Answer
$$\frac{{dy}}{{dx}} = - {\csc ^2}x$$
Work Step by Step
$$\eqalign{
& y = {\left( {\tan x} \right)^{ - 1}} \cr
& {\text{differentiate using the chain rule}} \cr
& \frac{{dy}}{{dx}} = - {\left( {\tan x} \right)^{ - 2}}\frac{d}{{dx}}\left( {\tan x} \right) \cr
& {\text{so}} \cr
& \frac{{dy}}{{dx}} = - {\left( {\tan x} \right)^{ - 2}}\left( {{{\sec }^2}x} \right) \cr
& {\text{simplifying}}{\text{, }} \cr
& \frac{{dy}}{{dx}} = - {\left( {\tan x} \right)^{ - 2}}{\sec ^2}x \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{{{\left( {\tan x} \right)}^2}}}\left( {{{\sec }^2}x} \right) \cr
& \frac{{dy}}{{dx}} = - \frac{{{{\sec }^2}x}}{{{{\tan }^2}x}} \cr
& \frac{{dy}}{{dx}} = - \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}}\left( {\frac{1}{{{{\cos }^2}x}}} \right) \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{{{\sin }^2}x}} \cr
& \frac{{dy}}{{dx}} = - {\csc ^2}x \cr} $$
