${\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right], - 1 < x < 1,x \ne 0$

Let $y = {\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right]$
Put ${x^2} = \cos 2\theta$

therefore,$y = {\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + \cos 2\theta } + \sqrt {1 - \cos 2\theta } }}{{\sqrt {1 + \cos 2\theta } - \sqrt {1 - \cos 2\theta } }}} \right)$

$= {\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + 2{{\cos }^2}\theta - 1} + \sqrt {1 - 1 + 2{{\sin }^2}\theta } }}{{\sqrt {1 + 2{{\cos }^2}\theta - 1} - \sqrt {1 - 1 + 2{{\sin }^2}\theta } }}} \right)$

$= {\tan ^{ - 1}}\left( {\frac{{\sqrt 2 \cos \theta + \sqrt 2 \sin \theta }}{{\sqrt 2 \cos \theta - \sqrt 2 \sin \theta }}} \right) = {\tan ^{ - 1}}\left[ {\frac{{\sqrt 2 (\cos \theta + \sin \theta )}}{{\sqrt 2 (\cos \theta - \sin \theta )}}} \right]$

$= {\tan ^{ - 1}}\left( {\frac{{\cos \theta + \sin \theta }}{{\cos \theta - \sin \theta }}} \right) = {\tan ^{ - 1}}\left( {\frac{{\frac{{\cos \theta + \sin \theta }}{{\cos \theta }}}}{{\frac{{\cos \theta - \sin \theta }}{{\cos \theta }}}}} \right)$

$= {\tan ^{ - 1}}\left( {\frac{{1 + \tan \theta }}{{1 - \tan \theta }}} \right)$
$= {\tan ^{ - 1}}\tan \left( {\frac{\pi }{4} + \theta } \right)$

$= \frac{\pi }{4} + \theta = \frac{\pi }{4} + \frac{1}{2}{\cos ^{ - 1}}{x^2}$

therefore,$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {\frac{\pi }{4}} \right) + \frac{d}{{dx}}\left( {\frac{1}{2}{{\cos }^{ - 1}}{x^2}} \right)$

$= 0 + \frac{1}{2} \cdot \frac{{ - 1}}{{\sqrt {1 - {x^4}} }} \cdot \frac{d}{{dx}}{x^2} = \frac{1}{2} \cdot \frac{{ - 2x}}{{\sqrt {1 - {x^4}} }} = \frac{{ - x}}{{\sqrt {1 - {x^4}} }}$

$\frac{dy}{dx}= \frac{{ - x}}{{\sqrt {1 - {x^4}} }}$

Find $\frac{{dy}}{{dx}}$ of each of the functions expressed in parametric form.