If $y = \log \left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right),$ then $\frac{{dy}}{{dx}}$ is equal to

We have, $y = \log \left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$
therefore,$\frac{{dy}}{{dx}} = \frac{1}{{\frac{{1 - {x^2}}}{{1 + {x^2}}}}} \cdot \frac{d}{{dx}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$

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

$= \frac{{ - 2x\left[ {1 + {x^2} + 1 - {x^2}} \right]}}{{\left( {1 - {x^2}} \right) \cdot \left( {1 + {x^2}} \right)}} = \frac{{ - 4x}}{{1 - {x^4}}}$