If \(y = \log_e(1 + x^{2})\), then the value of \(\displaystyle\frac{d^{2}y}{dx^{2}}\) at \(x = 0\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2
By the chain rule:
\( \displaystyle \frac{dy}{dx} = \frac{2x}{1 + x^{2}} \)
By the quotient rule:
\( \displaystyle \frac{d^{2}y}{dx^{2}} = \frac{2(1 + x^{2}) - 2x \cdot 2x}{(1 + x^{2})^{2}} = \frac{2 - 2x^{2}}{(1 + x^{2})^{2}} \)
At x = 0:
\( \displaystyle \frac{d^{2}y}{dx^{2}} = \frac{2 - 0}{1} = 2 \)