Can somebody walk me through how the steps $(10)$ and $(11)$ were carried out?
$(10)$ What happened with $\sin \theta$ after the substitution?
$(11)$ What's the name of the theorem which allows $f = \frac{d}{dx} \int f dx$ because I think that's what have been carried out in this step (Leibnitz Integral rule?)
link: http://physweb.bgu.ac.il/COURSES/PHYSICS_ExercisesPool/33_Electric_Field/e_33_2_132_s_TeX.pdf
The substitution $\cos \theta = t$ is applied. This gives that $-sin \theta d\theta = dt$, making the $\sin \theta$ disappear in equation (10), but adding a minus in front of the integral. Because of the substitution, $t$ runs from $\cos 0 = 1$ to $\cos \pi = -1$. Using the minus from $-\sin\theta d\theta$, the integration bounds are switched to their usual order: from $-1$ to $1$.
In line (11), we compute the derivative of $1/\sqrt{R^2 + z^2 - 2Rzt} = (R^2 + z^2 - 2Rzt)^{-1/2}$ with respect to $z$. This derivative equals $$-\frac{1}{2}\frac{1}{\sqrt[3]{R^2 + z^2 - 2Rzt}}(2z - 2Rt)$$ and then derivative and integral are switched.
The switching of derivative with respect to $z$ and integral with respect to $t$ is possible because of Leibniz Integral rule, since the integration bounds are independent of $z$.