Why in a directional derivative it has to be a unit vector

Question

Why when we compute the directional derivative we have to use the unit vector $\vec{\bf{u}}$? I know that using $2\vec{\bf{u}}$ would change the directional derivative as the second point is further apart. But why not use $\frac{\vec{\bf{u}}}{2}$ then? Wouldn't that gradient be "more precise"?

Answer 1

At the class we were given the definition of a directional derivative when the direction vector is a unit vector. Because of that I got confused. The real definition can be found here: https://en.wikipedia.org/wiki/Directional_derivative (Variation using only direction of vector for euclidean space). Here we can see that using the unit vector just simplifies things and is not necessary. It is very similar to 2D derivative as the distance between the two point $f(x+hv)-f(x)$ is the length of vector $\vec{\bf{v}}$.