Answer
$z(w) = \frac{-w}{(w^{2}-1)^{\frac{3}{2}}}$
Work Step by Step
$z(w) = \frac{1}{\sqrt (w^{2}-1)}$
Using the derivative definition:
$z'(w) = \lim\limits_{h \to 0}\frac{\frac{1}{\sqrt ((w+h)^{2}-1)}-\frac{1}{\sqrt (w^{2}-1)}}{h}$
$z(w) = \lim\limits_{h \to 0}\frac{(w^{2}-1) - ((w+h)^{2}-1)}{h(\sqrt((w+h)^{2}-1) \sqrt(w^{2}-1))(\sqrt ((w+h)^{2}-1) +\sqrt (w^{2}-1))}$
$z(w) = \lim\limits_{h \to 0}\frac{-(h+2w)}{(\sqrt((w+h)^{2}-1) \sqrt(w^{2}-1))(\sqrt ((w+h)^{2}-1) +\sqrt (w^{2}-1))}$
$z(w) = \frac{-(2w)}{2(w^{2}-1)^{\frac{3}{2}}}$
$z(w) = \frac{-w}{(w^{2}-1)^{\frac{3}{2}}}$
