If \( y=\displaystyle\frac{\log x}{x} \), then \( \displaystyle\frac{d^{2}y}{dx^{2}} \) is equal to (Consider \( \log_{e}x=\log x \)):
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{2\log x-3}{x^{3}} \)
By the quotient rule:
\( \displaystyle \frac{dy}{dx}=\frac{x\cdot\frac{1}{x}-\log x\cdot 1}{x^{2}} \)
\( \displaystyle =\frac{1-\log x}{x^{2}} \)
Applying the quotient rule again:
\( \displaystyle \frac{d^{2}y}{dx^{2}}=\frac{x^{2}\cdot\left(-\frac{1}{x}\right)-(1-\log x)\cdot 2x}{x^{4}} \)
\( \displaystyle =\frac{-x-2x+2x\log x}{x^{4}} \)
\( \displaystyle =\frac{x(2\log x-3)}{x^{4}} \)
\( \displaystyle =\frac{2\log x-3}{x^{3}} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{1-\log x}{x^{4}} \) — Incorrect. \( 1-\log x \) is the numerator of the first derivative; the second derivative has \( 2\log x-3 \) on top.
Option (2) \( \displaystyle\frac{2\log x-3}{x^{4}} \) — Incorrect. The numerator is \( x(2\log x-3) \); cancelling the common factor x with \( x^{4} \) leaves \( x^{3} \), not \( x^{4} \).
Option (3) \( \displaystyle\frac{\log x-3}{x^{3}} \) — Incorrect. The term \( -(1-\log x)\cdot 2x \) gives \( 2x\log x \), so the coefficient of \( \log x \) is 2, not 1.