If y = ( 1 + x^2 1 - x^2 ), then dy dx = - Vidyadip

MCQ

If $y = \sin \left( {{{1 + {x^2}} \over {1 - {x^2}}}} \right)$, then ${{dy} \over {dx}} = $

A
${{4x} \over {1 - {x^2}}}.\cos \left( {{{1 + {x^2}} \over {1 - {x^2}}}} \right)$
B
${x \over {{{(1 - {x^2})}^2}}}.\cos \left( {{{1 + {x^2}} \over {1 - {x^2}}}} \right)$
C
${x \over {(1 - {x^2})}}.\cos \left( {{{1 + {x^2}} \over {1 - {x^2}}}} \right)$
✓
${{4x} \over {{{(1 - {x^2})}^2}}}.\cos \left( {{{1 + {x^2}} \over {1 - {x^2}}}} \right)$

✓

Answer

Correct option: D.

${{4x} \over {{{(1 - {x^2})}^2}}}.\cos \left( {{{1 + {x^2}} \over {1 - {x^2}}}} \right)$

d
(d) $\frac{{dy}}{{dx}} = \cos \left( {\frac{{1 + {x^2}}}{{1 - {x^2}}}} \right){\rm{ }}\left[ {\frac{{(1 - {x^2})2x + (1 + {x^2})2x}}{{{{(1 - {x^2})}^2}}}} \right]$

$ = \frac{{4x}}{{{{(1 - {x^2})}^2}}}\cos \left( {\frac{{1 + {x^2}}}{{1 - {x^2}}}} \right)$.

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