The part that confuses me is the square being next to the $d$ vs being to the variable. My intuition tells me $d^2x$ is equivalent to $(dx)^2$ and $dx^2$ should be $d(x^2)$ (if that notation is valid)
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2$d^{2}y/dx^{2}$ is just the second derivative. So, it is the derivative of $dy/dx$. – Simon Terrington Jul 19 '20 at 10:58
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We have $$\frac{dy}{dx} = -\frac xy \\ \frac{d^2y}{dx^2} = \frac{y\cdot(-1)-(-x)\frac{dy}{dx}}{y^2}=\frac{-y+x\cdot -\frac xy}{y^2} =\frac{-y^2-x^2}{y^3}$$ Note that $d^2x \ne (dx)^2 \ne d(x^2) $, it is just a fancy way to say that we are taking the derivative of the derivative.
Vishu
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Oh, what a confusing notation. Is there any reason it is written like that? – user1754322 Jul 19 '20 at 11:12
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@user1754322 Not that I know of. Maybe you’d like to ask the person who invented this himself. – Vishu Jul 19 '20 at 11:18
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