newtonian gravity - Why is the force along $x$ direction not $0$ on bumpy surface?

Near earth the gravitational force is $F=-mg\hat{y}$ and the potential energy is given by $U=mgy$. The force points straight down with no horizontal $x$ component.

In this(don't have to click) video the professor shows a curve and clearly the potential energy varies with $x$ because the height varies with $x$:

$$U=mgy = mgf(x)$$
Good so far. But $\dfrac{\partial U}{\partial x} = -F_x = mgf'(x)$. Suddenly the $x$ component of the force is non zero. Since gravity is acting straight down, I'm not able to make sense how this vertical force can produce a horizontal component. Any help?

Answer

The thing you're missing is that the force of gravity isn't the only force involved.

The potential $U(y)=mgy$ is a function of $y$, not directly a function of $x$. In order to get a potential as a function of $x$, we need a function that gives us $y$ as a function of $x$, so we can plug in this function and define $U(x)$ as being equal to $U(y(x))=mgy(x)$.

This function $y(x)$ is decided by the shape of the ground. The ground is only important because it keeps things from falling further, which means it must exert a force that counters gravity, namely, the normal force. The function $y(x)$ is, indirectly, a description of the normal force, which means that the potential $U(x)$ incorporates both gravity and the normal force.

When you're at a point where $dU/dx\neq 0$, you can see based on the equations that $dy/dx\neq0$. Since the normal force is always perpendicular to the surface, $dy/dx\neq 0$ means that the normal force isn't completely vertical - it has a nonzero horizontal component. Therefore, the horizontal force you're asking about is the horizontal component of the normal force.