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If y=log⁡exxy = \dfrac{\log_e x}{x}, then d2ydx2\dfrac{d^2y}{dx^2} is :

Solution

✅ Correct Option: 4

By the quotient rule, dydx=1−log⁡exx2\dfrac{dy}{dx}=\dfrac{1-\log_e x}{x^2}. Differentiating again, d2ydx2=(−1/x)x2−(1−log⁡ex)(2x)x4=−3x+2xlog⁡exx4=2log⁡ex−3x3\dfrac{d^2y}{dx^2}=\dfrac{(-1/x)x^2-(1-\log_e x)(2x)}{x^4}=\dfrac{-3x+2x\log_e x}{x^4}=\dfrac{2\log_e x-3}{x^3}.

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