Thomas' Calculus 13th Edition

$$\frac{{dy}}{{dx}} = \frac{{1 - {x^2}}}{{{{\left( {1 + x} \right)}^4}}}$$
\eqalign{ & y = {\left( {\frac{{\sqrt x }}{{1 + x}}} \right)^2} \cr & {\text{simplify the function}} \cr & y = \frac{x}{{{{\left( {1 + x} \right)}^2}}} \cr & y = \frac{x}{{1 + 2x + {x^2}}} \cr & {\text{differentiate with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{x}{{1 + 2x + {x^2}}}} \right] \cr & {\text{use the quotient rule}} \cr & \frac{{dy}}{{dx}} = \frac{{\left( {1 + 2x + {x^2}} \right)\frac{d}{{dx}}\left( x \right) - x\frac{d}{{dx}}\left( {1 + 2x + {x^2}} \right)}}{{{{\left( {1 + 2x + {x^2}} \right)}^2}}} \cr & {\text{find the derivatives and simplify}} \cr & \frac{{dy}}{{dx}} = \frac{{\left( {1 + 2x + {x^2}} \right)\left( 1 \right) - x\left( {2 + 2x} \right)}}{{{{\left( {1 + 2x + {x^2}} \right)}^2}}} \cr & \frac{{dy}}{{dx}} = \frac{{1 + 2x + {x^2} - 2x - 2{x^2}}}{{{{\left( {1 + 2x + {x^2}} \right)}^2}}} \cr & \frac{{dy}}{{dx}} = \frac{{1 - {x^2}}}{{{{\left( {1 + x} \right)}^4}}} \cr}