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I'm pretty sure this is true, but haven't been able to figure out or find a proof, largely because I haven't been able to figure out what to Google for.

$$ \frac{d^2x}{dy^2} (\frac{dy}{dx})^2+ \frac{dx}{dy} \frac{d^2y}{dx^2}=0 $$

I would appreciate some help, even if it just a link to somewhere this is proven. (Or if it is not true, maybe a counterexample?)

Gerber
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  • Are we assuming $y = f(x)$ or $x = g(y)$ for some function $f$ or $g$, or working with some other function $h(x, y)$? – DylanSp Feb 24 '16 at 21:38

2 Answers2

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The relation is correct, provided that all is well interpreted. Unfortunately your notation may create confusion, and so let me suggest a more pedantic approach.

Let $y=f(x)$. Then $x=g(y)$, where $g$ is the inverse of $f$ (let's assume that it exists on a particular region). Then $1=g'(y)f'(x)$, in view of the chain rule, and so also $$ 0=g''(y)f'(x)+(g'(y))^2f''(x). $$ Hence, $$ \begin{split} \frac{d^2x}{dy^2} \left(\frac{dy}{dx}\right)^2+ \frac{dx}{dy} \frac{d^2y}{dx^2} &=g''(y)(f'(x))^2+g'(y)f''(x)\\ &=-(g'(y))^2f''(x)f'(x)+g'(y)f''(x)\\ &=-g'(y)f''(x)+g'(y)f''(x)=0. \end{split} $$

John B
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I think I figured it out. It is obtained by differentiating the following on both sides:

$$ \frac{dy}{dx}\frac{dx}{dy}=1 $$

Gerber
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