Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 25

Determine the derivative of the function $\displaystyle F(z) = \sqrt{\frac{z-1}{z+1}}$

$\begin{equation} \begin{aligned} F'(z) &= \frac{d}{dz} \left( \frac{z-1}{z+1} \right)^{\frac{1}{2}}\\ \\ F'(z) &= \frac{1}{2} \left( \frac{z-1}{z+1} \right)^{\frac{-1}{2}} \frac{d}{dz} \left( \frac{z-1}{z+1}\right)\\ \\ F'(z) &= \frac{1}{2} \left( \frac{z-1}{z+1} \right)^{\frac{-1}{2}} \left[ \frac{(z+1) \frac{d}{dz}(z-1)-(z-1)\frac{d}{dz}(z+1)}{(z+1)^2}\right]\\ \\ F'(z) &= \frac{1}{2} \left( \frac{z-1}{z+1} \right)^{\frac{-1}{2}} \left[ \frac{(z+1)(1)-(z-1)(1)}{(z+1)^2}\right]\\ \\ F'(z) &= \frac{1}{2} \left( \frac{z-1}{z+1} \right)^{\frac{-1}{2}} \left[ \frac{\cancel{z}+1 - \cancel{z} +1 }{(z+1)^2}\right]\\ \\ F'(z) &= \frac{1}{\cancel{2}} \left( \frac{z-1}{z+1} \right)^{\frac{-1}{2}} \left[ \frac{\cancel{2}}{(z+1)^2}\right]\\ \\ F'(z) &= \frac{(z-1)^{\frac{-1}{2}}}{(z+1)^{\frac{-1}{2}}} \left[ \frac{1}{(z+1)^2} \right]\\ \\ F'(z) &= \frac{(z-1)^{\frac{-1}{2}}}{(z+1)^{\frac{3}{2}}}\\ \\ F'(z) &= \frac{1}{(z+1)^{\frac{3}{2}} (z-1)^{\frac{-1}{2}}} \end{aligned} \end{equation}$