If \( y=\log_e 2x \), then the value of \( \displaystyle\frac{d^2y}{dx^2} \) at \( x=\displaystyle\frac{1}{2} \) is :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → -4
Since \( \log_e 2x=\log_e 2+\log_e x \), the constant loge 2 disappears on differentiation.
\( \displaystyle \frac{dy}{dx}=\frac{1}{x} \)
\( \displaystyle \frac{d^2y}{dx^2}=-\frac{1}{x^2} \)
At \( x=\frac{1}{2} \):
\( \displaystyle \frac{d^2y}{dx^2}=-\frac{1}{(1/2)^2}=-4 \)