Fundamental Thm of Calculus(two variables) When I'm studying the Calculus, I have a curious that
Let  $F(x,y) = \int_a^{x} \int_c^{y}f(u,v) dudv$, ($f \in C(D)$ $s.t.$ $D=[a,b]\times[c,d]$)
which means $f$ is continuous in D
Then What is $F_x(=\frac{\partial F}{\partial x})$???
Is it $F_x$ = $\int_c ^{y}f_x dy??$ (Just my thought :)
Any answer will be appreciated. Thanks.
 A: The answer is,
$$F_x(x, y) = \color{blue}{\int_c^y f(x, v) \, \mathrm{d}v}.$$
Why? Because of the fundamental theorem of calculus. Remember that, in order to calculate $F_x(x, y)$, we are considering the derivative of the function $F(\cdot, y)$, where $y$ is fixed and constant. The definition of the function $F(x, y)$ is the integral
$$F(x, y) = \int_a^x \color{red}{\int_c^y f(u, v) \, \mathrm{d}v}\, \mathrm{d}u,$$
which, when $y$ is considered fixed, is precisely in the form of the (one variable) fundamental theorem of calculus, which states that under the relevant conditions,
$$\frac{\mathrm{d}}{\mathrm{d}x} \int_a^x g(u) \, \mathrm{d}u = g(x).$$
In this case, we have
$$g(u) = \color{red}{\int_c^y f(u, v) \, \mathrm{d}v},$$
so the derivative with respect to $x$ is
$$g(x) = \color{blue}{\int_c^y f(x, v) \, \mathrm{d}v}.$$
A: You are integrating on the rectangular area $[a,x]\times[c,y]$. 
$$
F(x+h,y)=\int_a^{x+h}\int_b^yf(u,v)dudv\\
=\int_a^{x}\int_b^yf(u,v)dudv+\int_x^{x+h}\int_b^yf(u,v)dudv\\
\approx F(x,y)+h\int_b^yf(u,x)du.
$$
as $h\to 0$. So by the definition $F_x=\lim_{h\to 0}\frac{F(x+h,y)-F(x,y)}{h}$, you have
$$
F_x=\int_b^y f(u,x)du.
$$
Similarly, you can obatain
$$
F_y=\int_a^x f(y,v)dv.
$$
PS: You write $dudv$ in your question. I follow this order, but it is better to write $dvdu$
You can verify this by an example. Let
$$
f(u,v)=6uv^2,\\
\int_c^y 6uv^2 du=3(y^2-c^2)v^2.\\
F(x,y)=\int_a^{x}\int_b^y 6uv^2 dudv=(y^2-c^2)(x^3-a^3).
$$
Now, you can clearly see $F_x=\int_c^y 6ux^2 du$.
