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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?

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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.

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    $\begingroup$ You might be interested in this: terrytao.wordpress.com/2012/12/29/… $\endgroup$ Aug 24, 2013 at 17:59
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    $\begingroup$ +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. $\endgroup$ Aug 24, 2013 at 18:35
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We can unpack what is meant, mathematically, by $\dfrac {d^ny}{dx^n}$:

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

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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 $\delta x$.

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    $\begingroup$ Why isn't $(\mathrm{d}x)^n=\mathrm{d}^nx^n$, just as the d in the numerator gets extra treatment? $\endgroup$
    – qwertz
    Dec 2, 2016 at 15:43
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    $\begingroup$ @qwertz Because we are treating $dx$ as a single variable, not as a product between two variables (i.e. $d · x$, which is meaningless). $\endgroup$ Mar 24, 2021 at 15:53
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$$ \frac{d^2 y}{dx^2} = \frac{d}{dx}\frac{dy}{dx}=\frac{d\cdot dy}{dx \cdot dx} $$ Of course, formally.

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    $\begingroup$ You seem to be implying that the "d"s are multiplying. The top is actually d(dy), but it isn't totally true, either. $\endgroup$
    – johnnyb
    Jan 10, 2018 at 2:21
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As I already mentioned in the comment, I wrote one answer already on this topic, but now I’ll try to remake it to a more suitable case for this question.

To understand Leibniz notation, firstly, we need to understand what is meaning of "$d$" i.e. how it works: having differentiable function $f$ and point of its differentiability $x$ (i.e. having $f'(x)$) we create new, linear, function with respect to some new variable, suppose, $h$, by formula $(df(x))(h)=f'(x)\cdot h$. Here $(df(x))$ is name of new function. Usually we omit extra brackets and have more familiar view as $df(x)(h)=f'(x)\cdot h$. But, actually, after fixing $f$, we have two variables, so, formally is more clear representation $$(df)(x,h)=(df(x))(h)=df(x)(h)=f'(x)\cdot h\quad(1)$$ Before continue to second differential, let's look what makes "$d$" notation with identity function $I(x)=x$, because it's extensively using later: $$(dI)(x,h)=dI(x)(h)=dx(h)=I'(x)\cdot h= 1 \cdot h=h\quad(2)$$ Now we can use $(2)$ in $(1)$ and get $df(x)(h)=f'(x)\cdot h=f'(x)\cdot dx(h)$. Omitting variable $h$, we obtain well known $$df(x)=f'(x)\cdot dx\quad(3)$$ And from here we have representation of first derivative $\frac{df(x)}{dx}=f'(x)$, though formally it is $\frac{df(x)(h)}{dx(h)}=f'(x)$.

To take second differential with respect to $x$ let's firstly write $(1)$ in more convenience form as $(df)(x,h_1)=f'(x)\cdot h_1$. Now, again, we should take derivative with respect to $x$ from right side and consider linear function generally for new variable, suppose, $h_2$

$$d^2f=d(df)=d(df(x, h_1))(h_2)=f''(x)\cdot dx(h_1) \cdot dx(h_2)$$ So, formally $d^2f$ is function from $3$ variable $d^2f(x, h_1, h_2)$, linear with respect to last two. Taking $h_1=h_2=h$ and then omitting it, in same way as above, gives $$d^2f(x) = f''(x)\cdot (dx)^2$$ i.e. we obtain representation for second derivative $\frac{d^2f(x)}{(dx)^2}=f''(x)$, fully formal $\frac{d^2f(x, h_1, h_2)}{dx(h_1) \cdot dx(h_2)}=f''(x)$.

Continuation inductively gives, that $n$-th differential is function from $n+1$ variables $d^nf(x, h_1,\cdots,h_n)$, linear with respect to last $n$ variables i.e. $$d^nf(x, h_1,\cdots,h_n) = f^{(n)}(x)\cdot dx(h_1)\cdot \cdots \cdot dx(h_n)$$ taking, as earlier, $h_1= \cdots=h_n=h$ and then omitting it gives well known $$d^nf(x) = f^{(n)}(x)\cdot (dx)^n$$ so, we get $\frac{d^nf(x)}{(dx)^n} = f^{(n)}(x) $. Full formal expression

$$\frac{d^nf(x, h_1,\cdots,h_n)}{dx(h_1) \cdots dx(h_n)}=f^{(n)}(x)$$

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