Answer
$$\frac{{dy}}{{dx}} = \frac{{\left( {x - 1} \right){{\sec }^2}x - \tan x}}{{{{\left( {x - 1} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{\tan x}}{{x - 1}} \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = {D_x}\left[ {\frac{{\tan x}}{{x - 1}}} \right] \cr
& {\text{use the quotient rule for derivatives}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {x - 1} \right) \cdot {D_x}\left( {\tan x} \right) - \tan x \cdot {D_x}\left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^2}}} \cr
& {\text{solve derivatives}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {x - 1} \right){{\sec }^2}x - \tan x\left( 1 \right)}}{{{{\left( {x - 1} \right)}^2}}} \cr
& {\text{multiply}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {x - 1} \right){{\sec }^2}x - \tan x}}{{{{\left( {x - 1} \right)}^2}}} \cr} $$