In the interval left(frac{1}{e}, eright),a de | vedclass

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(D) To determine which function is decreasing in the interval $\left(\frac{1}{e}, e\right)$,we check the sign of the derivative $f'(x)$ for each funct

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$f(x)=x^{-x}$

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(D) To determine which function is decreasing in the interval $\left(\frac{1}{e}, e\right)$,we check the sign of the derivative $f'(x)$ for each function.For option $(A)$: $f(x) = \frac{\log x}{x}$,$f'(x) = \frac{1 - \log x}{x^2}$. For $x \in \left(\frac{1}{e}, e\right)$,$\log x$ ranges from $-1$ to $1$. Thus,$f'(x) > 0$ for $x For option $(B)$: $f(x) = x^2 \log x$,$f'(x) = 2x \log x + x = x(2 \log x + 1)$. For $x > \frac{1}{e}$,$f'(x) > 0$,so it is increasing.For option $(C)$: $f(x) = x \log x$,$f'(x) = \log x + 1$. For $x > \frac{1}{e}$,$\log x > -1$,so $f'(x) > 0$,so it is increasing.For option $(D)$: $f(x) = x^{-x}$. Let $y = x^{-x}$,then $\log y = -x \log x$. Differentiating with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = -(\log x + x \cdot \frac{1}{x}) = -(\log x + 1)$. Thus,$f'(x) = -x^{-x}(1 + \log x)$. In the interval $\left(\frac{1}{e}, e\right)$,$\log x > -1$,so $(1 + \log x) > 0$. Since $x^{-x} > 0$,$f'(x) < 0$. Therefore,$f(x) = x^{-x}$ is a decreasing function.

In the interval $\left(\frac{1}{e}, e\right)$,a decreasing function among the following functions is

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