TITLE: Intuition of multivariable chain rule
QUESTION [12 upvotes]: I was learning/reviewing the chain rule for multivariable calculus and was wondering why the multivariable calculus chain rule is a function of summation of products of derivatives rather than just product of derivatives, like its single variable counter part.
In particular I want to fix how I think of the chain rule and generalize my thoughts/intuition to multiple variables. 
Usually the way I used to remember the chain rule was by the usual "trick" of "canceling out" the middle dummy variables i.e. consider y = f(x(t)), then:
$$ \frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt} $$
and since the dx's cancel out, the chain rule works! Wohooo, super intuitive, easy to remember and even though its not mathematically rigorous, at least it sort of makes sense.
However, for multiple variables the equation looks very different. Consider $z = f(x(t), y(t) )$, then its chain rule derivative is:
$$ \frac{dz}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} $$
Even though there is some of the same "canceling" trick, the equation doesn't quite make as much intuitive sense to me or where it came from.
So what is the intuition behind this equation? Why is it a summation of products of derivatives? Anyone have a good way of generalizing the intuition from one variable to multiple variables? Or maybe we have to change our intuition in a significant way and that is fine, as long as its more useful for multiple variable calculus!

Quick comment, by intuition, I don't necessarily mean analogies to physics, but it can be to conceptual ideas in mathematics. So bringing in explanations say of real analysis and linear algebra that appeal the intuition/concepts of those areas are welcome! We all have different type of intuitions. :)

REPLY [5 votes]: $\frac{dz}{dt}$ measures what changes about $z$ when you change $t$ a little. Since $z = f(x(t), y(t))$, two things to the input of $z$ change when you change $t$ a little: $x$ changes a little, and $y$ changes a little. Both of these changes affect $z$ a little. The first term is the part of the total change in $z$ coming from $x$ changing, and the second term is the part of the total change in $z$ coming from $y$ changing. 
Symbolically (writing $\partial (-)$ for a small change in $(-)$):
$$z + \delta z = f(x(t + \delta t), y(t + \delta t)) \approx f(x(t) + \delta x, y(t) + \delta y) \approx f(x(t), y(t)) + \frac{\partial f}{\partial x} \delta x + \frac{\partial f}{\partial y} \delta y$$
where $\delta x \approx \frac{dx}{dt} \delta t$ and $\delta y \approx \frac{dy}{dt} \delta t$. 
The canceling trick is a very bad idea; it sort of works for ordinary calculus but as you can see it fails very badly for multivariable calculus. The problem is that partial derivatives do not behave anything like fractions. What they actually behave like are coefficients of a matrix (and the general form of the chain rule involves matrix multiplication); this will become much clearer if you take a linear algebra course.