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

MCQ

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

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

✓

Answer

Correct option: B.

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

b
(b) $y = (1 + {x^2}){\tan ^{ - 1}}x - x$

==> $\frac{{dy}}{{dx}} = (1 + {x^2}).\frac{1}{{(1 + {x^2})}} + {\tan ^{ - 1}}x(2x) - 1$

$= 2x{\tan ^{ - 1}}x.$

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