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Do rate of change of force with respect to $x$ $\frac{dF}{dx} = 0$ at the equilibrium position imply that it is the neutral equilibrium position?

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No. For instance, the force $F=-kx^3$ has $\frac{dF}{dx}|_{x=0}=0$, but is nevertheless a stable equilibrium.

Neutral equilibrium requires $\frac{dF}{dx}=0$, but $\frac{dF}{dx}=0$ does not imply neutral equilibrium. Higher-order derivatives need to be examined.

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    $\begingroup$ Damn, beaten to it! :-) $\endgroup$ – John Rennie Feb 3 '18 at 7:26
  • $\begingroup$ @JohnRennie I'm impressed that you almost beat me to it, and yours has a graph ;) $\endgroup$ – Chris Feb 3 '18 at 7:28
  • $\begingroup$ @Chris Thanks you, confusion got clear. I checked out many books but didn't get this!! $\endgroup$ – Umang Feb 3 '18 at 7:56
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$dF/dx=0$ implies that the position is locally neutral. You'd need to look at the higher order derivatives to see if it was globally neutral. For example consider this force distance curve:

enter image description here

This has $F=0$ and $F'=0$ at $x=0$, but it is obviously only locally neutral.

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