= Solution
The <generalized momentum> of this <Lagrangian> is
$$
p=\frac{\partial\mathcal L}{\partial\dot x}=\frac{m\dot x}{\sqrt{1-\dot x^2}},\qquad\frac{\partial\mathcal L}{\partial x}=-V'(x).
$$
The <Euler-Lagrange equation> therefore gives
$$
\boxed{\frac{d}{dt}\left(\frac{m\dot x}{\sqrt{1-\dot x^2}}\right)+V'(x)=0,\qquad \frac{m\ddot x}{(1-\dot x^2)^{3/2}}=-V'(x).}
$$
Because $\mathcal L$ has no explicit time dependence, the energy form of the <Beltrami identity> gives a conserved quantity $E=\dot x\,\partial\mathcal L/\partial\dot x-\mathcal L$. Here
$$
E=\frac{m\dot x^2}{\sqrt{1-\dot x^2}}+m\sqrt{1-\dot x^2}+V(x)=\boxed{\frac{m}{\sqrt{1-\dot x^2}}+V(x)}.
$$
For a direct check of <conservation of energy>, differentiate:
$$
\frac{dE}{dt}=\dot x\left[\frac{m\ddot x}{(1-\dot x^2)^{3/2}}+V'(x)\right]=0.
$$
This is the rest-plus-kinetic energy of a <relativistic particle>, in units with light speed one, plus its <potential energy>.
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