Question
यदि $y = {x^2} + {x^{\log x}},$ तो $\frac{{dy}}{{dx}} = $
==> $\frac{{dy}}{{dx}} = 2x + {x^{\log x}}\left( {2{{\log }_e}x.\frac{1}{x}} \right)$
$ = \frac{{2({x^2} + {x^{\log x}}{{\log }_e}x)}}{x}$.
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