The Maximum value of $\left(\frac{1}{x}\right)^x$ is

Finding the maximum of $f(x) = (1/x)^x$

Rewriting as $f(x) = x^{-x}$ and taking the natural logarithm gives $\ln y = -x \ln x$, where $y = f(x)$.

Differentiate both sides with respect to $x$, yielding:

$\frac{1}{y} \frac{dy}{dx} = -(\ln x + 1) \quad \text{via product rule on} \quad -x \ln x$

$\frac{dy}{dx} = -y(\ln x + 1)$

Substitute $y = x^{-x}$ to obtain the derivative:

$\frac{dy}{dx} = -x^{-x}(\ln x + 1)$

Setting $\frac{dy}{dx} = 0$ for critical points, and noting $x^{-x} > 0$ for $x > 0$, solve:

$\ln x + 1 = 0 \quad \Rightarrow \quad \ln x = -1 \quad \Rightarrow \quad x = e^{-1}$

The first derivative test confirms a local maximum at $x = e^{-1}$: derivative is positive for $x < e^{-1}$ and negative for $x > e^{-1}$.

The maximum value is:

$f\left(\frac{1}{e}\right) = e^{1/e}$

Given the options, only $\mathrm{e}^{\mathrm{e}}$ is a constant, while B, C, and D depend on $x$, suggesting a potential inconsistency with the calculated result $e^{1/e}$.