Question Detail
Question
Obtain $\frac{d}{dx} \left[ \ln\left(\frac{x^2}{x^2+1}\right) \right]$.
Correct Answer
Option A is the correct answer.
Detailed Explanation
First, use logarithm properties to simplify the expression:$\ln\left(\frac{x^2}{x^2+1}\right) = \ln(x^2) - \ln(x^2+1) = 2\ln x - \ln(x^2+1)$.Now, differentiate with respect to $x$:$\frac{d}{dx} \left[ 2\ln x - \ln(x^2+1) \right] = 2\left(\frac{1}{x}\right) - \left(\frac{1}{x^2+1}\right) \cdot \frac{d}{dx}(x^2+1)$$= \frac{2}{x} - \frac{2x}{x^2+1}$.Combine the fractions:$= \frac{2(x^2+1) - 2x(x)}{x(x^2+1)} = \frac{2x^2 + 2 - 2x^2}{x(x^2+1)} = \frac{2}{x(x^2+1)}$.
Hint
Use logarithm properties to simplify the expression before differentiating. Remember that $\ln(a/b) = \ln a - \ln b$ and $\ln(a^n) = n \ln a$.