why can we get $\frac{\partial q_t(x,y)}{\partial t}=\Delta q_t(x,y)?$ If for the transition density $p_n(x,y)=P^{x}(X_n=y)$ of discrete time random walk $X$ on a general garph $G$, we have
$$q_t(x,y)=\sum_{n=0}^{\infty}\frac{e^{-t}t^n}{n!}p_n(x,y), \quad t\ge0$$, why can we get
$$\frac{\partial q_t(x,y)}{\partial t}=\Delta q_t(x,y)?$$
If we differentiate both sides of $q_t$, we get
$$\frac{\partial q_t(x,y)}{\partial t}=-q_t(x,y)+\sum_{n=1}^{\infty}\frac{e^{-t}t^{n-1}}{(n-1)!}p_n(x,y).$$
Why the RHS would be $\Delta q_t(x,y)?$
 A: What is actually happening here is the usual decomposition of a continuous time Markov chain into a jump chain and holding times. In this case the average holding time at each state is $1$, and the $n$th term in the summation of $q_t(x,y)$ is the probability to go from $x$ to $y$ by making exactly $n$ jumps up to time $t$. The (suitably normalized) graph Laplacian appears as a result of the fact that as $\Delta t \to 0$ the probability that more than one jump occurs in time $[t,t+\Delta t]$ goes to zero asymptotically more rapidly than $\Delta t$.
In detail, if you want to differentiate at a later time, you have
$$q_{t+h}(x,y)=\sum_z q_t(x,z) \sum_{n=0}^\infty e^{-h} \frac{h^n}{n!} p_n(z,y).$$
This says the probability to go from $x$ to $y$ in time $t+h$ is the sum of the probability to go from $x$ to $z$ in time $t$ weighted by the probability to go from $z$ to $y$ in time $h$.
Now you differentiate that with respect to $h$ and set $h=0$. So you get
$$\frac{\partial q_t}{\partial t}(x,y)=\left. \sum_z q_t(x,z) \left ( -q_h(z,y) + \sum_{n=1}^\infty e^{-h} \frac{h^{n-1}}{(n-1)!} p_n(z,y) \right ) \right |_{h=0} \\
= \sum_z q_t(x,z) \left ( -q_0(z,y) + p_1(z,y) \right ).$$
