Need clarification on pH titrimetry .
Since $pH = -log(a_{[H^+]})$, $[H^+]_{initial}$ is irreverent when discussing the equivalence point. You can always show a full titration curve over all pH for a given acid-base titration regardless of initial proton concentrations. Therefore, equivalence point is independent of initial concentration. That being said, the fact that the equivalence point occurs when $\frac{pH}{dV}$ is at a maximum  has to do with the shape of the pH vs Volume graph. Since titration curves look similar to $y = tan^{-1}(x)$, the maximum slope will occur at the equivalence point (that would be (0,0) for inverse tangent). This can be further confirmed by taking the second derivative and comparing where the graph crosses 0 and comparing to the maximum of $\frac{pH}{dV}$.

As for the derivative, using a derivative calculator to ensure the correct answer, I got:

$2[H^+]-(\frac{C_{A}V_{A}}{V_{A}+V_{B}})+(\frac{C_{B}V_{B}}{V_{A}+V_{B}})$ which simplifies to $2[H^+]+\frac{C_{B}V_{B}-C_{A}V_{A}}{V_{A}+V_{B}}$. A simple way to determine if these are the same is to use a CAS (computer algebra system) and divide the answer provided by one of these answers. If the result is 1, the difference is only in simplification.

I hope this answers the question adequately.