I a variable-centric approach to algebra, one generally works with a collection of interdependent variables.
For example, consider modeling a problem where you have a point that loops counter-clockwise around the unit circle at a constant rate. One might introduce three separate variables:
- $t$, the current time
- $x$, the first coordinate of the point's position
- $y$, the second coordinate of the point's position
and these are interrelated by various equations:
$$ x = \cos(t) \qquad y = \sin(t) \qquad x^2 + y^2 = 1 $$
There is a gadget called a "differential", which we write as $\mathrm{d}x$, which captures the idea of "the rate at which $x$ varies", in an absolute way.
Now, a differential is not a number; it's a new kind of mathematical gadget. Introductory calculus classes generally don't teach it; instead they want a way to do calculus that avoids working with them.
It turns out that differentials satisfy analogs of the basic derivative laws; the differentials of the above equations are
$$ \mathrm{d}x = -\sin(t) \, \mathrm{d}t
\qquad \mathrm{d}y = \cos(t) \, \mathrm{d}t
\qquad 2x \, \mathrm{d}x + 2y \, \mathrm{d}y = 0 $$
The main thing to observe here is that, in this situation, the differentials are all proportional to one another; we can meaningfully ask for their ratios, and get an informative result that doesn't involve differentials at all in their expression. We can solve these equations to get
$$ \frac{\mathrm{d}x}{\mathrm{d}t} = -\sin(t)
\qquad \frac{\mathrm{d}y}{\mathrm{d}t} = \cos(t)
\qquad \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{x}{y} $$
Since we're comparing the proportion between the rate at which two variables change, these derivatives are, respectively, the "derivative of $x$ with respect to $t$", "derivative of $y$ with respect to $t$", and "derivative of $y$ with respect to $x$".
In these terms, the chain rule is simply the algebraic property of chaining proportions. And you can check, for example,
$$ \frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}x}{\mathrm{d}t} = \left(-\frac{x}{y}\right) \cdot (-\sin(t)) = \frac{\cos(t)}{\sin(t)} \cdot \sin(t) = \cos(t) $$
Now, to connect it all back to the familiar definition of derivative in terms of functions it turns out that you have the following theorem:
Theorem: If $x$ and $y$ are related by the equation $y = f(x)$ where $f$ is differentiable, then $$\frac{\mathrm{d}y}{\mathrm{d}x} = f'(x) $$