Show that $F_x(x,y) = P(x,y), \ F_y(x,y) = Q(x,y)$. In Complex Variables and Applications by Brown and Churchill it comes: 

When the point $(x_0,y_0)$ is kept fixed and $(x , y)$ is allowed to vary throughout a simply connected domain $D$, the integral represents a single-valued function $$F(x,y) = \int_{(x_0,y_0)}^{(x,y)} P(s,t) ds+ Q(s,t) dt$$ of $x$ and $y$ whose first-order partial derivatives are given by the equations $$F_x(x,y) = P(x,y), \ \ F_y(x,y) = Q(x,y).$$ 

My question is how $F_x(x,y) = P(x,y), \  F_y(x,y) = Q(x,y)$ hold? 
I searched MSE for similar question, but no much similarities with other questions and Leibniz integral rule couldn't be much help: For, there are variables in upper bound (actually two), partial derivative with respect to one variable makes the other function zero!! and confusion of different variables 'inside' the integral.   
 A: Fix a point $(x,y)$. Since the domain $D$ is open, there is some small disk $\Delta$ centered at $(x,y)$ which is entirely contained in $D$. The line segment $(a,y) \to (x,y)$ is thus contained in $D$ for some constant $a$ sufficiently close to $x$. Since we have independence of the integration path, choose a path which starts by going from $(x_0,y_0)$ to $(a,y)$, followed by the line segment $(a,y) \to (x,y)$. We then have
$$F(x,y)=\int_{(x_0,y_0)}^{(a,y)} P(s,t) \,\mathrm{d} s+ Q(s,t)\, \mathrm{d}t+\int_{(a,y)}^{(x,y)} P(s,t) \,\mathrm{d} s+ Q(s,t)\, \mathrm{d}t . $$
Taking the partial derivative w.r.t. $x$ we notice that the first term vanishes, meaning
$$F_x(x,y)=\frac{\partial}{\partial x} \int_{(a,y)}^{(x,y)} P(s,t) \,\mathrm{d} s+ Q(s,t)\, \mathrm{d}t. $$
This integral can be rewritten using the parameterization $s=u,t=y$, for $a \leqslant u \leqslant x$, giving
$$F_x(x,y)=\frac{\partial}{\partial x} \int_a^x P(u,y) \mathrm{d} u. $$
The Fundamental Theorem of Calculus then gives $F_x=P$. A similar approach gives $F_y=Q$.
