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Let $x$ depend on $t$. $\dot{x}$ is derivative x over t. I want to know is there some formulas to simplify $\frac{d\dot{x}}{dx}$? Any hint or thought is appreciated. Thank you!

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why does somebody down vote? If you down vote, then tell why! – ashim Jan 30 '12 at 5:09
There seems no such formula, observing several instances such as $x = \mathrm{const}$, $x = e^{at}$, $x = \tan t$, etc. – Sangchul Lee Jan 30 '12 at 5:44
@sos440: There is such formula. Was that you who downvoted? It's a perfectly good question. – anon Jan 30 '12 at 6:27
@anon : I'm not the one who downvoted. It was already downvoted when I began writing my comment. I acknowledge, however, that I misunderstood capoluca's question and that I tried to figure out there is no such function $F$ satisfying $d\dot{x}/dx = F(x, \dot{x})$. – Sangchul Lee Jan 30 '12 at 7:52

2 Answers 2

up vote 8 down vote accepted

Put succinctly,

$$\large \frac{\mathrm{d}\dot{x}}{\mathrm{d}x}=\frac{(\mathrm{d}\dot{x}/\mathrm{d}t)}{(\mathrm{d}x/\mathrm{d}t)}=\ddot{x}/\dot{x}.$$

With more exposition: set $\Delta x=x(t+\Delta t)-x(t)$ and $\Delta \dot{x}=\dot{x}(t+\Delta t)-\dot{x}(t)$. Substitution gives

$$\large\lim\limits_{\Delta x\to0}\frac{\Delta \dot{x}}{\Delta x}=\lim\limits_{\Delta t\to0}\frac{\Delta \dot{x}}{\Delta x}=\lim\limits_{\Delta t\to 0}\frac{\Delta\dot{x}/\Delta t}{\Delta x/\Delta t}=\ddot{x}/\dot{x}.$$

The substitution is valid because $\Delta t\to0$ as we let $\Delta x\to0$.

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Thanks, I like your explanation. – ashim Jan 30 '12 at 18:55

$$\frac {d\dot{x}}{dx}=\frac{d}{dx}\left(\frac{dx}{dt}\right)=\frac{d}{dt}\left(\frac{dx}{dt}\right)\cdot \frac{dt}{dx}=\frac{d}{dt}(\dot{x})\cdot \frac{1}{\dot{x}}=\frac{\ddot{x}}{\dot{x}}$$

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