Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Isn't the whole point of differential notation that, when the behind the scenes machinery is built up, you can treat them like fractions? – Neal Aug 30 '13 at 0:12
@Neal: Maybe Leibnitz had that idea in mind when he used the notation without worries, but that was almost 4 centuries ago I think. Do you think that he can cancel $\partial{x_i}$ and $dx_i$ to get $df = n \partial{f}$? – some1.new4u Aug 30 '13 at 0:15
He cannot cancel in the sense that he's not cancelling fractions. However the point of the notation is to cancel as if you are working with fractions. Its understood that these are not fractions but differentials but as Neal points out, that becomes behind the scenes and the idea behind the notation is to let us put that aside and perform calculations and manipulations more efficiently. – frogeyedpeas Aug 30 '13 at 0:20
Your best option is to explain how dy and dx are meaningless by themselves (just 0). – frogeyedpeas Aug 30 '13 at 0:21
If he thinks that differentials are just like fractions, then how does he do that with second derivatives? – imranfat Aug 30 '13 at 1:08
up vote 14 down vote accepted

A standard example is the equation $PV = T$. Note that

$$P = \frac{T}{V} \implies \frac{\partial P}{\partial V} = -\frac{T}{V^2}$$ $$V = \frac{1}{P}T \implies \frac{\partial V}{\partial T} = \frac{1}{P}$$ $$T = PV \implies \frac{\partial T}{\partial P} = V$$ so $$\frac{\partial P}{\partial V} \frac{\partial V}{\partial T} \frac{\partial T}{\partial P} = -\frac{T}{V^2}\frac{1}{P}V = -\frac{T}{PV} = -1.$$

Edit: There's also the chain rule. If $f$ is a function of two variables, say $f(u,v)$, where both $u$ and $v$ are themselves functions of two variables (say $u=u(x,y)$ and $v=v(x,y)$), then the chain rule is

$$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x}.$$

If we could just cancel the $\partial u$'s and $\partial v$'s, we'd get the absurd $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial x}$.

Admittedly, this is not the most conceptual explanation, but I imagine it'll convince quite a few high school (and college) students.

share|cite|improve this answer
Ah, the negative 1 rule from my engineering math book. Briliant !!! – imranfat Aug 30 '13 at 1:22
Although it is a cute example, I feel like it will just leave a gap somewhere. As in, why didn't the cancellations work out? What's really going on in that example? It's clear that many identities fall apart allowing for cancellations already. I think the crux lies in the right interpretation of partial differentiation: $\partial T / \partial P$ controls how $T$ changes to small changes in $P$, leaving other variables fixed. So already for $(\partial V / \partial T)(\partial T / \partial P)$, you may not in general recover $\partial V / \partial P$ (And in fact you don't)... – Evan Sep 1 '13 at 2:16
... so when we write $\partial T$, as a quantity by itself it makes no sense because it is tied to a particular variable downstairs. On the other hand, the notation $dT$ (well, okay, for the one-variable case) has no ambiguity because there is only one variable of dependence. I think this is a crucial difference to why cancellations work in one case but not the other. Just thought this is worth pointing out. (For instance, for parametric $(x(t),y(t))$, cancellations do work out when computing $dy/dx$ at a point $t$ from $dx/dt,dy/dt$) – Evan Sep 1 '13 at 2:18
@Evan: I like your point about $\partial T$ being "tied to a variable downstairs." However, I'd go further and say that it's not just this particular example that leaves a gap in students' understanding, but the entire notational setup. I mean, until one can actually define rigorously what $dx$ and $df$ mean, there's no use in asking whether or not we can cancel these symbols. Symbols are meaningless until we give them concrete meaning. – Jesse Madnick Sep 1 '13 at 6:19
@JesseMadnick All good points. It is very interesting that differential forms make them work, though pretty inaccessible. For practical purposes I think nobody should be taking the full limit to $0$ and stick with approximations. For instance the OP's definition of df I very strongly object to, but has good intentions (approximation type term good, full limit to 0 bad, the friend may as well ask why isn't df just 0??) – Evan Sep 1 '13 at 7:07

I think you can use the following strategy:

Your differential

$df = f_1 dx_1 + \ldots f_n dx_n$

Shows how $f$ changes to small changes in the coordinates. However, these coordinates can change independently of one another, so it is important to reason about how much each one is changing by...

In the formula

$df/dt = f_1 dx_1/dt + \ldots + f_n dx_n/dt$,

Note you are asking how $f(x(t))$ changes with $t$. However, for $x=g(t)$, each direction is changing at different rates described by $dx_i/dt$. So what would one be cancelling anyway?

As an example, you can use different examples of $g$ to illustrate this point. Use $g$ that only changes in the $x_i$ direction. When that happens, every other term but $f_i dx_i/dt$ disappears.

The point is, when you generalize to higher dimensions, you have to consider each independent variable separately for the differential. In fact, he might have asked from the beginning why not $df = n \partial f$, whatever that means? Hope this helps point in a right direction.

Edit: Adding along to this theme, it might be illuminating to show him how directional derivatives work, because it again illustrates the same point.

share|cite|improve this answer

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.

share|cite|improve this answer

In your example, I like to think of the differential $df$ (evaluated at a point) as a linear map from $\mathbb{R}^n$ to $\mathbb{R}$, which in turn you can think of as a row vector, the row vector being $$\left[\frac{\partial f}{\partial x_1}, \dots, \frac{\partial f}{\partial x_n} \right] = \frac{\partial f}{\partial x_1}dx_1 + \dots + \frac{\partial f}{\partial x_n}dx_n.$$ Then the $dx_i$ are just the basis elements, e.g. $$dx_1 = [1, 0, \dots, 0].$$ This approach then more easily generalizes to functions $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$.

share|cite|improve this answer
Unfortunately that is too abstract for a high school student I think. – some1.new4u Aug 30 '13 at 0:51
You may be be right. But without some linear algebra, there is not much point to trying to learn multivariable calculus. – John M Aug 30 '13 at 1:01
Let me put it this way: Derivatives are all about finding the best linear approximation to a function. Since this is a linear relationship, there is always a linear map involved. Spending time on that concept will pay off hugely later when you later are trying to explain why the chain rule works. Ultimately, the "why" is always more important than the "how". – John M Aug 30 '13 at 1:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.