d dx [(1 + x^2 ) ^ - 1 x] = - Vidyadip

MCQ

${d \over {dx}}[(1 + {x^2}){\tan ^{ - 1}}x] = $

A
$x\,{\tan ^{ - 1}}x$
B
$2\,{\tan ^{ - 1}}x$
✓
$2x\,{\tan ^{ - 1}}x + 1$
D
$x\,{\tan ^{ - 1}}x + 1$

✓

Answer

Correct option: C.

$2x\,{\tan ^{ - 1}}x + 1$

c
(c) $\frac{d}{{dx}}[(1 + {x^2}){\tan ^{ - 1}}x] = 1 + 2x{\tan ^{ - 1}}x$.

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