Fundamental theorem of calculus with improper integral Suppose $f(x,y)$ is continuous everywhere except where $y=x$. Does FTC apply in a scenario where $x$ is given as a boundary? If so, how? Here's a general example.
$$\frac{d}{dx}\int_x^bf(x,y) dy$$
There is a related post here that doesn't answer this question. The specific integral I'm looking at is more like
$$\frac{d}{dx}\int_x^\infty \frac{g(y)}{\sqrt{y^2-x^2}} dy$$
where it is known that the original integral (prior to differentiation) converges for all $x\ge 0$. I'm tentatively approaching the derivative as
$$\frac{d}{dx}\int_x^c\frac{g(y)}{\sqrt{y^2-x^2}} dy + \lim_{b\to\infty}\int_c^b\frac{d}{dx}\frac{g(y)}{\sqrt{y^2-x^2}} dy$$
 A: I ended up solving my specific problem by introducing a substitution to set the limits of integration to something nicer. 
\begin{align}
\frac{d}{dx}\int_x^\infty \frac{g(y)}{\sqrt{y^2-x^2}} dy 
& =\frac{d}{dx}\int_0^\infty \frac{g(y)}{\sqrt{u}} \frac{du}{2y} 
&\text{ using: } u = y^2 - x^2
\end{align}
Since $u$ is a dummy variable, I don't see an issue with interchanging the order of operations. From there, one can apply the chain rule, and then undo the substitution.
\begin{align}
\frac{d}{dx}\int_0^\infty \frac{1}{\sqrt{u}} \frac{g(y)}{2y} du
&= \int_0^\infty \frac{1}{\sqrt{u}} \frac{d}{dx}\left(\frac{g(y)}{2y}\right)du 
\\
&= \int_0^\infty \frac{1}{\sqrt{u}} \frac{d}{dy}\left(\frac{g(y)}{2y}\right)\frac{dy}{dx} du 
\\
&= \int_x^\infty \frac{1}{\sqrt{y^2-x^2}} \frac{d}{dy}\left(\frac{g(y)}{2y}\right)\frac{x}{y} (2ydy) 
\\
&= x\int_x^\infty \frac{1}{\sqrt{y^2-x^2}} \frac{d}{dy}\left(\frac{g(y)}{y}\right) dy
\end{align}
So, in conclusion:
\begin{align}
\frac{d}{dx}\int_x^\infty \frac{g(y)}{\sqrt{y^2-x^2}} dy 
&= x\int_x^\infty \frac{1}{\sqrt{y^2-x^2}} \frac{d}{dy}\left(\frac{g(y)}{y}\right) dy
\end{align}
