show that the maximum value of (1/x)^x is e^1/e

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$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ \mathrm{y}={\left(\frac{1}{\mathrm{x}}\right)}^{\mathrm{x}} \\ \mathrm{Taking} \log \mathrm{on} \mathrm{both} \mathrm{sides}, \mathrm{we} \mathrm{get}: \\ \log \left(\mathrm{y}\right)=\log \left\{{\left(\frac{1}{\mathrm{x}}\right)}^{\mathrm{x}}\right\} \\ \Rightarrow \log \left(\mathrm{y}\right)=\mathrm{xlog}\left(\frac{1}{\mathrm{x}}\right) \left[\mathrm{As} \log \left({\mathrm{m}}^{\mathrm{n}}\right)=\mathrm{nlogm}\right] \\ \Rightarrow \log \left(\mathrm{y}\right)=\mathrm{xlog}\left({\mathrm{x}}^{-1}\right) \\ \Rightarrow \log \left(\mathrm{y}\right)=-\mathrm{xlog}\left(\mathrm{x}\right) \\ \mathrm{Differentiating} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x}, \mathrm{we} \mathrm{get}: \\ \Rightarrow \frac{1}{\mathrm{y}}.\frac{\mathrm{dy}}{\mathrm{dx}}=-\left[\mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{logx}\right)+\mathrm{logx}.\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}\right)\right] \\ \Rightarrow \frac{1}{\mathrm{y}}.\frac{\mathrm{dy}}{\mathrm{dx}}=-\left[\mathrm{x}.\frac{1}{\mathrm{x}}+\mathrm{logx}.1\right] \\ \Rightarrow \frac{1}{\mathrm{y}}.\frac{\mathrm{dy}}{\mathrm{dx}}=-\left[1+\mathrm{logx}\right] ;\left(\mathrm{i}\right) \\ \mathrm{For} \mathrm{maxima}/ \mathrm{minima}, \frac{\mathrm{dy}}{\mathrm{dx}}=0 \\ \Rightarrow -\left[1+\mathrm{logx}\right]=0 \\ \Rightarrow 1+\mathrm{logx}=0 \\ \Rightarrow \mathrm{logx}=-1 \\ \Rightarrow \mathrm{x}={\mathrm{e}}^{-1} \\ \Rightarrow \mathrm{x}=\frac{1}{\mathrm{e}} \\ \mathrm{Now} \mathrm{again} \mathrm{differentiate} \left(\mathrm{i}\right) \mathrm{with} \mathrm{repsect} \mathrm{to} \mathrm{x}: \\ \frac{1}{\mathrm{y}}.\frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{dx}}+\frac{\mathrm{dy}}{\mathrm{dx}}.\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\mathrm{y}}\right)=-\frac{\mathrm{d}}{\mathrm{dx}}\left[1+\mathrm{logx}\right] \\ \Rightarrow \frac{1}{\mathrm{y}}.\frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{dx}}+\frac{\mathrm{dy}}{\mathrm{dx}}.\left\{-\frac{1}{{\mathrm{y}}^2}\right\}.\frac{\mathrm{dy}}{\mathrm{dx}}=-\left(0+\frac{1}{\mathrm{x}}\right) \\ \mathrm{Put} \frac{\mathrm{dy}}{\mathrm{dx}}=0 \mathrm{and} \mathrm{x}=\frac{1}{\mathrm{e}} \\ \Rightarrow \frac{1}{\mathrm{y}}.\frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{dx}}+0=-\frac{1}{{\displaystyle \frac{1}{\mathrm{e}}}} \\ \Rightarrow \frac{1}{\mathrm{y}}.\frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{dx}}=-\mathrm{e} \\ \mathrm{Hence} \frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{dx}}<0 \mathrm{when} \mathrm{x}=\frac{1}{\mathrm{e}}, \mathrm{which} \mathrm{means} \mathrm{x}=\frac{1}{\mathrm{e}} \mathrm{is} \mathrm{point} \mathrm{of} \mathrm{local} \mathrm{maxima}. \\ \mathrm{y}={\left(\frac{1}{\mathrm{x}}\right)}^{\mathrm{x}} \\ \mathrm{Hence} \\ {\mathbf{y}}_{\mathbf{max}}\mathbf{=}{\left(\frac{\mathbf{1}}{{\displaystyle \frac{\mathbf{1}}{\mathbf{e}}}}\right)}^{\frac{\mathbf{1}}{\mathbf{e}}}\mathbf{=}{\mathbf{e}}^{\frac{\mathbf{1}}{\mathbf{e}}}\mathbf{ } \end{gathered}$$
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