I have a confusion when trying to get the result of the expression below, $$ I = \frac{\partial}{\partial y}\left(\int_r^y \frac{1}{\sqrt{y^2-s^2}} ds \right). $$ All variables are real and $y>r$.
In my first attempt, I tried to substitute $g(y,s)=1/\sqrt{y^2-s^2}$ and I got the expression below $$I = \frac{\partial}{\partial y} \int_r^y g(y,s) ds = g(y,y) + \int_r^y \frac{\partial g(y,s)}{\partial y} ds$$
which gave me an undefined result, since $g(y,y)$ contains a division by zero.
However, if I did the integration before taking its derivative, I got this result: $$ I = \frac{\partial}{\partial y} \left[\cos^{-1}(r/y) \right]=\frac{r}{y\sqrt{y^2-r^2}}.$$ In my first attempt, I got undefined result. But at the latter time, I got a different result. What is wrong with this?
Hints:
The function $$s~\mapsto~ \frac{\partial g(y,s)}{\partial y} $$ has a non-integrable singularity $\propto (y-s)^{-\frac{3}{2}}$ for $s\to y^{-}$, so the theorem for differentiation under the integral sign does not apply.
On the other hand, the un-differentiated integrand $s\mapsto g(y,s)$ has an integrable singularity $\propto (y-s)^{-\frac{1}{2}}$ for $s\to y^{-}$, so that it's possible to integrate first.