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If $y=x^{x e^{x}}, \frac{d y}{d x}=y \cdot g(x)$, then $g(x)=$
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Verified Answer
The correct answer is:
$\left[e^{x}+e^{x}(x+1) \log x\right]$
We have $y=x^{x e^{x}}$
$\therefore \log y=x e^{x} \log x$
$\therefore \frac{1}{y} \frac{d y}{d x}=e^{x} \log x+\frac{x e^{x}}{x}+x e^{x} \log x$
$=e^{x} \log x+e^{x}+x e^{x} \log x$
$\therefore \frac{d y}{d x}=y\left[e^{x}+e^{x} \log x(1+x)\right]$
$\therefore \log y=x e^{x} \log x$
$\therefore \frac{1}{y} \frac{d y}{d x}=e^{x} \log x+\frac{x e^{x}}{x}+x e^{x} \log x$
$=e^{x} \log x+e^{x}+x e^{x} \log x$
$\therefore \frac{d y}{d x}=y\left[e^{x}+e^{x} \log x(1+x)\right]$
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