Equilibrium and the derivative of potential energy In his Classical Mechanics popular lectures ( Lecture 3, at the beginning)  , Susskind illustrates the idea of a stationary quantity using an example involving the notion of equilibrium.
Link : https://www.youtube.com/watch?v=3apIZCpmdls
He states that " object M is in equilibrium at point x" is equilalent to " the derivative of V(x) - potential energy function - with respect to x is equal to 0".
This implies ( as is explained in a short MIT video) that , using the Fundamental Theorem of Calculus, the derivative of potential energy is related to force; but I cannot manage to see exactly how.
Thanks in advance.
 A: If you have a force [in 1D - the generalization in 3D is straightforward but you need vector calculus] $F(x)$ which is conservative then the potential energy is defined as
$$V(x) = - \int _{x_0}^{x} F(x')dx'$$
where $x_0$ is an arbitrary starting point.
This means, that, by the fundamental theorem of calculus, you can "eliminate" the integral by taking the derivative of the potential
$${dV \over dx} = -F(x)$$
so the condition that the derivative of the energy is $0$ means
$${dV\over dx} = 0 = F(x)$$
meaning that the total force acting on your mass is $F(x)=0$ i.e. the mass is at equilibrium (no forces are acting).
If you then wonder whether the equilibrium is stable (i.e., if you move a bit you come back where you started) then you need to look at the second derivative of the potential energy. You can check more details here.
A: Work is defined as $F\times d$. If your force varies, then $W = \int F \cdot dx$, and you can interpret an integral as a sum of many tiny lengths. Work increases energy. When a force like gravity does work, it is increasing the kinetic energy by losing potential energy, since total energy is conserved.
If you imagine a hill, the slope of the hill tells you how much force pushes you downhill. Hills are useful analogies because the height of a hill is linearly proportional to its potential energy.
