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Q.
The maximum value of $\frac{\ln x}{x}$ in $(2, \infty)$ is
Application of Derivatives
Solution:
Let $y=\frac{\ln x}{x}$
$\frac{d y}{d x}=\frac{x \cdot \frac{1}{x}-\ln x \cdot 1}{x^2}=\frac{1-\log x}{x^2}$
For maxima, put $\frac{d y}{d x}=0$
$ \Rightarrow \frac{1-\ln x}{x^2}=0 \Rightarrow x=e$
Now, $ \frac{d^2 y}{d x^2}=\frac{x^2\left(-\frac{1}{x}\right)-(1-\ln x) 2 x}{\left(x^2\right)^2}$
At $x = e,\frac{d^2y}{dx^2} < 0$
$\therefore$ The maximum value at $x=e$ is $y=\frac{1}{e}$