Consider this graph of some point's position vector's endpoint - a function of time - on a Cartesian plane:
This is a perfectly normal path - after all, in real world there is nothing keeping point from taking a sharp turn like this. However, in mathematical physics, we consider position vector as a differentiable function, where $\dfrac{df(t)}{dt}=(\dfrac{df_x(t)}{dt},\dfrac{df_y(t)}{dt},\dfrac{df_z(t)}{dt})$.
In the graph I posted, the component $f_x$ seems to be undifferentiable at point $f_x(t)$, since there its goes through a cusp.
Is there something I'm not seeing and there is a derivative of $f_x$ at $t$, or is this point really undifferentiable?
And if it is, why do we define position vector as a differentiable function?
Yes, the function $f_x$ is not differentiable at that corner point, unless the particle (or whatever it is) stops at the point. Consider the trajectory $t\mapsto(t^3,|t|^3)$. It makes a sharp turn but the function is differentiable everywhere. But this is only possible if the particle stops at the corner.
Why do you define position vector as a differentiable function, then? You shouldn't. Typically piecewise differentiable is reasonable for physical problems: the particle travels along nice differentiable trajectories, but there can be sharp turns where differentiability fails.
There is also a way to define the derivative of $f_x$ in your example, but it would not be a function. It would be a distribution. This generalization is not always useful, though, and it can be technically more confusing than just working with piecewise smooth trajectories.