If \(y = \tan^{-1} x\), then the value of \(\displaystyle\frac{d^{2}y}{dx^{2}}\) at \(x = 1\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(-\displaystyle\frac{1}{2}\)
The first derivative is the standard one:
\( \displaystyle \frac{dy}{dx} = \frac{1}{1 + x^{2}} = (1 + x^{2})^{-1} \)
Differentiating again by the chain rule:
\( \displaystyle \frac{d^{2}y}{dx^{2}} = -(1 + x^{2})^{-2} \cdot 2x = -\frac{2x}{(1 + x^{2})^{2}} \)
At x = 1:
\( \displaystyle \frac{d^{2}y}{dx^{2}} = -\frac{2}{2^{2}} = -\frac{1}{2} \)