If \( y=\log_e\left(\displaystyle\frac{x^2}{e^2}\right) \), then \( \displaystyle\frac{d^2y}{dx^2} \) is equal to:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{-2}{x^2} \)
By the laws of logarithms,
\( \displaystyle y=\log_e x^2-\log_e e^2 \)
\( \displaystyle =2\log_e x-2 \)
\( \displaystyle \frac{dy}{dx}=\frac{2}{x} \)
\( \displaystyle \frac{d^2y}{dx^2}=-\frac{2}{x^2} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{-1}{x} \) — Incorrect. The second derivative is \( -\frac{2}{x^2} \); the power of x must go down to \( x^{-2} \).
Option (2) \( \displaystyle\frac{-1}{x^2} \) — Incorrect. The factor 2 from \( \log_e x^2=2\log_e x \) is lost.
Option (4) \( \displaystyle\frac{2}{x^2} \) — Incorrect. The sign is wrong: the derivative of \( \frac{2}{x}=2x^{-1} \) is \( -2x^{-2} \).