Answer
$\frac{{dy}}{{dx}} = \frac{1}{2(1 + {x^2})}$
Work Step by Step
$$\eqalign{
& y = {\tan ^{ - 1}}\left( {x - \sqrt {1 + {x^2}} } \right) \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\left( {x - \sqrt {1 + {x^2}} } \right)} \right] \cr
& {\text{Use }}\frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}u} \right] = \frac{1}{{1 + {u^2}}}\frac{{du}}{{dx}},{\text{ let }}u = x - \sqrt {1 + {x^2}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{1 + {{\left( {x - \sqrt {1 + {x^2}} } \right)}^2}}}\frac{d}{{dx}}\left[ {x - \sqrt {1 + {x^2}} } \right] \cr
& {\text{Computing the derivative}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{1 + {{\left( {x - \sqrt {1 + {x^2}} } \right)}^2}}}\left( {1 - \frac{{2x}}{{2\sqrt {1 + {x^2}} }}} \right) \cr
& {\text{Simplify}} \cr
&= \frac{1}{{1 + {{\left( {x - \sqrt {1 + {x^2}} } \right)}^2}}}\left( {1 - \frac{x}{{\sqrt {1 + {x^2}} }}} \right) \cr
&= \frac{1}{{1 + {{\left( {x - \sqrt {1 + {x^2}} } \right)}^2}}}\left( {\frac{{\sqrt {1 + {x^2}} - x}}{{\sqrt {1 + {x^2}} }}} \right) \cr
&= \frac{{\sqrt {1 + {x^2}} - x}}{{\sqrt {1 + {x^2}} \left[ {1 + x^2-2x\sqrt{1+x^2}+1+x^2 }\right] }} \cr
&= \frac{{\sqrt {1 + {x^2}} - x}}{{2\sqrt {1 + {x^2}} ({1 + x^2-x\sqrt{1+x^2}} })} \cr
&= \frac{{\sqrt {1 + {x^2}} - x}}{{2(1 + {x^2}) (\sqrt{1 + x^2}-x} } \cr
&= \frac{{1}}{{2(1 + {x^2)}}} \cr} $$
