Check the second variation is positive for brachistochrone problem.

Question

When studying the brachistochrone problem, I was asked to check the second variation and verifying the cycloid is, indeed, a minimal path. I know the second variation is to check that: $$\frac{\partial^2 f}{\partial z'^2}\ge0,~\frac{\partial^2 f}{\partial z^2}-\frac{d}{dx}\frac{\partial^2 f}{\partial z'\partial z}>0.$$ In the brachistochrone problem, the function $f=\sqrt{\frac{1+y'^2}{2g(y_1-y)}}$, and $z=-y>0$. By substituting y with z and simply considering the starting point to be the origin, and reducing the coefficient, I have $f=\sqrt{\frac{1+z'^2}{z}}$. Then $\frac{\partial f}{\partial z'}=\frac{z'}{\sqrt{z(1+z'^2)}}$, $\frac{\partial^2 f}{\partial z^2}=\frac{3f}{4z^2}$,$\frac{\partial^2 f}{\partial z'\partial z}=-\frac{1}{2z}\frac{\partial f}{\partial z'}$, I wonder how can I calculate $\frac{d}{dx}\frac{\partial^2 f}{\partial z'\partial z}$. Thanks in advance!