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

If $y=x^{x e^{x}}, \frac{d y}{d x}=y \cdot g(x)$, then $g(x)=$
  1. $\left[e^{x}+e^{x}(x+1) \log x\right]$
  2. $\left[e^{x}-e^{x} \cdot x \cdot(1+\log x)\right]$
  3. $\left[e^{x}+e^{x} \cdot x \cdot(1+\log x)\right]$
  4. $\left[e^{x}(x+1) \log x\right]$

Solution

We have $y=x^{x e^{x}}$ $\therefore \log y=x e^{x} \log x$ $\therefore \frac{1}{y} \frac{d y}{d x}=e^{x} \log x+\frac{x e^{x}}{x}+x e^{x} \log x$ $=e^{x} \log x+e^{x}+x e^{x} \log x$ $\therefore \frac{d y}{d x}=y\left[e^{x}+e^{x} \log x(1+x)\right]$

Asked in: MHT CET 2020 (20 Oct Shift 2)

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