If we have a vector, $\,U=U\left(x_1,x_2,x_3\right)$, in the coordinate axis $\left(x_1,x_2,x_3\right)$, then why does the following differential relation hold?
$$ dU= \left(\frac{\partial U}{\partial x_1}\right)dx_1 + \left(\frac{\partial U}{\partial x_2}\right)dx_2 + \left(\frac{\partial U}{\partial x_3}\right)dx_3 $$
Help would be appreciated – I'd like to see the proof of why this holds in the vector case.
Given a vector valued function $u:\>(\Omega\subset {\mathbb R}^n)\to{\mathbb R}^m$ and a point $p\in\Omega$ the differential of $u$ at $p$ is a linear map $$du(p) :\quad T_p\to T_{u(p)}, \qquad X\mapsto du(p).X\ ,$$ which is defined by $$u(p+X)-u(p)=du(p).X+o\bigl(|X|\bigr)\qquad(X\to0)\ .\tag{1}$$ Since $p$ will be fixed in the sequel I shall just write $du$ for this map. Particular such functions are the coordinate functions $$x_i:\quad x=(x_1,\ldots, x_n)\mapsto x_i=\langle x,e_i\rangle\qquad(1\leq i\leq n)\ .$$ For these one has $$x_i(p+X)-x_i(p)=p_i+X_i-p_i=X_i\ ,$$ so that, trivially, $$dx_i(X)=X_i\qquad(1\leq i\leq n)\ .$$ Now $$du.X=du.\left(\sum_{i=1}^n X_i e_i\right)=\sum_{i=1}^n X_i\ \ du.e_i=\sum_{i=1}^n du.e_i \ \ dx_i(X)\ .\tag{2}$$ It remains to find out what the $du.e_i$ are. Putting $X:=t\>e_i$ in $(1)$ we have $$u(p+ t e_i)-u(p)=t\ du.e_i +o\bigl(|t|\bigr)\qquad (t\to0)\ ,$$ which is the same as $$du.e_i=\lim_{t\to0}{u(p+ t e_i)-u(p)\over t}=:\left({\partial u\over\partial x_i}\right)_p\ .$$ Plugging this into $(2)$ we obtain $$du.X=\sum_{i=1}^n \left({\partial u\over\partial x_i}\right)_p\ dx_i(X)\ ,$$ and since this is true for all $X\in T_p$ we may as well write $$du=\sum_{i=1}^n \left({\partial u\over\partial x_i}\right)_p\ dx_i\ .$$ Note that $\left({\partial u\over\partial x_i}\right)_p$ is a vector, namely the $i^{\rm th}$ column vector of the Jacobian matrix of $u$ at $p$.