differentiable at $x \in U$, if there is a linear map $f'(x) \colon \mathbf R^d \to \mathbf R$ such that 
  $$ f(x+h) = f(x) + f'(x)h + o(|h|), \quad h\to 0 $$

The linear map $f'(x)$, which is called the differential (or derivative) of $f$ at $x$, as in the $1$d-case describes the tangential surface, which is the image of $h \mapsto f(x) + f'(x)h$. 
The connection with the partial derivatives is through 

Theorem. Let $U \subseteq \mathbf R^d$ be open and $f \colon U \to \mathbf R$ and $x \in U$. If all partial derivatives exist in some neighbourhood of $x$ and are continuous, then $f$ is differentiable at $x$ with derivative
  $$ f'(x)h = \sum_{i=1}^d \partial_i f(x) h_i, \qquad h \in \mathbf R^d $$

