Answer
$-\frac{2a^2z}{a^4-z^4}$
Work Step by Step
$\frac{d}{dz}\ln\sqrt{\frac{a^2-z^2}{a^2+z^2}}$
In this case, it is much easier to simplify the expression (using the logarithm laws on page 439) before taking the derivative.
$=\frac{d}{dz}\frac{1}{2}\ln\frac{a^2-z^2}{a^2+z^2}$
$=\frac{d}{dz}\frac{1}{2}(\ln(a^2-z^2)-\ln(a^2+z^2))$
$=\frac{1}{2}(\frac{1}{a^2-z^2}\frac{d}{dz}(a^2-z^2)-\frac{1}{a^2+z^2}\frac{d}{dz}(a^2+z^2))$
$=\frac{1}{2}(\frac{-2z}{a^2-z^2}-\frac{2z}{a^2+z^2})$
$=-\frac{z}{a^2-z^2}-\frac{z}{a^2+z^2}$
$=-\frac{z}{a^2-z^2}*\frac{a^2+z^2}{a^2+z^2}-\frac{z}{a^2+z^2}*\frac{a^2-z^2}{a^2-z^2}$
$=-\frac{a^2z+z^3}{a^4-z^4}-\frac{a^2z-z^3}{a^4-z^4}$
$=\frac{-(a^2z+z^3)-(a^2z-z^3)}{a^4-z^4}$
$=\frac{-a^2z-z^3-a^2z+z^3}{a^4-z^4}$
$=-\frac{2a^2z}{a^4-z^4}$
