I am working on a specific example where the metric I am using is the $AdS_4$ metric whose ricci scalar $R=-12/l^2$ for some characteristic scale $l$: $$ds^2=-\cosh^2\left(\frac{\rho}{l}\right)dt^2+d\rho^2+l^2\sinh^2\left(\frac{\rho}{l}\right)(d\theta^2+\sin^2\theta~d\phi^2)$$ And more specifically, I am looking at a physical system where a particle of mass $m$ moves freely in this $AdS_4$ spacetime. So of course we have our action$$S_1=-m\int~d\tau\sqrt{\cosh^2\left(\frac{\rho}{l}\right)\dot t^2-\dot\rho^2-l^2\sinh^2\left(\frac{\rho}{l}\right)(\dot\theta^2+\sin^2(\theta)\dot\phi^2)}$$ where dot denotes differentiation against $\tau$ and $\tau$ is arbitrary parametrisation. Let us denote the entire expression containing the square root as $\sqrt{A}$ for simplification purposes. The Euler Lagrange equation of this action on $\rho$, one could check, is given by$$\frac{d}{d\tau}\left(-\frac{\dot\rho}{\sqrt{A}}\right)=\frac{1}{\sqrt{A}}\cdot\frac{1}{l}\sinh\left(\frac{\rho}{l}\right)\cosh\left(\frac{\rho}{l}\right)\left(\dot t^2-l^2(\dot\theta^2+(\sin^2\theta)\dot\phi^2)\right)$$
On a general note, if we have an action of the form $$S_A= S(f)$$ where $f$ is some function of spacetime points, and we consider a second action $$S_B=S(f^2)$$ we have that if $\delta S_A=0$ then $\delta S_B=2f\cdot\delta S_A=0$ as well. So having $S_1$ as written above, we should have a classically equivalent action $S_2$ given by $$S_2=\frac{m}{2}\int d\tau \left(-\cosh^2\left(\frac{\rho}{l}\right)\dot t^2+\dot\rho^2+l^2\sinh^2\left(\frac{\rho}{l}\right)(\dot\theta^2+\sin^2(\theta)\dot\phi^2\right)$$ The problem is if we write down the E.O.M for $\rho$ from this action $S_2$, as one may check, given by $$\frac{d}{d\tau}\left(-\dot\rho\right)=\frac{1}{l}\sinh\left(\frac{\rho}{l}\right)\cosh\left(\frac{\rho}{l}\right)\left(\dot t^2-l^2(\dot\theta^2+(\sin^2\theta)\dot\phi^2)\right)$$ it is not quite the same as the E.O.M we got from $S_1$. To have them equivalent to one another we require that $ \frac{dA}{d\tau}=0$ but I can hardly see why this is true.
