Explanation: Let $\;\; y= \left(\;\frac{1}{x}\right)^x$Then, $\;\; \log y = x \log \left(\;\frac{1}{x}\right) = -x \log x$ Differentiating both sides w.r.t. $x$ $\therefore \;\; \;\frac{1}{y} \;\frac{dy}{dx}\; = - \left[x \cdot \;\frac{1}{x} + \log x\right]$ $\;=-(1+\log x) \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot (i)$On differentiating again eq. (ii), we get $\;\frac{1}{y} \;\frac{d^2 y}{dx^2} - \;\frac{1}{y^2} \left(\;\frac{dy}{dx}\right)^2 = \;\frac{-1}{x} \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot (ii)$From eq. (ii), we get $\;\frac{dy}{dx}\; = -y(1+\log x) \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot (iii)$ $\;= - \left(\;\frac{1}{x}\right)^x (1+\log x)$ For maximum or minimum values of $y$, put $\;\frac{dy}{dx}=0$Therefore, $\left(\;\frac{1}{x}\right)^x (1+\log x)=0$However, $\left(\;\frac{1}{x}\right)^x \neq 0$ for any value of $x$. Therefore$1+\log x\;=0$$\Rightarrow \;\; \log x\; = -1 \Rightarrow x = e^{-1} \Rightarrow x = \;\frac{1}{e}$ When $x = \;\frac{1}{e}$, from eq. (iii)$\;\frac{1}{y} \;\frac{d^2 y}{dx^2} - 0\; = -e$$\Rightarrow \;\; \;\frac{d^2 y}{dx^2}\; = -e (e)^{\frac{1}{e}} < 0$Hence, $y$ is maximum when $x = \;\frac{1}{e}$ and maximum value of $y = e^{\frac{1}{e}}$