If $y = {\tan ^{ - 1}}x,$ then find $\frac{{{d^2}y}}{{d{x^2}}}$ in terms of $y$ alone

We have, $y = {\tan ^{ - 1}}x$ [on differentiating w.r.t. $\left. x \right]$
therefore,$\frac{{dy}}{{dx}} = \frac{1}{{1 + {x^2}}}$ [again differentiating w.r.t. $\left. x \right]$

Now, $\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}{\left( {1 + {x^2}} \right)^{ - 1}}$

$= - 1{\left( {1 + {x^2}} \right)^{ - 2}} \cdot \frac{d}{{dx}}\left( {1 + {x^2}} \right)$

$= - \frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}} \cdot 2x$

$= \frac{{ - 2{\rm{ tan }}y}}{{{{\left( {1 + {{\tan }^2}y} \right)}^2}}}$

$= \frac{{ - 2\tan y}}{{{{\left( {{{\sec }^2}y} \right)}^2}}}$

$= - 2\frac{{\sin y}}{{\cos y}} \cdot {\cos ^2}y \cdot {\cos ^2}y$

$= - \sin 2y \cdot {\cos ^2}y$

Verify the Rolle's theorem for each of the functions in following questions.