Pr. Susskind tries a " easy" derivation of Euler-Lagrange's equation in this video : https://www.youtube.com/watch?v=3apIZCpmdls&t=4086s .
His method is to turn the equation for the action into a discrete sum :
$\Large A = \epsilon \sum L ( x_i , \frac {x_{i+1} - x_i} {\epsilon}) , \text {over} \space i$
where the fraction is the discrete version of velocity, and $\epsilon$ is the discrete version of $dt$ ; $\Large L $ is the "lagrangian" ( a functional that depends both on position and velocity) .
Next, as his final goal is to set $ \delta A = 0$ , he tries to obtain an expression for the derivative of the discrete version of action .
He applies the rule $\delta F(x, y) = (\partial F/ \partial x ) dx + (\partial F/ \partial y )dy$ .
The derivative ( discrete version ) he is looking for will be a sum of derivatives, one for every value of $i$, so he concentrates on a generic $x_i$.
My question is a computational one : why does a factor of $1/\epsilon$ appear in the expression of one of the differentials, namely, the partial derivative of $\Large L$ with respect to velocity under its discrete version ( discrete velocity that is represented by the fraction : $\frac {x_{i+1} - x_i} {\epsilon})$.
Note : the published version of the course does not help me since the derivation contains an analogous factor of $1/\Delta t$ that I cannot explain any more to myself ( The Theoretical Minimum, volume 1, p. 113)
The derivation of Euler's equation begins at : 32:23.
The moment at which this ( strange to me) $1 / \epsilon$ factor is : 41:30.
He's stating that velocity is "discretely" proporcional to the inverse of the elapsed time. So chain rule yields that factor. In other words, $$\frac{\partial v}{\partial x_i}=\frac{\partial}{\partial x_i}\left(\frac{x_i-x_{i-1}}{\varepsilon}\right)=\frac{1}{\varepsilon}$$
EDIT:
Susskind writes the Lagrangean as $$\mathscr{L}(x_i,v_i)$$ right? And he asks you to believe him when he says that if you differentiate this and carry out the sum, you'll recover the EL equations. So see $\mathscr{L}$ as function of $x_i$, and $v_i$ itself as a function of the $x_i$. So when you differentiate the Lagrangian with respect to $x_i$, the chain rule kicks in: $$\frac{d}{dx_i}\mathscr{L}(x_i,v_i(x_i))=\frac{\partial \mathscr{L}}{\partial x_i}+\frac{d v_i}{dx_i}\frac{\partial \mathscr{L}}{\partial v_i}$$ But what is $\frac{d v_i}{dx_i}$? Well, as a function of $x_i$, $$v_i=\frac{x_i-x_{i-1}}{\varepsilon}$$ If you treat $x_{i-1}$ as a constant (or even as another variable, but in that case you'd have to switch the d's to $\partial$'s), then this derivative is precisely $$\frac{1}{\varepsilon}$$