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This is a bit of a strange question, but I couldn't find an answer elsewhere and decided post my own question.

I'm currently studying differential equations and had a question regarding the Leibniz notation for higher-order DE's. I'll just give an example of what is puzzling me as it is clearer and faster.

If we want the second order derivative of an equation $f(x)$, I noticed that it is written as:

$$\frac{d^2 f(x)}{dx^2}$$

Why is the $d$ not squared in the denominator? Checking the Wikipedia article for the second derivative has shown me that:

$$ \begin{align}\frac{d^2}{dx^2}[x^n] & = \frac{d}{dx} \frac{d}{dx} [x^n] \\\ & = \frac{d}{dx}[nx^{n - 1}] \\\ & = n\frac{d}{dx}[x^{n - 1}] \\\ & = n(n - 1)x^{n - 2} \end{align}$$

In order for the first line to make sense, shouldn't the left-hand side be written as

$$\frac{d^2}{d^2 x^2}[x^n]$$

so that there are also the same number of $d$'s?

Sean
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    Related: https://math.stackexchange.com/questions/25102/why-is-the-2nd-derivative-written-as-frac-mathrm-d2y-mathrm-dx2 – Hans Lundmark Sep 17 '19 at 08:53

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This is because $dx$ is a single symbol. Even though it looks like a $d$ and an $x$, this is not the case (at least in this context).

Arthur
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  • Thanks for the answer! One question I have is that if $dx$ is "a single symbol," then why is $x$ altered while $d$ is not? Similarly, does this mean that the $d$ in the numerator is conceptually different from the $d$ in the denominator? – Sean Sep 17 '19 at 04:50
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    @Seankala $x$ isn't altered, $dx$ is: $(dx)^2=dx\cdot dx =dx^2$. As for the rest... Really, $\frac{df}{dx}$ is the limit of a fraction $\frac{\Delta f}{\Delta x}$. Here, the numerator and denominator are just single symbols. There is no $\Delta$ in there, only $\Delta f$ and $\Delta x$. The notation $\frac{df}{dx}$ reflects this origin. That's where it comes from. Everything else is just convenient notational convention, and nothing else (which is to say, I think you're trying to read meaning into the symbols which isn't really there). – Arthur Sep 17 '19 at 05:13
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    (cont.) That being said, I can, off the top of my head, think of two reasons why we don't consider the $d$ and the $f$ in the numerator as tightly "stuck together" as the $d$ and the $x$ in the denominator. One is that pretending that $\frac{df}{dx}$ is a fraction allows us to come up with $\frac d{dx}$ to denote a derivative operator. Again, notational convenience. There is no actual meaning to the fraction-like form. The other is that the second derivative is$$\frac{d\frac{df}{dx}}{dx}$$and again, for notational convenience we pretend it is a fraction and write it as $\frac{d^2f}{dx^2}$. – Arthur Sep 17 '19 at 05:17
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    So yes, the $d$ in the numerator does behave differently in that that's the one that "interacts" with things around it. The $dx$ in the denominator is relatively fixed and unchanging. (I said in the first comment that $dx$ is altered. That's not right. I meant that $dx$ can get "company" in the denominator by a second $dx$ if we're taking a second derivative. None of them are "altered". There is just two of them.) – Arthur Sep 17 '19 at 05:21