Decoding Leibniz notation – Slava Akhmechet

I wrote this for myself to understand the Leibniz notation. Prerequisites for this post are the definition of the derivative and the Lagrange notation. If you don’t understand these yet, please study them first.

So…

You may have already seen something like $\frac{dy}{dx}$

This flexibility makes the notation very useful in science and engineering, but also makes it difficult to learn. I explore it here to make learning easier.

Historical motivation

We start with the historical interpretation, where the notation began. Leibniz didn’t know about limits. He thought the derivative is the value of the quotient

${\displaystyle \frac{f(x+h)-f(x)}{h}}$

when $h$ is “infinitesimally small”. He denoted this infinitesimally small quantity of $h$ by $dx$, and the corresponding difference $f(x+dx)-f(x)$

${\displaystyle \frac{df(x)}{dx}=f^{\mathrm{\prime }}}$

Intuitively, we can think of $d$ in a historical context as “delta” or “change”. Then we can interpret this notation as Leibniz did– a quotient of a tiny change in $f(x)$ and a tiny change in $x$. But this explanation comes with two important disclaimers.

First, $d$ is not a value. If it were a value, you could cancel out $d$’s in the numerator and the denominator. But you can’t. Instead think of $d$ as an operator. When applied to $f(x)$ or $x$, it produces an infinitesimally small quantity. Alternatively you can think of $df(x)$ and $dx$ as one symbol that happens to look like multiplication, but isn’t.

Second, $\frac{df(x)}{dx}$

${\displaystyle {\frac{df(x)}{dx}\mid }_{x=a}=f^{\mathrm{\prime }}(a)}$

Writing all that is a pain and in practice people rarely do it this way, but we’ll get to that in a minute.

Modern interpretation

To summarize, the full and unambiguous Leibniz notation is:

${\displaystyle \frac{df(x)}{dx}=f^{\mathrm{\prime }}\text{and}{\frac{df(x)}{dx}\mid }_{x=a}=f^{\mathrm{\prime }}(a)}$

In modern mathematics real numbers do not have a notion of infinitesimally small quantities. Thus in a modern interpretation we treat $\frac{df(x)}{dx}$

Second derivative

A question arises for how to express the second (or nth) derivative in the Leibniz notation. Let $g(x)=$

${\displaystyle \frac{d\left(\frac{df(x)}{dx}\right)}{dx}=f^{\mathrm{\prime }\mathrm{\prime }}}$

Of course this is too verbose and no one wants to write it this way. This is where the vagaries begin. For convenience people use the usual algebraic rules to get a simpler notation, even though formally everything is one symbol and you can’t actually do algebra on it:

${\displaystyle \frac{d\left(\frac{df(x)}{dx}\right)}{dx}=\frac{d^{2}f(x)}{d{x}^2}}$

Two questions arise here.

First, why $dx^2$

Another probably more honest way to answer this question is to recall that this isn’t real algebra– we just use a simularcum of algebra out of convenience. But convenience is a morally flexible thing, and people decided to drop parentheses because they’re a pain to write. So $(dx)^2$

Second, we said before that $df(x)$ can be thought of as one symbol. Then what is this $d^2$

Liberties and ambiguities

There are a few more liberties people take with the Leibniz notation. Let $f(x)=x^2$

${\displaystyle \frac{df(x)}{dx}\text{or}\frac{d{x}^2}{dx}}$

Here $\frac{d{x}^2}{dx}$

Suppose we wanted to state what the derivative of $f$ at a point $a$ is. In Lagrange notation we say $f’(a)=2a$

${\displaystyle {\frac{df(x)}{dx}\mid }_{x=a}=2a}$

But this is obviously a pain, so people end up taking two liberties. First, everyone drops the vertical line that denotes the application at $a$. So in practice the form above becomes:

${\displaystyle \frac{df(x)}{dx}=2x}$

This shouldn’t “compile” because $=f’$

Second, people decided that writing $\frac{df(x)}{dx}$

To summarize what we have so far:

${\displaystyle {\frac{df(x)}{dx}\mid }_{x=a}=2a\text{becomes}\frac{df}{dx}=2x}$

Even more liberties

You’d think that we already pushed the notation past all limits of propriety, but scientists and engineers manage to push it even further. Consider the following simple problem. A circle’s radius is growing at 1 inch per second. How quickly is the area of the circle growing? Let’s solve it with Lagrange’s notation first.

The area of a circle is $A=r^2$

${\displaystyle A^{\mathrm{\prime }}(r(t))=(A\circ r{)}^{\mathrm{\prime }}(t)=\pi 2r(t)r^{\mathrm{\prime }}(t)=2\pi r(t)\text{ in/s}}$

Thus at a given time $t$ the area is increasing at the rate of $2r$

Now here’s the rub. In science and engineering most values are somehow related to other values, and nearly everything is related to time. Explicitly defining functions makes even simple relationships (like the one above) complicated to write down. So people dispense with denoting functions explicitly, and just treat these quantities as functions. In practice, the Leibniz notation for the equation above is something like this:

${\displaystyle \frac{dA}{dt}=\frac{dA}{dr}\cdot \frac{dr}{dt}=2\pi r\text{ in/s}}$

We’re not explicitly defining or mentioning functions anywhere, but immediately proceed with the understanding that the variables $A$ and $r$ are really functions.

As a matter of studying advice, I spent hours trying to understand exactly why anyone might want to do this and how the mechanics work, until I sat down to do a bunch of simple related rates problems, at which point abusing the notation in this way quickly became the most natural thing in the world. So if you’re stuck, go solve a bunch of simple problems and then come back here. Hopefully by then everything will make a lot more sense.

Sep 30, 2024