If $f(x)=\frac{x}{\log x}$, then $f(x)$ is increasing in

If $f(x)=\frac{x}{\log x}$, then $f(x)$ is increasing in
  1. $(0, \infty)$
  2. $(e, \infty)$
  3. $(-\infty, 0)$
  4. $[\mathrm{e}, \infty)$

Solution

$y=\frac{x}{\log x} \Rightarrow \frac{d y}{d x}=\frac{\log x \times 1-x \times \frac{1}{x}}{(\log x)^2}=\frac{(\log x)-1}{(\log x)^2}$ $f(x)$ is increasing in $[e, \infty)$

Asked in: MHT CET 2022 (07 Aug Shift 2)

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