MCQ
જો $f(x) = {\cot ^{ - 1}}\left( {{{{x^x} - {x^{ - x}}} \over 2}} \right)\,$ તો $f'(1) = . . .$
- ✓$-1$
- B$1$
- C$\log \,\,2$
- D$ - \log \,2$
$\therefore$ $y = f(x) = {\cot ^{ - 1}}\left( {\frac{{{{\tan }^2}\theta - 1}}{{2\tan \theta }}} \right)$
$= {\cot ^{ - 1}}( - \cot 2\theta ) = \pi - {\cot ^{ - 1}}(\cot 2\theta )$
==> $y = \pi - 2\theta = \pi - 2{\tan ^{ - 1}}({x^x})$
$\frac{{dy}}{{dx}} = \frac{{ - 2}}{{1 + {x^{2x}}}}.{x^x}(1 + \log x)$
==> $f'(1) = - 1$.
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