If y = frac{log x}{x},then the value of x^2 f | vedclass

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(A) Given $y = \frac{\log x}{x}$.First,find the first derivative $\frac{dy}{dx}$ using the quotient rule:$\frac{dy}{dx} = \frac{x(\frac{1}{x}) - \log

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$0$

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(A) Given $y = \frac{\log x}{x}$.First,find the first derivative $\frac{dy}{dx}$ using the quotient rule:$\frac{dy}{dx} = \frac{x(\frac{1}{x}) - \log x(1)}{x^2} = \frac{1 - \log x}{x^2}$.Next,find the second derivative $\frac{d^2y}{dx^2}$:$\frac{d^2y}{dx^2} = \frac{x^2(-\frac{1}{x}) - (1 - \log x)(2x)}{x^4} = \frac{-x - 2x + 2x \log x}{x^4} = \frac{-3x + 2x \log x}{x^4}$.Now,substitute these into the expression $x^2 \frac{d^2y}{dx^2} + 3x \frac{dy}{dx} + y$:$= x^2 \left( \frac{-3x + 2x \log x}{x^4} \right) + 3x \left( \frac{1 - \log x}{x^2} \right) + \frac{\log x}{x}$$= \frac{-3x + 2x \log x}{x^2} + \frac{3 - 3 \log x}{x} + \frac{\log x}{x}$$= \frac{-3 + 2 \log x}{x} + \frac{3 - 3 \log x}{x} + \frac{\log x}{x}$$= \frac{-3 + 2 \log x + 3 - 3 \log x + \log x}{x} = \frac{0}{x} = 0$.Thus,the value is $0$ for all $x > 0$.

If $y = \frac{\log x}{x}$,then the value of $x^2 \frac{d^2 y}{d x^2} + 3x \frac{d y}{d x} + y$ at the point $(\sqrt[3]{e}, \frac{\log(\sqrt[3]{e})}{\sqrt[3]{e}})$ is

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