If y = x^2 + x^{log x},then frac{dy}{dx} = | vedclass

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(C) Given $y = x^2 + x^{\log x}$.Let $u = x^2$ and $v = x^{\log x}$.Then $y = u + v$,so $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$.For $u = x^2$,

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$\frac{2(x^2 + \log x \cdot x^{\log x})}{x}$

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(C) Given $y = x^2 + x^{\log x}$.Let $u = x^2$ and $v = x^{\log x}$.Then $y = u + v$,so $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$.For $u = x^2$,$\frac{du}{dx} = 2x$.For $v = x^{\log x}$,taking log on both sides: $\log v = \log x \cdot \log x = (\log x)^2$.Differentiating with respect to $x$: $\frac{1}{v} \frac{dv}{dx} = 2 \log x \cdot \frac{1}{x}$.So,$\frac{dv}{dx} = v \cdot \frac{2 \log x}{x} = x^{\log x} \cdot \frac{2 \log x}{x}$.Thus,$\frac{dy}{dx} = 2x + \frac{2 \log x \cdot x^{\log x}}{x}$.Taking $\frac{2}{x}$ as common: $\frac{dy}{dx} = \frac{2(x^2 + \log x \cdot x^{\log x})}{x}$.

If $y = x^2 + x^{\log x}$,then $\frac{dy}{dx} = $

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