1.The derivative of f(x)=xtan−1xf(x) = x^{\tan^{-1} x}f(x)=xtan−1x with respect to g(x)=sec−1(12x2−1)g(x) = \sec^{-1} \left( \frac{1}{2x^2 - 1} \right)g(x)=sec−1(2x2−11) isa.121−x2xtan−1x[logx1+x2+tan−1xx]\frac{1}{2} \sqrt{1 - x^2} x^{\tan^{-1} x} \left[ \frac{\log x}{1 + x^2} + \frac{\tan^{-1} x}{x} \right]211−x2xtan−1x[1+x2logx+xtan−1x]b.−121−x2xx−a−1-\frac{1}{2} \sqrt{1 - x^2} x^{x^{-a^{-1}}}−211−x2xx−a−1 [log(tan−1x)+x(1+x2)tan−1x][\log (\tan^{-1} x) + x (1 + x^2) \tan^{-1} x][log(tan−1x)+x(1+x2)tan−1x]c.−2tan−1logx1+x2+tan−1xx1−x2\frac{-2 \tan^{-1} \frac{\log x}{1 + x^2} + \frac{\tan^{-1} x}{x}}{\sqrt{1 - x^2}}1−x2−2tan−11+x2logx+xtan−1xd.−121−x2xtanx−1-\frac{1}{2} \sqrt{1 - x^2} x^{\tan x^{-1}}−211−x2xtanx−1 [logx1+x2+tan−1xx]\left[ \frac{\log x}{1 + x^2} + \frac{\tan^{-1} x}{x} \right][1+x2logx+xtan−1x]Login to continueOnly logged in users canattempt or see the solution.
