Why is this derivative of an integral calculated in this way? Could anyone please give me an insight as to why the following derivative is correct?
The function to derivate is
$$f(x) = -ax +b\int_{-\infty}^xF(\xi)d\xi$$
The result is
$$f'(x) = -a + bF(x)$$
Particularly, I do not understand how the integral is resolved to $bF(x)$. What rule am I missing?
 A: Well first, the differentiation is not correct unless you make some more assumptions about $F$; you really need to tell us what you know about $F$ here.
For example, let $a=0$, $b=1$ and $$F(\xi)=\begin{cases}
0,&(\xi<0),
\\1,&(\xi\ge0).\end{cases}$$Then $f'(0)$ does not exist. (Note that this $F$ is a CDF...)
But the result in question is just a special case of FTC if you assume (of course) that $\int_{-\infty}^0F$ exists and that $F$ is continuous. Contrary to one of the comments, the fact that it's an improper integral is no problem here:


FTC. Suppose $F$ is continuous on $[a,b]$ and $I(x)=\int_a^x F(\xi)$ for $x\in [a,b]$.  Then $I'=F$ on $(a,b)$.


Now assuming that $\int_{-\infty}^0F$ exists and that $F$ is continuous, write $$I(x)=\int_{-\infty}^x F=\int_{-\infty}^0+\int_0^x=I_1(x)+I_2(x).$$Now $I_1$ is constant, and FTC shows that $I_2'(x)=F(x)$.
A: Hint:
Assume the antiderivative of $F$ to be $G$.
Now what is the derivative of
$$-ax+b(G(x)-G(-\infty))\ ?$$ 
A: The fundamental theorem of Calculus says essentially that $\frac{d}{dx}\int_a^x f(z)dz = f(x)$, so the function that has as its variable the upper limit of some integral over a function is differentiable with value $f(x)$. (normally stated for finite intervals, but valid too for integrable functions on infinite ones, as here)
A: We have
$$
f(x) = -ax +b\int_{-\infty}^0F(\xi)d\xi +b\int_{0}^xF(\xi)d\xi
= -ax +f(0) +b\int_{0}^xF(\xi)d\xi
$$
The claim now follows from the fundamental theorem of calculus if $F$ is continuous.
