Derivative of an Integral whose integrand is discontinuous Let, $g(x) =\int_{a}^{x} f(t) dt$ be an integral function. 
What can we say about $g'(c)$ when, 
a) $f$ is removable discontinuous at $c \in[a,x]$?
b) $f$ is infinite discontinuous at $c\in [a,x]$? 
c) $f$ is jump discontinuous at $c\in [a,x]$?
Here is what I think. 
a) Given, $$f(x) = \begin{cases} {x^3} & \text{if $x\in \mathbb{R}  - \{2\} $} \\ 10 & \text{if $x=2$} \end{cases}$$
Here, $x=2=c$ is a point of removable discontinuity. The right-hand derivative of $g(x)$ at $c$ is given as, $$g'(c) =\lim_{h\to 0^{+} } \frac{g(c+h)-g(c)}{h}= \lim_{h\to 0^{+} }\frac{\int_{c}^{c+h}f(t) dt}{h}$$
The left-hand derivative of $g(x)$ at $c$ is given as, $$g'(c) = \lim_{h\to 0^{+} } \frac{g(c-h)-g(c)}{-h}= \lim_{h\to 0^{+} }\frac{\int_{c-h}^{c}f(t) dt}{-h}$$
We see that the left and right derivatives converge at 8. This means that $g'(2)=8$. We know that, "for continuous integrands", $g'(x)=f(x)$. Here, $f(2)\neq 8$. This tells me that the derivative does not exist at $x=2$. 
Am I thinking in the right direction? 
 A: To expand on my comment which covers only cases a) and c) I show that the Fundamental Theorem of Calculus can be applied in case a) also. Let $f$ have a removable discontinuity at $c$ and let $h(x) =f(x) $ for all $x\neq c$ and $h(c) =\lim_{x\to c} f(x) $ so that $h$ is continuous at $c$. Now changing integrand at a finite number of points does not change the integral so $$g(x) =\int_{a} ^{x} f(t) \, dt=\int_{a} ^{x} h(t) \, dt$$ and by Fundamental Theorem of Calculus we have $g'(c) =h(c) =\lim_{x\to c} f(x) $. 
For case b) the integral does not make sense as the integrand is unbounded. For that you will need to deal with improper Riemann integral and then if the integral exists the derivative $g'(c) $ will also be infinite. You can see for example the case when $$g(x) =\arcsin x, f(x) =1/\sqrt{1-x^{2}},a=0,c=1$$
For case c) the derivative $g'(c) $ does not exist (based on my comment, the left and right derivatives are unequal).
There is another case d) which you have not considered where $f$ has oscillatory discontinuity at $c$. For that case you have the favorite example $$f(x) =\cos(1/x),f(0)=0,a=c=0, g'(0)=0$$ It is possible in this case that $g'(c)$ may not exist, but I don't have an example with me right now. 
