For a simple one dimensional example of the relationship between force and potential energy, assume that the potential energy of a particle is given by the equation
![{\displaystyle U(x)=-{\frac {A}{x}}\left[1+Be^{-x/c}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6809a53033675893a5b3c7d9b1aa6c114203d466)
The relationship between force and potential energy is
For our 1D example, where the potential energy is dependent only on the position, 
Taking the derivative yields
![{\displaystyle {\frac {dU}{dx}}={\frac {A}{x^{2}}}\left[1+Be^{-x/c}\right]-{\frac {A}{x}}\left[-{\frac {B}{C}}e^{-x/c}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4d247c168da8c1fd8a5dd03a6c4bee998006684)
Therefore, the force on the particle is governed by the equation
![{\displaystyle F=-{\frac {A}{x^{2}}}\left[1+Be^{-x/c}\right]+{\frac {A}{x}}\left[-{\frac {B}{C}}e^{-x/c}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/763450d124ff6bc33ab55fa7558e7c3d22030c6a)
If we further assume that the constant C is so much larger than x, the force will simplify. Rearranging to get
![{\displaystyle F=-{\frac {A}{x^{2}}}\left[1+Be^{-x/c}-{\frac {Bx}{C}}e^{-x/c}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/50796171716ece731a993c0865cefb3689cf9db0)
Because 
, 
, we get
![{\displaystyle F=-{\frac {A}{x^{2}}}\left[1+B-{\frac {Bx}{C}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8bf116cb317439c616d14a2296adc982bba43e42)
Further simplifying 
gives the force under this assumption
![{\displaystyle F=-{\frac {A}{x^{2}}}\left[1+B\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0dceec0c918f38510e7051caf99f419e441d2124)
