Answer
$$\frac{{dy}}{{dx}} = \frac{1}{{\ln 8}}\csc x\sec x$$
Work Step by Step
$$\eqalign{
& y = {\log _8}\left| {\tan x} \right| \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\log }_8}\left| {\tan x} \right|} \right] \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\left( {\ln 8} \right)\tan x}}\frac{d}{{dx}}\left[ {\tan x} \right] \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\left( {\ln 8} \right)\tan x}}\left( {{{\sec }^2}x} \right) \cr
& {\text{Simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\ln 8}}\csc x\sec x \cr} $$
