If \( e^y(x+1)=1 \), then \( \displaystyle\frac{d^2y}{dx^2} \) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{1}{(x+1)^2} \)
Since \( e^y=\frac{1}{x+1} \), y can be written explicitly:
\( \displaystyle y=-\log(x+1) \)
Differentiating twice:
\( \displaystyle \frac{dy}{dx}=-\frac{1}{x+1} \)
\( \displaystyle \frac{d^2y}{dx^2}=\frac{1}{(x+1)^2} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{-1}{x+1} \) — Incorrect. This is \( \frac{dy}{dx} \), the first derivative, not the second.
Option (2) \( \displaystyle\frac{-1}{(x+1)^2} \) — Incorrect. The sign is wrong: differentiating \( -(x+1)^{-1} \) gives \( +(x+1)^{-2} \).
Option (3) \( \displaystyle\frac{1}{x+1} \) — Incorrect. This is \( -\frac{dy}{dx} \); the second derivative has \( (x+1)^2 \) in the denominator.