# Leibniz notation for high-order derivatives

What is the reason for the positioning of the superscript $n$ in an $n$-order derivative $\frac{d^ny}{dx^n}$? Is it just a convention or does it have some mathematical meaning?

The $n$th derivative is the $n$-fold composition of the first derivative operator:

$$\frac{d^n y}{d x^n} = \left(\frac{d}{dx}\right)^n(y).$$ So a simple justification is to just treat the $d/dx$ as a fraction and distribute the exponent to the top and bottom. This is "valid" in a certain sense - if you look at the finite approximations to the second derivative for example, you have

$$\frac{d^2 y}{dx^2} \approx \frac{1}{\delta x}\delta\left(\frac{\delta y}{\delta x}\right) = \frac{\delta(\delta y)}{(\delta x)^2}$$

where $\delta f = f(x+\delta x) - f(x)$ is the change in $f(x)$ when you change $x$ by $\delta x$.

• Why isn't $(\mathrm{d}x)^n=\mathrm{d}^nx^n$, just as the d in the numerator gets extra treatment? – qwertz Dec 2 '16 at 15:43
• @qwertz Because we are treating $dx$ as a single variable, not as a product between two variables (i.e. $d · x$, which is meaningless). – Daniel Muñoz Parsapoormoghadam Mar 24 at 15:53

Several people have already posted answers saying it's $\left(\dfrac{d}{dx}\right)^n y$, so instead of saying more about that I will mention another aspect.

Say $y$ is in meters and $x$ is in seconds; then in what units is $\dfrac{dy}{dx}$ measured? The unit is $\text{meter}/\text{second}$. The infinitely small quantities $dy$ and $dx$ are respectively in meters and seconds, and you're dividing one by the other.

So in what units is $\dfrac{d^n y}{dx^n}$ measured? The thing on the bottom is in $\text{second}^n$ (seconds to the $n$th power); the thing on top is still in meters, not meters to the $n$th power. The "$d$" is in effect unitless, or dimensionless if you like that word.

I don't think it's mere chance that has resulted in long-run survival of a notation that is "dimensionally correct". But somehow it seems unfashionable to talk about this.

• You might be interested in this: terrytao.wordpress.com/2012/12/29/… – Baby Dragon Aug 24 '13 at 17:59
• +1: I think it is very appropriate to talk about "dimensional analysis" considerations in elementary calculus. (For instance, I want every university student to have a clear understanding of of area, volume, velocity etc. much more than I want them to be able to differentiate and integrate.) There are many opportunities to make this point. I usually propose the false product rule $(fg)' = f'g'$ and then point out that it is not even dimensionally correct, whereas the more complicated-looking (true!) product rule is. – Pete L. Clark Aug 24 '13 at 18:35

We can unpack what is meant, mathematically, by $\dfrac {d^ny}{dx^n}$:

$$\frac{d^n y}{dx^n} = \frac{d}{dx}\left(\dfrac d{dx}\left(\frac d{dx}\cdots\left(\frac{dy}{dx}\right)\right)\right)=\frac{\underbrace{d\cdot d \cdots \cdot d}_{n\;\text{times}}y}{\underbrace{dx \cdot dx\cdots dx}_{n\;\text{times}}}$$

$$\frac{d^2 y}{dx^2} = \frac{d}{dx}\frac{dy}{dx}=\frac{d\cdot dy}{dx \cdot dx}$$ Of course, formally.

• You seem to be implying that the "d"s are multiplying. The top is actually d(dy), but it isn't totally true, either. – johnnyb Jan 10 '18 at 2:21