6. Application of derivatives — ગણિત ધોરણ 12 સાયન્સ — Question

==>$\frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{{{x^2}}} - \frac{{\log x}}{{{x^2}}} = \frac{{1 - \log x}}{{{x^2}}}$

==>$\frac{{dy}}{{dx}} = {x^{1/x}}\left( {\frac{{1 - \log x}}{{{x^2}}}} \right)$

Now, ${x^{1/x}} > 0$for all x and $\frac{{1 - \log x}}{{{x^2}}} > 0$ in $ (1, e) $ and

$\frac{{1 - \log x}}{{{x^2}}} < 0$ in $(e,\infty )$

$\therefore$ $f(x)$ is increasing in $ (1, e) $ and decreasing in $(e,\,\infty ).$