Is the integral from $x$ to $y$ of a continuous function differentiable? 
Let $f(t)$ be integrable and continuous function on $[a,b]$. Let ${ F(x,y) = \int_{x}^{y}{f(t)\,dt} }$. Show that $F(x,y$) is differentiable on the rectangle $[a,b] \times [a,b]$.

I tried to prove that
 ${ \frac{{ f(x_0+ \triangle x, y_0+ \triangle y)  - f(x_0, y_0) - \frac{\partial f} {\partial x}  (x_0, y_0)\triangle x - \frac{\partial f} {\partial y}  (x_0, y_0)\triangle y}}{ \sqrt{(\triangle x)^2 + (\triangle x)^2}} \to 0}$ as $(\triangle x, \triangle y) \to (0,0)$ by the differential definition.
But how do I found the partial derivatives? and even if I'm able to find them, how do I estimate the whole expression? 
 A: First of all I'll show differentiability on $(a,b) \times (a,b)$, because defining differentiability on a closed set can be tricky.

What you need to use is the Fundamental theorem of calculus and the (multivariable) chain rule. Temporarily define the function $\psi:[a,b] \to \Bbb{R}$ by
\begin{align}
\psi(x) = \int_a^xf
\end{align}
Since $f$ is assumed to be continuous, the FTC implies that $\psi$ is differentiable on $(a,b)$, and that $\psi' = f$. 
Next, define the projection functions $\pi_1: \Bbb{R}^2 \to \Bbb{R}$, by $\pi_1(x,y) = x$, and similarly, $\pi_2$ is projection onto second factor. Then, the function $F$ you're given can be written as a difference of compositions:
\begin{align}
F = \psi \circ \pi_2 - \psi \circ \pi_1
\end{align}
Now, note that both $\pi_i$'s are linear maps, and hence they are differentiable on $(a,b) \times (a,b)$. Notice that $F$ is a difference of a composition of differentiable functions. Hence by the chain rule and sum rule, $F$ is also differentiable on $(a,b) \times (a,b)$. 

The argument above proves differentiability of $F$. Now, the actual derivative matrix $F'(x,y)$ is just a $1 \times 2$ matrix which can be computed using the chain rule and sum rule:
\begin{align}
F'(x,y) &= \psi'(\pi_2(x,y)) \cdot (\pi_2)'(x,y) - \psi'(\pi_1(x,y)) \cdot (\pi_1)'(x,y) \\\\
&= f(y) \cdot \begin{pmatrix} 0 & 1 \end{pmatrix} - f(x) \cdot \begin{pmatrix} 1 & 0 \end{pmatrix} \qquad \text{since $\psi'=f$ by FTC}\\\\
&=
\begin{pmatrix}
-f(x) & f(y)
\end{pmatrix}
\end{align}

If you try to prove this directly from the difference quotient, you'll pretty much end up proving the chain rule in a very special case along with the single variable fundamental theorem of calculus (and very likely give an incorrect/incomplete proof along the way if you're not careful). If you need to review any of those proofs, you should do so, but after learning them, you should just invoke them without re-proving them.
