Question
यदि $y = {x^2}\log x + \frac{2}{{\sqrt x }},$ तो $\frac{{dy}}{{dx}} = $
$\frac{{dy}}{{dx}} = 2x\log x + x - {x^{ - 3/2}} = x + 2x\log x - \frac{1}{{{x^{3/2}}}}$.
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