The maximum value of frac{log x}{x}, 0

Question

(D) Let $y = \frac{\log x}{x}$.Find the derivative with respect to $x$ using the quotient rule:$\frac{dy}{dx} = \frac{x \cdot \frac{1}{x} - \log x \cd

Vedclass · Accepted Answer

$e^{-1}$

Vedclass · Answer

(D) Let $y = \frac{\log x}{x}$.Find the derivative with respect to $x$ using the quotient rule:$\frac{dy}{dx} = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2}$.For critical points,set $\frac{dy}{dx} = 0$:$1 - \log x = 0 \Rightarrow \log x = 1 \Rightarrow x = e$.To verify the maximum,find the second derivative:$\frac{d^2y}{dx^2} = \frac{x^2(-\frac{1}{x}) - (1 - \log x)(2x)}{x^4} = \frac{-x - 2x + 2x \log x}{x^4} = \frac{2 \log x - 3}{x^3}$.At $x = e$,$\frac{d^2y}{dx^2} = \frac{2(1) - 3}{e^3} = -\frac{1}{e^3} The maximum value is $y(e) = \frac{\log e}{e} = \frac{1}{e} = e^{-1}$.

The maximum value of $\frac{\log x}{x}, 0 < x < \infty$ is

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