In classical mechanics we often define the action as the quantity
$$ \int_{0}^{T} \left[ T - V \right] dt$$
Which in many applications is some variant of
$$ \int_{0}^{T} \left[ \frac{1}{2}m \left( x' \right)^2 - V(x) \right] dt. $$
The usual justification for the principle of least action is the observation that if you take integrand above and put it into the euler lagrange equations you get back Newton's law.
I.E. if you believe
$$ \frac{\partial L}{\partial x} - \frac{d}{dt} \left( \frac{\partial L}{\partial x'} \right) + \frac{d^2}{dt^2} \left(\frac{\partial L}{\partial x''} \right) - ... = 0$$
With $L = \frac{1}{2}m \left( x' \right)^2 - V(x) $ you will find
$$ - \frac{dV}{dx} = mx'' $$
(i.e $F = ma$).
So this is old news that we consider quite well understood but then I realized the following, suppose we try to minimize this action instead:
$$ \int_{0}^{T} \left[ \frac{1}{2}mxx'' + V(x) \right] dt $$
I.E. $L = \frac{1}{2}mxx'' + V(x) $. If we plug this into the euler lagrange equations we ALSO end up deriving
$$ - \frac{dV}{dx} = mx'' $$
Via
$$ \frac{\partial }{\partial x}[\frac{1}{2}mxx'' + V] + \frac{d^2}{dt^2}\frac{\partial}{\partial x''}\left[ \frac{1}{2}mxx'' + V \right] = 0 \rightarrow \frac{1}{2}mxx'' + \frac{dV}{dx} + \frac{1}{2}mxx'' = 0 \rightarrow mxx'' + \frac{dV}{dx} = 0 \rightarrow F = -\frac{dv}{dx}$$
I found this very curious, I recognize the physical significance of $(mx'')*x$ as the classical expression for work (Force x distance). But is there any deeper physical significance to this second lagrangian, or is this just a curious mathematical oddity/not a useful problem solving tool. Can this second lagrangian be used in place of the first in other contexts (ex: in the Feynman path integral).
So it seems that $\frac{1}{2}mxx'' - V$ is a conserved quantity. (I came to this conclusion after checking only one example involving a newtonian gravitational field between two bodies at 2 locations, so maybe this is wrong.)