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

MCQ

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

A
${1 \over {1 + {x^2}}}$
✓
$ - {1 \over {1 + {x^2}}}$
C
${2 \over {1 + {x^2}}}$
D
$ - {2 \over {1 + {x^2}}}$

✓

Answer

Correct option: B.

$ - {1 \over {1 + {x^2}}}$

b
(b) $y = {\cot ^{ - 1}}\left( {\frac{{1 + x}}{{1 - x}}} \right)$

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

$ = - \frac{{2{{(1 - x)}^2}}}{{2(1 + {x^2})(1 - {x^2})}} = - \frac{1}{{1 + {x^2}}}$.

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