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Let us have function $y(x)$ and $x = \varphi(t)$. Then parametric derivative should be: $$y'_x = \dfrac{y'_t}{x'_t}.$$

The parametric derivative of the second order should be: $$y''_{xx} = (y'_x)'_x = \dfrac{(y'_x)'_t}{x'_t} = \dfrac{\left(\dfrac{y'_t}{x'_t}\right)'_t}{x'_t} = \dfrac{\dfrac{y''_{tt} \cdot x'_t - y'_t \cdot x''_{tt}}{(x'_t)^2}}{x'_t} = \dfrac{y''_{tt} \cdot x'_t - y'_t \cdot x''_{tt}}{(x'_t)^3}.$$

And the parametric derivative of the third order should be $$y'''_{xxx} = (y''_x)'_x = \dfrac{(y''_x)'_t}{x'_t} = \dfrac{\left(\dfrac{y''_{tt} \cdot x'_t - y'_t \cdot x''_{tt}}{(x'_t)^3}\right)'_t}{x'_t} = \dfrac{\dfrac{\left( y'''_{ttt} \cdot x'_t - y'_{t} \cdot x'''_{ttt} \right) \cdot (x'_t) - 3 \left( y''_{tt} \cdot x'_t - y'_t \cdot x''_{tt} \right) x''_{tt}}{(x'_t)^4}}{x'_t} = \dfrac{\left( y'''_{ttt} \cdot x'_t - y'_{t} \cdot x'''_{ttt} \right) \cdot (x'_t) - 3 \left( y''_{tt} \cdot x'_t - y'_t \cdot x''_{tt} \right) x''_{tt}}{(x'_t)^5}.$$

Is it correct?

Andrew
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    What is going on with $y_{ttt}$? Not at all sure what that means here. Comparing to standard usages, I would interpret it as $\frac{\partial^3 y}{\partial t^3}$ but that isn't meaningful here – FShrike May 15 '23 at 20:20

1 Answers1

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Some of what you have written is incorrect. For example, you have written, in various places,$$y_x’’$$ when you actually mean $y_{xx}’’$

Nevertheless, your final expression is correct (in the notation you have chosen).

You might consider writing $y’$ to denote differentiation of $y$ with respect to $t$ or even $\dot{y}$ (using classical mechanics notation) to make it easier to read.

David Quinn
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