If \(y = \displaystyle\frac{\log_e x}{x}\), then \(\displaystyle\frac{d^2y}{dx^2} =\)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{2\log_e x-3}{x^3}\)
By the quotient rule:
\( \displaystyle \frac{dy}{dx} = \frac{x \cdot \frac{1}{x} - \log_e x \cdot 1}{x^2} \)
\( \displaystyle = \frac{1 - \log_e x}{x^2} \)
Differentiating again by the quotient rule:
\( \displaystyle \frac{d^2y}{dx^2} = \frac{x^2\left(-\frac{1}{x}\right) - (1 - \log_e x)(2x)}{x^4} \)
\( \displaystyle = \frac{-x - 2x + 2x\log_e x}{x^4} \)
\( \displaystyle = \frac{x\left(2\log_e x - 3\right)}{x^4} \)
\( \displaystyle = \frac{2\log_e x - 3}{x^3} \)