TITLE: How to convince a high school student that differentials don't work like fractions in general?
QUESTION [10 upvotes]: It all started when I tried to convince a 10th grader that if $f$ is a function defined on $\mathbb{R}^n$ the differential is defined by:
$\large \displaystyle df = \frac{\partial{f}}{\partial{x_1}}dx_1 + \frac{\partial{f}}{\partial{x_2}}dx_2 + \cdots  \frac{\partial{f}}{\partial{x_n}}dx_n$
and if $x_i = g_i(t)$ then:
$\large\displaystyle \frac{df}{dt} = \frac{\partial{f}}{\partial{x_1}}\frac{dx_1}{dt} + \frac{\partial{f}}{\partial{x_2}}\frac{dx_2}{dt} + \cdots  \frac{\partial{f}}{\partial{x_n}}\frac{dx_n}{dt}$
As he's a 10th grader, he's supposed to think of $df$ as a small change in the value of $f$ caused by a small change in $(x_1,...,x_n)$.
I have defined $df$ for a differentiable function $f: \mathbb{R} \to \mathbb{R}$ in the following naive but intuitive way and he has happily accepted this definition:
$\large \displaystyle df = \lim_{\Delta{x} \to 0} \Delta{y}$ where $\large \Delta{y} = f'(x)\Delta{x} + \epsilon(\Delta{x})\Delta{x}$ and  $\large \epsilon(\Delta{x})$ is a function of $\large \Delta{x}$ that compensates the error for turning $\large f'(x) = \displaystyle \lim_{\Delta{x} \to 0}\frac{\Delta{y}}{\Delta{x}}$ into an equality and by definition we have $\large \displaystyle \lim_{\Delta{x} \to 0}\epsilon(\Delta{x}) = 0$

Using that definition, I convinced him why the differential of a multivariable function is generalized to higher dimensions that way. But I failed to convince him why it's not a good idea to cancel $\partial{x_i}$ in the denominator with $dx_i$ just like we're dealing with fractions. I'm also afraid of proving the chain rule for him by dividing $\Delta{t}$ and then letting $\Delta{t} \to 0$. I'm looking for an easy explanation, suitable for a high school student, that convinces him why differentials shouldn't be looked at as fractions contrary to what many students think in high school.

REPLY [2 votes]: You seem to have two problems:

Your student is being misled by traditional but awful notation.
You want a nice conceptual explanation for the chain rule, rather than a nasty technical one.

I think both problems can be solved by taking a more modern approach.

I like the route Evan and John M suggested, introducing differentials via directional derivatives. I usually define $df(v)$ as the rate at which $f$ changes when you move through the domain with velocity $v$. I find that students are generally happy to accept this intuitive definition without more details. It's pretty obvious that $df(\alpha v)$ should always be equal to $\alpha df(v)$. Moreover, if $f$ is "nice enough," then $df(v + w) = df(v) + df(w)$. Notice that, in the Fréchet approach, this additivity property isn't proven from other facts—it's part of the definition of "nice enough"!
If we have a basis $e_1, \ldots, e_n$ for the domain of $f$, it's often useful to compute $df(v)$ in coordinates as
$$df(v) = df(v_1 e_1 + \ldots + v_n e_n)$$
$$= v_1 df(e_1) + \ldots + v_n df(e_n).$$
This is the first expression in your question, rewritten in more coherent notation—in particular, there's nothing that looks like a fraction.
Now, suppose we want to find the rate of change of $f$ as we move along a path $\gamma(t)$. If our "velocity through time"—the rate at which the clock is running—is $\epsilon$, then our velocity through space is $d\gamma(\epsilon)$, so the rate of change of $f$ is
$$df(d\gamma(\epsilon)).$$
Expanding $\gamma$ in coordinates as $\gamma_1 e_1 + \ldots + \gamma_n e_n$, we get
$$df(d\gamma(\epsilon)) = df(d\gamma_1(\epsilon) e_1 + \ldots + d\gamma_n(\epsilon) e_n)$$
$$= df(e_1) d\gamma_1(\epsilon) + \ldots + df(e_n) d\gamma_n(\epsilon),$$
your second equation—again without anything that looks like a fraction.

If the basis $e_1, \ldots, e_n$ comes from a coordinate system $x_1, \ldots, x_n$, it's a nice exercise to figure out that $dx_\mu(v_1 e_1 + \ldots + v_n e_n) = v_\mu$, so
$$df(v) = v_1 df(e_1) + \ldots + v_n df(e_n)$$
$$= df(e_1) dx_1(v) + \ldots + df(e_n) dx_n(v).$$
We can then write
$$df = df(e_1) dx_1 + \ldots + df(e_n) dx_n,$$
with the understanding that the argument is supposed to distribute over all the terms on the right-hand side.
If you want to turn your lesson into a Victorian-era costume drama, you can introduce the shorthand $\tfrac{\partial f}{\partial x_\mu} = df(e_\mu)$, yielding the familiar coordinate expression for $df$. I would stress that, like giant sideburns and tight-laced corsets, this fraction-like notation is not necessarily meaningful, or particularly good for you.