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I was trying to derive Euler Lagrange Form which looks like

$$\frac{\partial L}{\partial q}=\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}$$

I was using a function which I got from here,

$$S(q)=\int_{t_0}^{t_1} L(q,\dot{q},t)dt$$

I was using Feynman technique since I know that rather than simple integration.

$$\frac{dS}{dt}=\int_{t_0}^{t_1} (\frac{\partial L}{\partial q}\dot{q}+\frac{\partial L}{\partial \dot{q}}\frac{\partial }{\partial t}\dot{q} )dt$$

Total time derivative of that function is $0$ so, writing $\frac{dS}{dt}=0$ $$0=\dot{q}(\frac{\partial L}{\partial q}+\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot{q}})$$

$$ \frac{\partial L}{\partial q}+\frac{\partial }{\partial t}\frac{\partial L}{\partial \dot{q}}=0$$

I believe there's difference between total time derivative and short ($\frac{\partial}{\partial t}$) time derivative. Without caring of that I found that my equation is closely related to Euler Lagrange. We have only difference between sign.

Arctic Char
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    You have the term $\frac{\partial}{\partial t} \dot q$. You will need an integration by part to get $\dot q$. Doing it correct you recover the usual EL equation. Also, $\frac{dS}{dt}$ seems to be a bad choice: $t$ is the variable of the function $q$. – Arctic Char Oct 03 '21 at 09:03
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    $\int_{t_0}^{t_1} \frac{\partial L}{\partial \dot{q}}\frac{\partial }{\partial t}\dot{q} dt = -\int_{t_0}^{t_1} \frac{\partial }{\partial t}\frac{\partial L}{\partial \dot{q}}\dot{q} dt$. – Arctic Char Oct 03 '21 at 09:31
  • @ArcticChar Could you please explain the line? I never thought like that... :O – Unknown Oct 03 '21 at 09:35

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