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Q.
The maximum value of $\frac{\log\,x}{x}$ in $(2,\infty)$ is
Application of Derivatives
Solution:
Let $y = \frac{log\,x}{x}$ and
$\therefore \frac{dy}{dx} = \frac{1-log\,x}{x^{2}}$ and
$\frac{d^{2}y}{dx^{2}} = \frac{x^{2}\left(-\frac{1}{x}\right)-\left(1-log\,x\right)2x}{x^{4}}$
$= \frac{-1-\left(1-log\,x\right)2}{x^{3}}$
$= \frac{2\,log\,x-3}{x^{3}}$
$y$ is Max. or Min. when $\frac{dy}{dx} = 0$
$\Rightarrow log\, x = 1$
$\Rightarrow x = e$ For $x = e$, $\frac{d^{2}y}{dx^{2}} < 0$
$\therefore y$ is Max. of $x = e$
$\therefore $ Max. of $y = \frac{log\,e}{e} = \frac{1}{e}$