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I know the second derivative of y wrt x: $\frac{d^2y}{dx^2}$ means $\frac{d}{dx}(\frac{dy}{dx})$, but is there a mathematical reason you square the $d$ in the numerator but the $x$ in the denominator?

I've wondered if it's because the $d$ in the denominator represents some arbitrary infinitely tiny amount, and the $d$ in the numerator is that same $d$, only squared to account for the second derivative.

Does that make sense, and/or am I missing something significant about derivatives?

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    $\begingroup$ $dx^2$ really means $(dx)^2$. So you have the right idea about what it should represent. $\endgroup$
    – Kaynex
    Commented Dec 8, 2015 at 3:45

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$\frac{d}{dx}$ is a linear operator on the space of differentiable functions. When we write $\frac{dy}{dx}$, we are really saying $\frac{d}{dx}(y)$, where $y$ represents some function of $x$. Doing this twice, we get $$\frac{d}{dx}\frac{d}{dx}(y)=\left(\frac{d}{dx}\right)^2(y)=\frac{d^2y}{dx^2}$$

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  • $\begingroup$ I think that $C^1$ is more commonly taken to be the set of continuously differentiable functions. The map $d/dx$ can act on a larger space of functions. This isn't so important for the question, though. $\endgroup$ Commented Dec 8, 2015 at 3:43
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This is just a widely accepted abuse of notation. If we didn't abuse notation, we would instead write $\frac{d^2 y}{(dx)^2}$.

It's important to keep in mind that this notation of $\frac{dy}{dx}$, called Leibniz notation, is chosen to help you remember certain formulas (the chain rule) but isn't really rigorous on its own. You can certainly get yourself into some trouble by just pushing these symbols around like you would fractions!

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  • $\begingroup$ But wouldn't that mean $\frac{d}{dx}*\frac{dy}{dx}$ which is not the same as $\frac{d}{dx}(\frac{dy}{dx})$? $\endgroup$
    – tethernova
    Commented Dec 8, 2015 at 3:35
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    $\begingroup$ @Bye_World, you could argue that Leibniz notation is abusive and not rigorous, and I'd agree to an extent. But often the point of choosing notation in math is to highlight some particular feature of what is going on, and to that end, I'd argue that Leibniz notation serves its purpose well. $\endgroup$ Commented Dec 8, 2015 at 3:38
  • $\begingroup$ @tethernova, the notation is chosen so that applying the derivative operator $\frac{d}{dx}$ is written multiplicatively, so the two things you wrote are the same. $\endgroup$ Commented Dec 8, 2015 at 3:41

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