Let $G$ be a weighted complete graph with n vertices, where weights are distinct variables. Let $x_{ij}$ denote the weight of the edge $ij$. Show that $\partial k(G)/\partial x_{ij} = k(G/ij)$ where $G/ij$ is the contraction of G by the edge $ij$ and $k$ of graph refers to the number of spanning trees.
In above statement, can't imagine how the spanning number of graph $k$ could be differentiated by the weight of an edge.
The image set of $k$ would be discrete space but is it possible to differentiate it with an variable $\in \Bbb R$ ?
The statement of the problem should be true if we define $k(G)$, for a weighted graph $G$, to be $$ k(G) = \sum_{T} \prod_{ij \in E(T)} x_{ij} $$ where the sum ranges over all spanning trees $T$. That is, the weight of a spanning tree is the product of the weights of its edges, and $k(G)$ is the total weight of all spanning trees of $G$. (This reduces to the number of spanning trees in an arbitrary graph if we set $x_{ij} = 1$ for edges and $x_{ij}=0$ for non-edges.)