find the derivative of the integral Prove that the following integral $F(x)$ is differentiable for every $x \in \mathbb{R}$ and calculate its derivative.
$$F(x) = \int\limits_0^1 e^{|x-y|} \mathrm{d}y$$
I don't know how to get rid of the absolute value in the integral
any ideas?

Let $\phi_y(x) = e^{|x-y|}$. 
The mean value theorem shows that $|e^a-e^b| \le e^{\max(a,b)} |a-b|$, and
also, $||a|-|b|| \le |a-b|$, hence
we have
$|{\phi_y(x+h)-\phi_y(x) \over h}| \le e^{\max(|x-y|,|x+h-y|)} |h|$ for $h \neq  0$.
If $\phi_y(x) = e^{|x-y|}$ then $\phi_y'(x) = e^{|x-y|}$ for $x>y$ and $\phi_y'(x) = -e^{|x-y|}$ for $x <y$.
If we fix $x$ and take $|h| < 1$, then we have $e^{\max(|x-y|,|x+h-y|)} \le M$ for some $M$ and all $y \in [0,1]$, hence the quotient
${\phi_y(x+h)-\phi_y(x) \over h}$ is uniformly bounded by $M$ and
${\phi_y(x+h)-\phi_y(x) \over h} \to \phi_y'(x)$ for ae. $y$. The dominated
convergence theorem gives
$F'(x) = \int_0^1 \phi_y'(x) dy$, hence we have
$F'(x) = \int_0^1 \operatorname{sgn}(x-y)\phi_y(x) dy = \int_0^1 \operatorname{sgn}(x-y)e^{|x-y|} dy$.
