If y=x^{x e^{x}},frac{d y}{d x}=y cdot g(x),t | vedclass

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(C) Given $y = x^{x e^{x}}$.Taking logarithm on both sides,we get $\log y = x e^{x} \log x$.Differentiating both sides with respect to $x$ using the p

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$e^{x}(1 + \log x + x \log x)$

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(C) Given $y = x^{x e^{x}}$.Taking logarithm on both sides,we get $\log y = x e^{x} \log x$.Differentiating both sides with respect to $x$ using the product rule:$\frac{1}{y} \frac{d y}{d x} = \frac{d}{d x}(x e^{x}) \cdot \log x + (x e^{x}) \cdot \frac{d}{d x}(\log x)$.$\frac{1}{y} \frac{d y}{d x} = (e^{x} + x e^{x}) \log x + (x e^{x}) \cdot \frac{1}{x}$.$\frac{1}{y} \frac{d y}{d x} = e^{x} \log x + x e^{x} \log x + e^{x}$.$\frac{1}{y} \frac{d y}{d x} = e^{x}(1 + \log x + x \log x)$.Since $\frac{d y}{d x} = y \cdot g(x)$,we have $g(x) = e^{x}(1 + \log x + x \log x)$.

If $y=x^{x e^{x}}$,$\frac{d y}{d x}=y \cdot g(x)$,then $g(x)=$

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