# The maximum value of $\left ( \frac{1}{x} \right )^x$ is: A. $e$ B. $e^e$ C. $e^{\frac{1}{e}}$ D. $\left ( \frac{1}{e} \right )^{\frac{1}{e}}$

Let $y=\left(\frac{1}{x}\right)^{x}$
Take logarithm on both side
$\\ \log y=\log \left(\frac{1}{x}\right)^{x} \\\log y=x \log \frac{1}{x}$
Applying first derivative and get

$\frac{\mathrm{d}(\log \mathrm{y})}{\mathrm{dx}}=\frac{\mathrm{d}\left(\mathrm{x} \log \frac{1}{\mathrm{x}}\right)}{\mathrm{dx}}$

Apply product rule and get
$\\\frac{\mathrm{d}(\log \mathrm{y})}{\mathrm{dx}}=\mathrm{x} \frac{\mathrm{d}\left(\log \frac{1}{\mathrm{x}}\right)}{\mathrm{dx}}+\log \frac{1}{\mathrm{x}} \cdot \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dx}}$
Applying first derivative and get

\begin{aligned} &\frac{1}{y} \cdot \frac{d y}{d x}=x \cdot \frac{1}{\frac{1}{x}} \cdot \frac{d\left(\frac{1}{x}\right)}{d x}+\log \frac{1}{x}, 1\\ &\frac{1}{y} \cdot \frac{d y}{d x}=x^{2} \cdot \frac{d\left(x^{-1}\right)}{d x}+\log \frac{1}{x}\\ &\frac{1}{y} \cdot \frac{d y}{d x}=x^{2} \cdot(-1) \cdot x^{-1-1}+\log \frac{1}{x}\\ &\frac{1}{y} \cdot \frac{d y}{d x}=x^{2} \cdot(-1) \cdot x^{-2}+\log \frac{1}{x}\\ &\frac{1}{y} \cdot \frac{d y}{d x}=-1+\log \frac{1}{x}\\ &\Rightarrow \frac{d y}{d x}=y\left(-1+\log \frac{1}{x}\right)\\ &\text { Substitute value of } y \text { from (i) and get }\\ &\Rightarrow \frac{d y}{d x}=\left(\frac{1}{x}\right)^{x} \cdot\left(-1+\log \frac{1}{x}\right) \end{aligned}

Now we find critical point by equating (i) to 0

\begin{aligned} &\left(\frac{1}{x}\right)^{x} \cdot\left(-1+\log \frac{1}{x}\right)=0\\ &\Rightarrow\left(-1+\log \frac{1}{x}\right)=0 \quad\left(\frac{1}{x}\right)^{x} \text { can't be equal to } 0\\ &\Rightarrow\left(\log \frac{1}{x}\right)=1\\ &\text { But } 1=\log \mathrm{e}^{1}\\ &\Rightarrow\left(\log \frac{1}{x}\right)=\log e\\ &\text { Equate the terms }\\ &\Rightarrow \frac{1}{\mathrm{x}}=e\\ &\Rightarrow x=\frac{1}{e} \end{aligned}

Therefore f(x) has a stationary point at $x=\frac{1}{e}$.

i.e the maximum value of $f\left (\frac{1}{e} \right )=e^{\frac{1}{e}}$

So, the correct answer is option C.
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