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The WolframMathworld website has a page containing the derivation for the brachistochrone with and without friction.

I am able to follow the derivation for the with-friction case up to the point of writing out the Euler-Lagrange equation from the functional derived from the physics of the situation. This is done in step (29) of the demonstration in the linked website.

I am asking for an algebraic demonstration of how applying the Euler-Lagrange equation to the integrand in step (28) results in the expression in step (29).

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Let $f(x,y,\xi) := \sqrt{\frac{1 + \xi^2}{2g(y - \mu x)}}$. Then $$t = \int f(x,y,y')\,dx$$ and the Euler-Lagrange equation is given by $$\frac{\partial f}{\partial y} - \frac{d}{dx}\frac{\partial f}{\partial \xi} = 0.$$ We now compute these derivatives $$\frac{\partial f}{\partial y} = -\sqrt{\frac{1 + \xi^2}{2g}}\frac{1}{2(y - \mu x)^{3/2}},$$ $$\frac{\partial f}{\partial \xi} = \frac{\xi}{\sqrt{(1 + \xi^2)2g(y - \mu x)}},$$ and evaluate them at $(x,y(x),y'(x))$. The Euler-Lagrange equation can then be rewritten as $$\frac{d}{dx}\left(\frac{y'(x)}{\sqrt{(1 + y'(x)^2)2g(y(x) - \mu x)}}\right) = \sqrt{\frac{1 + y'(x)^2}{2g}}\frac{1}{2(y(x) - \mu x)^{3/2}}.$$ Equation (29) follows upon expanding the derivative on the left hand side in the previous equation. The computations are a little lengthy, but I'm sure you can fill in the details.

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