Question
$\frac{d}{{dx}}({x^{{{\log }_e}x}}) = $
==> ${\log _e}y = {\log _e}x{\log _e}x = {({\log _e}x)^2}$
==> $\frac{1}{y}\frac{{dy}}{{dx}} = 2{\log _e}x.\frac{1}{x}$
$\therefore \frac{{dy}}{{dx}} = 2{x^{({{\log }_e}x - 1)}}{\log _e}x$.
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