If y = frac{log x}{x},then find the value of | vedclass

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

Vedclass · Accepted Answer

$-3$

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(B) Given $y = \frac{\log x}{x}$.First,find the first derivative $\frac{dy}{dx}$ using the quotient rule:$\frac{dy}{dx} = \frac{x \cdot \frac{d}{dx}(\log x) - \log x \cdot \frac{d}{dx}(x)}{x^2} = \frac{x \cdot \frac{1}{x} - \log x}{x^2} = \frac{1 - \log x}{x^2}$.Now,find the second derivative $\frac{d^2 y}{dx^2}$ by differentiating $\frac{dy}{dx} = \frac{1 - \log x}{x^2}$ with respect to $x$:$\frac{d^2 y}{dx^2} = \frac{x^2 \cdot \frac{d}{dx}(1 - \log x) - (1 - \log x) \cdot \frac{d}{dx}(x^2)}{(x^2)^2}$$\frac{d^2 y}{dx^2} = \frac{x^2 \cdot (-\frac{1}{x}) - (1 - \log x) \cdot (2x)}{x^4} = \frac{-x - 2x + 2x \log x}{x^4} = \frac{-3x + 2x \log x}{x^4} = \frac{-3 + 2 \log x}{x^3}$.At $x = 1$:$\frac{d^2 y}{dx^2} = \frac{-3 + 2 \log 1}{1^3} = \frac{-3 + 2(0)}{1} = -3$.

If $y = \frac{\log x}{x}$,then find the value of $\frac{d^2 y}{d x^2}$ at $x = 1$.

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