Let $U: \mathbb{R} \times \mathbb{R}^N \to \mathbb{R}$ be a smooth function. In particular, we know that $U$ is of the form $$ U(x, y_1, \ldots, y_N) = F_x ( y_1, \ldots, y_N), $$ where $F_x: \mathbb{R}^N \to \mathbb{R}$ is also a smooth function, for each $x \in \mathbb{R}$. (Not sure whether this is relevant though.)
I am wondering whether the following is true:
$$\partial_{x_i} U(x_i, x_1, x_2, \ldots, x_N) = \partial_{x_i} U(x_i, \mathbf{z} ) \bigg|_{\mathbf{z} = ( x_1, x_2, \ldots, x_N)} + \partial_{x_i} U( \alpha, x_1, x_2, \ldots, x_N) \bigg|_{\alpha=x_i}.$$
This seems intuitive, but I can't be able to apply the chain rule to this case. Differentiation from first principles doesn't help much: $$ \begin{eqnarray} && \frac{ U(x_i+h, x_1, x_2, \ldots, x_i+h, x_{i+1}, \ldots, x_N) - U(x_i, x_1, x_2, \ldots, x_N) }{h} \\ & = & \frac{ U(x_i+h, x_1, x_2, \ldots, x_i+h, x_{i+1}, \ldots, x_N) - U(x_i+h, x_1, x_2, \ldots, x_N) }{h} \\ && + \frac{ U(x_i+h, x_1, x_2, \ldots, x_N) - U(x_i, x_1, x_2, \ldots, x_N) }{h}. \end{eqnarray} $$ The second term becomes the first term on the RHS of the equation in the box, upon taking limits. However, the first term is not quite the second term on the RHS of the equation in the box, since we have $x_i+h$ instead of $x_i$. Any ideas?
Try to check it this way:
Define $t(x_i)=x_i$ and $s(x_i)=x_i$ And as $x_j,j\neq i$ are independent of $x_i$, we have a total derivative.
$$\frac{\mathrm dU(t(x_i), x_1, \ldots s(x_i),\dots x_N)}{\mathrm dx_i}=\partial_{t}U(t(x_i),x_1\dots x_n)\frac{\mathrm dt}{\mathrm dx_i}+\partial_{s}U(x_i,x_1\dots s(x_i)\dots x_n)\frac{\mathrm ds}{\mathrm dx_i}$$