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