d dx ^ - 1 [ x - x x + x ] = - Vidyadip

MCQ

${d \over {dx}}{\tan ^{ - 1}}\left[ {{{\cos x - \sin x} \over {\cos x + \sin x}}} \right] = $

A
${1 \over {2\,\,(1 + {x^2})}}$
B
${1 \over {1 + {x^2}}}$
C
$1$
✓
$-1$

✓

Answer

Correct option: D.

$-1$

d
(d) $\frac{d}{{dx}}{\tan ^{ - 1}}\left[ {\frac{{\cos x - \sin x}}{{\cos x + \sin x}}} \right]$

$ = \frac{d}{{dx}}{\tan ^{ - 1}}\left[ {\tan \left( {\frac{\pi }{4} - x} \right)} \right] = - 1$.

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