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Q. The maximum value of $\frac{log_ex}{x}$, if x > 0 is

KCETKCET 2020

Solution:

$(y = \frac{log_e \ x}{x})$

$⇒ \left(\frac{dy}{dx}=\frac{1 - log_e x}{x^2}\right)$

For maxima, $dy/dx = 0$

$⇒ 1 - log_e x = 0$

$⇒ x = e$

$dy/dx$ changes sign from positive to negative at $x = e$

$∴ y_{\text{max}} = 1/e$