TITLE: Differentials Definition
QUESTION [16 upvotes]: Please define differentials rigorously such that they give a consistency to their use in the following links. 
I have read 

Is $\frac{\textrm{d}y}{\textrm{d}x}$ not a ratio?
What is the practical difference between a differential and a derivative? 
Differential of a function at Wikipedia.
If $\frac{dy}{dt}dt$ doesn't cancel, then what do you call it? 
Leibniz's notation at Wikipedia.
Exact differential equation at Wikipedia.
Moment of inertia at Wikipedia.
Center of mass at Wikipedia, etc.

REPLY [7 votes]: The seemingly endless number of vague conceptions of "differentials" bothered me a great deal when I first learned calculus -- so much so that I spent an inordinate amount of time thinking about all this.  You say you want rigor?  OK, then let's be precise.
As I see it, there are three genuinely distinct ways of making the notion of "differentials" precise.  However, there are a number of caveats, which I'll mention at the end.
(1) As "infinitesimal quantities."
This notion can be made precise via Abraham Robinson's "non-standard analysis."  I won't go into this further, except to reiterate that this is different from the next two ideas.
(2) As a "total derivative."
This is the notion explained by Mate Kosor.  The precise definition is as follows:

Definition: Let $f \colon \mathbb{R}^n \to \mathbb{R}^m$ be a function.  The (total) differential (or total derivative) of $f$ at a point $p \in \mathbb{R}^n$ (if it exists) is the (unique) linear map $df_p \colon \mathbb{R}^n \to \mathbb{R}^m$ such that
  $$\lim_{h \to 0} \frac{|f(p +h) - f(p) - df_p(h)|}{|h|} = 0.$$
  We say that $f$ is differentiable at $p \in \mathbb{R}^n$ if such a linear map $df_p \colon \mathbb{R}^n \to \mathbb{R}^m$ exists.

It can be proven (I will not do this) that the above definition is well-defined -- that is, there is at most one such linear map $\lambda_p \colon \mathbb{R}^n \to \mathbb{R}^m$ satisfying
$$\lim_{h \to 0} \frac{|f(p +h) - f(p) - df_p(h)|}{|h|} = 0.$$
It is a fact (which I will also not prove) that if $f \colon \mathbb{R}^n \to \mathbb{R}^m$ is differentiable, then (with respect to the standard bases) the matrix of the linear map $df_p \colon \mathbb{R}^n \to \mathbb{R}^m$ is the Jacobian matrix:
$$df_p = \begin{pmatrix}
\frac{\partial f^1}{\partial x^1} & \cdots & \frac{\partial f^1}{\partial x^n} \\
\vdots & & \vdots \\
\frac{\partial f^m}{\partial x^1} & \cdots & \frac{\partial f^m}{\partial x^n}
\end{pmatrix}$$
Note that in the case of real-valued functions $f \colon \mathbb{R}^n \to \mathbb{R}$, we have
$$df_p(h) = \begin{pmatrix}
\frac{\partial f}{\partial x^1} & \cdots & \frac{\partial f}{\partial x^n}
\end{pmatrix}
\begin{pmatrix}
h_1 \\
\vdots \\
h_n
\end{pmatrix} = \frac{\partial f}{\partial x^1}h_1 + \cdots + \frac{\partial f}{\partial x^n}h_n.$$
I highlight this special case in order to draw an analogy with the classical equation
$$dy = \frac{\partial f}{\partial x^1}dx^1 + \cdots + \frac{\partial f}{\partial x^n}dx^n.$$
Note also the case of a function $f \colon \mathbb{R} \to \mathbb{R}$.  In this case, the (total) differential of $f$ is the $1 \times 1$ matrix
$$df_p = (f'(p)).$$
That is, $df_p(h) = f'(p) h$.
(3) Via differential forms.
This is the notion referenced in the first paragraph of Christian Blatter's answer (but not the rest).

Definition: A differential $k$-form on $\mathbb{R}^n$ is a smooth function whose inputs are points $p \in \mathbb{R}^n$ and whose outputs are alternating $k$-multilinear functions $\omega_p \colon T_p\mathbb{R}^n \times \cdots \times T_p\mathbb{R}^n \to \mathbb{R}$.
Here, $T_p\mathbb{R}^n$ is the tangent space of $\mathbb{R}^n$ at the point $p \in \mathbb{R}^n$.

Examples: (I have to go now, but I will fill this in later.)
Exterior differentiation and the letter $d$: (I have to go now, but I will fill this in later.)
Caveat: Unfortunately, there are a couple of instances where the notation "$ds$", "$dS$," "$dA$," or "$dV$" is not actually $d$ of anything.  This is the case for a certain class of differential forms called "volume forms," and (to my understanding) is the topic of Chrstian Blatter's answer.
(In the case of un-oriented manifolds, this notation failure can also be rectified by means of "densities."  I will not go into this here.)