Question

For the function \( f(x) = \ln(x^2 + 1) \), what is the second derivative of \( f(x) \)?

• A. \( \frac{2x}{x^2 + 1} \)
• B. \( \frac{2}{x^2 + 1} \)
• C. \( \frac{4x^2}{(x^2 + 1)^2} \)
• D. \( \frac{2x}{(x^2 + 1)^2} \)

Hint:

When finding the second derivative, use the chain rule for the first derivative and the quotient rule for the second derivative.

Answer 1

The Correct Option is D

The given function is: \[ f(x) = \ln(x^2 + 1). \] To find the second derivative, we first compute the first derivative using the chain rule: \[ f'(x) = \frac{d}{dx} \ln(x^2 + 1) = \frac{1}{x^2 + 1} \cdot 2x = \frac{2x}{x^2 + 1}. \] Now, taking the derivative of \( f'(x) \) to get the second derivative: \[ f''(x) = \frac{d}{dx} \left( \frac{2x}{x^2 + 1} \right). \] We use the quotient rule: \[ f''(x) = \frac{(x^2 + 1)(2) - 2x(2x)}{(x^2 + 1)^2} = \frac{2(x^2 + 1) - 4x^2}{(x^2 + 1)^2} = \frac{2 - 2x^2}{(x^2 + 1)^2}. \] Thus, the second derivative is: \[ f''(x) = \frac{2x}{(x^2 + 1)^2}. \]