Question
यदि $y = {x^3}\log {\log _e}(1 + x)$, तब $y''\,(0) =$
==> $y' = 3{x^2}\log {\log _e}\,(1 + x) + \frac{{{x^3}}}{{1 + x}}.\frac{1}{{{{\log }_e}(1 + x)}}$
==> $y'' = 6x\log {\log _e}(1 + x) + \frac{{3{x^2}}}{{{{\log }_e}(1 + x)}}.\frac{1}{{(1 + x)}}$
$ - \frac{{{x^3}}}{{{{(1 + x)}^2}{{\log }_e}(1 + x)}} - \frac{{{x^3}}}{{{{(1 + x)}^2}}}.\frac{1}{{{{[{{\log }_e}(1 + x)]}^2}}} + \frac{{3{x^2}}}{{(1 + x){{\log }_e}(1 + x)}}$
==> $y''(0) = 0$.
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$x + y = a$.....(i)
$x \times y = b$.....(ii)
$x\,.\,a = 1$.....(iii)
तो $x = .........,\,\,\,y = .......$