If $\bf{X}$ is a function of time only i.e. $\bf{X} = \bf{X}\left(t\right)$, then, is the following development true?
$d\bf{X}=\frac{\partial \bf{X}}{\partial t}dt$.
$\Rightarrow$ $\frac{d\bf{X}}{dt}=\frac{\partial \bf{X}}{\partial t}$.
$\Rightarrow$ $d\bf{X} = \partial{\bf{X}}$.
Partial and total will be the same if there is only one independent variable.
Note if there are many independent variables, total derivative is made up of all the partial derivatives. Example, if $X$ is function of two variables, i.e. $X(u,v)$ then $$ dX = \frac{\partial X}{\partial u} du + \frac{\partial X}{\partial v} dv $$
If it turns out that $u$ and $v$ are not really independent, and say $v = v(u)$ then you get $$ dX = \frac{\partial X}{\partial u} du + \frac{\partial X}{\partial v} \frac{dv}{du} du = \left[ \frac{\partial X}{\partial u} + \frac{\partial X}{\partial v} \frac{dv}{du}\right] du = \frac{dX}{du} du $$
Sometimes it is a matter of semantics. Do you treat $u$ and $v$ as independent and then substitute $v=v(u)$ or you do it from the beginning is often a matter of convenience.
