Question
यदि $y = (1 + {x^2}){\tan ^{ - 1}}x - x,$ तब $\frac{{dy}}{{dx}} = $
==> $\frac{{dy}}{{dx}} = (1 + {x^2}).\frac{1}{{(1 + {x^2})}} + {\tan ^{ - 1}}x(2x) - 1$
$= 2x{\tan ^{ - 1}}x.$
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