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Let's say I have:

$$ y = -\frac{1}{\tan(x)} $$

How do you get a relation between $\mathrm{d}^2x$ and $\mathrm{d}^2y$ ?

What I know:

$$\mathrm{d}y = \frac{\mathrm{d}x}{\sin(x)^2} $$

Axel
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2 Answers2

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Hint: $$y^{(n)}(x)=\frac{d^ny(x)}{dx^n} \iff d^ny(x)=y^{(n)}(x)\space dx^n.$$

Anton Vrdoljak
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$$y=-\frac{1}{\tan(x)}=-\cot(x)$$ and $$\frac{dy}{dx}=\csc^2(x)$$

To find $\frac{\,d^2y}{\,dx^2}$, differentiate $\csc^2(x)$ once more to get $$\frac{\,d^2y}{\,dx^2}=\frac{\,dy}{\,dx}\csc^2(x)=-2\csc^2(x)\cot(x)\Longrightarrow d^2y=-2\csc^2(x)\cot(x)dx^2\tag{*}$$

Then, $$y=-\cot(x)\Longrightarrow x=\cot^{-1}(-y) $$ so now, $$\frac{dx}{dy}=\frac{1}{1+y^2}\Longrightarrow \frac{d^2x}{dy^2}=-\frac{2y}{(1+y^2)^2}\Longrightarrow d^2x=-\frac{2y}{(1+y^2)^2}dy^2\tag{**}$$

From $(*)$, we have

$$d^2y=2\csc^2(x)ydx^2 \Longrightarrow y=\frac{d^2y}{dx^2}\frac{\sin^2(x)}{2}$$

Substitute $y$ into $(**)$ for your relation.

Not really sure what meaning or significance it has; it seems like an abuse of notation. This would only work on certain constrained domains of $y=-\cot(x)$

C Squared
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  • Thank you, I will close my post. Somehow I realized that maybe this question was not relevant enough. Thanks for your time. – Axel Aug 31 '20 at 07:23
  • @Axel Just curious, what was your problem and how were you approximating $d^2x$? – C Squared Aug 31 '20 at 07:37
  • Well that was rather simple, $x$ was "oscillating lightly" between a position $x_0$. So I could consider that $1/\sin(x)^2$ was almost constant. Making that approximation I could say that $\mathrm{d}^2y=\mathrm{csc}(x_0)^2 \mathrm{d}^2x$. – Axel Aug 31 '20 at 07:54