Continuity and Differentiability — Maths STD 12 Science — Question

$\therefore \frac{{dy}}{{dx}} = \frac{1}{{1 + {x^2}}}$

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

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

$= \frac{{\left( {1 + {x^2}} \right) \times 0 - 2x}}{{{{\left( {1 + {x^2}} \right)}^2}}} = \frac{{ - 2x}}{{{{\left( {1 + {x^2}} \right)}^2}}}$