Coordinate transformation for coordinate basis vectors

Consdier a coordinate transformation $x^arightarrow x^{‘a}$ on a manifold with coordinate basis vectors $textbf{e}_a$ and $textbf{e}^{‘}_a$.

The relationship between the coordinate basis vectors $textbf{e}_a$ and $textbf{e}^{‘}_a$ is given by
$$textbf{e}^{‘}_b=frac{partial x^a}{partial x^{‘b}} textbf{e}_a.$$

One way to prove this is to consider the infinitesimal displacement vector $dtextbf{s}$ between two points on the manifold given by $$dtextbf{s}=dx^atextbf{e}_a=dx^{‘b}textbf{e}^{‘}_{b}$$ and using $dx^a=(partial x^a/partial x^{‘b})dx^{‘b}$ to get

$${partial x^aover partial x^{‘b}}dx^{‘b}textbf{e}_a=dx^{‘b}textbf{e}^{‘}_b$$
$$({partial x^aover partial x^{‘b}}textbf{e}_a-textbf{e}^{‘}_b)dx^{‘b}=0$$
from where we get $$textbf{e}^{‘}_b=frac{partial x^a}{partial x^{‘b}} textbf{e}_a.$$

The jump from the second last equation to the last equation confuses me. All the second last equation says is that the sum $({partial x^aover partial x^{‘b}}textbf{e}_a-textbf{e}^{‘}_b)dx^{‘b}$ sums to zero. How can we say that $dx^{‘b}$ is non-zero and hence $({partial x^aover partial x^{‘b}}textbf{e}_a-textbf{e}^{‘}_b)=0$?

Also, even if $dx^{‘b}$ is non zero, the terms $({partial x^aover partial x^{‘b}}textbf{e}_a-textbf{e}^{‘}_b)$ need not necessarily all equal to zero for the sum $({partial x^aover partial x^{‘b}}textbf{e}_a-textbf{e}^{‘}_b)dx^{‘b}$ to sum to zero.