On Understanding "Differentials"

关于“微分”的理解 苏剑林 · 2012-12-14 · Questions calculus derivative

I've never wanted to dwell too much on textbook definitions, because I know that a genuine understanding of a definition needs to be gradually absorbed through long-term practice in application. So the only thing we really need to do is get on with our research.

However, a few friends recently asked me about my understanding of differentials — for instance, "does dx have to be very small?" — and so I decided to write down my thoughts on it here.

Something closely tied to differentials, and something we're all quite familiar with, is of course the "increment," such as $\Delta y$, $\Delta x$, and so on — an increment can clearly be arbitrarily large (as long as the independent variable stays within the domain). Now consider a function $y=f(x)$: how does the differential of the function arise? It's because directly studying the increment of a function is rather cumbersome, so we introduce the differential dy: when $\Delta x$ is small, it represents the leading term of the increment: $\Delta y=dy+o(\Delta x)=A \Delta x+o(\Delta x)$, where A is some constant.

We usually say that $o(\Delta x)$ is a higher-order infinitesimal compared with $\Delta x$, but this result is not very intuitive. $o(\Delta x)$ is usually written as $O(\Delta x^2)$, and it means that there exists some positive constant k such that $\Delta y-dy$ can be bounded within $k \Delta x^2$.

Now, $dy$ is the leading term of $\Delta x$, and since $\Delta x$ can be arbitrarily large, dy can of course also be arbitrarily large. As for the independent variable x itself, it can be regarded as the function $y=f(x)=x$, so its differential is obviously $dx=\Delta x$, with A=1; and we already know that for other functions, $A=f'(x)$, i.e., $dy=f'(x)dx=f'(x)\Delta x$. Clearly, as long as the derivative is nonzero, dy can be arbitrarily large, because $\Delta x$ can take any value.

Many people think that dx and dy must always be small, mainly because of the following rule:

$$\begin{aligned}f'(x)=\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} \\ f'(x)=\frac{dy}{dx}\end{aligned}$$

so dx and dy correspond to $\Delta x \to 0$ and $\Delta y \to 0$. That's how I understood it at first too, and it seemed fine — but it was only after studying university calculus, and later trying to make sense of concepts in quantum mechanics, that I confirmed I had been wrong. The understanding given above is the correct one. If someone asks why $f'(x)=\frac{dy}{dx}$, it's much cleaner to answer simply that this is because the very definition of dy is $dy=f'(x)dx$! That way we don't get things backwards.

Some readers might raise the following objection: for the composite function $y=f(g(x)),u=g(x)$, we know by the chain rule that:

$$\begin{aligned}dy=f'(u)u' dx=f'(u)u' \Delta x \\ dy=f'(u)du=f'(u) \Delta u\end{aligned}$$

Doesn't that give us $\Delta u=u' \Delta x$? Isn't that an obvious contradiction?

Indeed, this is a contradiction that's easy to spot. But it's really just because our notation is too crude. You see, the differential of a function depends on which variable is treated as independent. When x is the independent variable, we have

$$\Delta y=dy+o(\Delta x)$$

but when u is the independent variable, we instead get

$$\Delta y=dy+o(\Delta u)$$

Since u and x are not equal, how could we possibly assume that the "dy" in these two expressions is the same thing? At this point, I imagine many readers who had just started to see the light might feel confused again: if dy differs depending on the variable, why doesn't the textbook use different notation to distinguish them? This is actually a matter of an unspoken convention — much like how we use a prime mark to denote the derivative of a function, $f'(x)$, without specifying explicitly which variable we're differentiating with respect to; we simply take it for granted that whatever appears inside the parentheses is the variable in question. Likewise, although dy varies depending on which variable it's paired with, whenever it's actually used it's always accompanied by dx, du, or some other differential — and that pairing effectively specifies the independent variable. Moreover, even though the various forms of dy differ from one another, the difference between them is only a second-order infinitesimal, so using the same notation for all of them doesn't actually cause any confusion.