A version of the Fundamental Theorem of Calculus for two variables Let $f(x,y)$ be differentiable in the rectangle $R=[a,b]\times[c,d]$, show that the function

$\displaystyle F(x,y)=\int_{a}^{x} f(t,y)  \, dt$

is also differentiable in $R$ and that

$\displaystyle \frac{\partial F}{\partial x}=f(x,y)$
$\displaystyle \frac{\partial F}{\partial y}=\int_{a}^{x} \frac{ \partial f} {\partial y}(t,y)  \, dt$

So far, I've been trying by saying that since $f$ is differentiable then $f$ is continuous in $R$ an then trying to use this somehow in the definition of differentiability.
I was suggested to add and subtract $F(x,y+k)$ (From the definition of differentiability I guess?) and then to apply the Mean Value Theorem for Integrals, but to far I haven't been able to prove what I need.
Do you know how to prove it? Or maybe you know the name of this theorem so I can research on it.
 A: Before starting, I believe there is something to say about differentiability; the formulae in the OP can be proven if the partial derivatives of $f$ are continuous at all point $(x,y)$ in the compact $[a,b]\times [c,d]$. Therefore, in what follows "differentiable" means "belonging to class $C^1([a,b]\times [c,d])$".  
On the second formula; we begin by writing
$$\frac{F(x,y+h)-F(x,y)}{h}:=\int_{a}^{x}\frac{f(t,y+h)-f(t,y)}{h}dt =
\int_{a}^{x}\frac{\partial f}{\partial y}(t,y)dt+\int_{a}^{x}\left[\frac{f(t,y+h)-f(t,y)}{h}-\frac{\partial f}{\partial y}(t,y)\right]dt =
\int_{a}^{x}\frac{\partial f}{\partial y}(t,y)dt+\int_{a}^{x}\left[\frac{\partial f}{\partial y}(t,y+\theta h)-\frac{\partial f}{\partial y}(t,y)\right]dt $$
by the mean value theorem (as suggested in the OP), with $\theta=\theta(t,y,h),$  $0<\theta<1$.
Now we use the fact that $f\in C^1([a,b]\times [c,d])$: this implies that also $\frac{\partial f}{\partial y}$ is continuous on $[a,b]\times[c,d]$ and therefore $ivi$ uniformly continuous (quickly: a continuous function on a compact subset of $\mathbb R^n$ is uniformly continuous on that subset). In symbols:
$$\forall \epsilon >0,~ \exists \delta=\delta(\epsilon): ~|h|<\delta\Rightarrow 
|\frac{\partial f}{\partial y}(t,y+\theta h)-\frac{\partial f}{\partial y}(t,y)|<\epsilon $$
In summary
$$|\frac{F(x,y+h)-F(x,y)}{h} - \int_{a}^{x}\frac{\partial f}{\partial y}(t,y)dt |\leq \int_a^x |\frac{\partial f}{\partial y}(t,y+\theta h)-\frac{\partial f}{\partial y}(t,y)|dt\leq \epsilon (x-a)\leq \epsilon(b-a), $$
for all $x\in[a,b]$.
