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In the Swedish equivalent of a calculus class, I came across various notations for derivatives, including: Lagrange's $f'(x)$, Newton's $\dot{x}$, and Leibniz's $\frac{dy}{dx}$. For second-order these become $f''(x)$, $\ddot{x}$, and $\frac{d^2y}{dx^2}$ respectively (as you already may know) and it is the last notation $\frac{d^2y}{dx^2}$ that has left my mind boggled, since I haven't found any visualization of it nor finding why just the $d$ is squared in the numerator.

My intuition when I first saw it was that it might make more sense to square the entire fraction as you take the quotient of $y$ and $x$ twice. However that is just what my gut felt like, and I found no further clarification in the textbook.

So to summarize, if anyone has a neat explanation or some sort of visualization of why the notation is as it is, I would appreciate it.

Adrian Wallin
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3 Answers3

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Up to my knowledge it is more an operational reading, i.e. kind of

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

what truely is meant there.

--- rk

Dr. Richard Klitzing
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There is no definitive answer to this, it's just notation. The best explanation I can think of is this: $d^2y$ basically means $d(dy)$. So, if $dy=y'dx$, then $$d^2y=d(dy)=d(y'dx)=dx\,dy'=y''dx\,dx=y''dx^2$$ Thus, dividing both sides by $dx^2$ yields

$$y''=\frac{d^2y}{dx^2}$$

In fact, this reasoning can be made rigorous by considering $dy$ as a linear form, but for the purpose of this question it is not necessary to delve into such formalizations.

NtLake
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  • If $d$ denotes the total derivative and $dx$ the $1$-form, then dividing by $dx$ does not make sense... The space of differential forms is not a field, its a ring when equipped with the wedge product... – Dr. Tony Tony Chopper Nov 22 '24 at 14:35
  • @Dr.TonyTonyChopper $dx$ is a real number here. $dy$ (meaning $d(y)$) is defined as $dy(dx)=y'dx$ and $d^2y$ as the quadratic form associated with $d(dy)(dx_1,dx_2)=dx_1,y''dx_2$, that is $d^2y(dx)=y''dx^2$. It is fine to divide by a nonzero real number, as $dx^2$ is here. – NtLake Nov 22 '24 at 14:53
  • @Dr.TonyTonyChopper You can identify the 1-form $\mathrm d x$ with a map from vector fields to functions. Then dividing a 1-form by another 1-form is a perfectly sensible thing to do, as long as you don't plug in vectors for which the form in the denominator vanishes. – Nico Zimmer Nov 22 '24 at 15:41
  • @NicoZimmer that is true, but then ur restricting urself to all invertible elements in the ring $C^k(M)$ if $k$ is an natural number and $M$ a manifold. So it does make sense to divide a differential form when we consider the following domain for the form; $\mathcal{D} = {X\in \mathfrak{X}(M): \omega_x(X_x)\neq 0 \ \forall x\in M}$. However, this takes away many functions which are differentiable, so writing $d(f)$ for such functions makes sense, however "dividing" by $dx$ does not make sense then. So in general, I stick to my initial comment. With adjustments one can "divide" by $dx$... – Dr. Tony Tony Chopper Nov 22 '24 at 16:01
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The Leibniz notation is just a lazy way to denote multiple derivatives. We start with the first derivative $\frac{dy}{dx}=\frac d{dx}y$. This refers to the limit of the difference quotient or more abstractly, the differential operator evaluated at the function $y$. The second derivative is just the differential operator evaluated at the first derivative $\frac d{dx}\frac{d}{dx}y$. If we were to treat those as fractions and multiply them we should get $\frac{d^2y}{(dx)^2}=\frac{d^2y}{dx^2}$ where we suppress the parentheses as we do in $(\text{cm})^2=\text{cm}^2$ where $\text c=10^{-2}$ for example. You can in fact express derivatives as powers of the first derivative operator applied to the function.

Stardust
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