If \(y=\displaystyle\frac{1}{x+1}\), then \(\displaystyle\frac{d^2y}{dx^2}\) at \(x = 2\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{2}{27}\)
Write \(y = (x+1)^{-1}\) and differentiate twice with the power rule:
\( \displaystyle \frac{dy}{dx} = -(x+1)^{-2} \)
\( \displaystyle \frac{d^2y}{dx^2} = 2(x+1)^{-3} \)
\( \displaystyle = \frac{2}{(x+1)^3} \)
At x = 2:
\( \displaystyle \frac{d^2y}{dx^2} = \frac{2}{3^3} = \frac{2}{27} \)