## Calculus: Early Transcendentals 8th Edition

$H'(z) = \frac{2a^2~z}{z^4-a^4}$
$H(z) = ln~\sqrt{\frac{a^2-z^2}{a^2+z^2}}$ We can differentiate the function: $H'(z) = \frac{1}{\sqrt{\frac{a^2-z^2}{a^2+z^2}}}\cdot \frac{d}{dx}(\sqrt{\frac{a^2-z^2}{a^2+z^2}})$ $H'(z) = \sqrt{\frac{a^2+z^2}{a^2-z^2}}\cdot \frac{1}{2}\frac{1}{\sqrt{\frac{a^2-z^2}{a^2+z^2}}}\cdot\frac{d}{dx}(\frac{a^2-z^2}{a^2+z^2})$ $H'(z) = \frac{1}{2}\cdot \frac{a^2+z^2}{a^2-z^2}\cdot \frac{-2z(a^2+z^2)-(a^2-z^2)(2z)}{(a^2+z^2)^2}$ $H'(z) = \frac{1}{a^2-z^2}\cdot \frac{-2za^2}{a^2+z^2}$ $H'(z) = \frac{2a^2~z}{z^4-a^4}$