${\tan ^{ - 1}}x$

Let $y = {\tan ^{ - 1}}x$
$\Rightarrow$ $\cfrac{{dy}}{{dx}} = \cfrac{1}{{(1 + {x^2})}}$

therefore, $\cfrac{{{d^2}y}}{{d{x^2}}} = \cfrac{{(1 + {x^2}) \cdot 0 - 1 \cdot (2x)}}{{{{(1 + {x^2})}^2}}} = \cfrac{{ - 2x}}{{{{(1 + {x^2})}^2}}}$