MCQ
$\frac{d}{{dx}}{\tan ^{ - 1}}\frac{{1 - x}}{{1 + x}} = .......$
- ✓$\frac{{ - 1}}{{1 + {x^2}}}$
- B$\frac{1}{{1 + {x^2}}}$
- C$\frac{{1 + x}}{{1 - x}}$
- D$\frac{2}{{1 + {x^2}}}$
$\frac{d}{{dx}}{\tan ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right) = \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}x} \right]$
$ = 0 - \frac{1}{{1 + {x^2}}}$
$ = \frac{{ - 1}}{{1 + {x^2}}}$
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