# 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.

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 $$F_y=Q$$.