Question Detail

To obtain the derivative of $y = \ln\left(\frac{x^2}{x^2+1}\right)$, we can first use the logarithm property $\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)$ and $\ln(a^n) = n\ln(a)$:$y = \ln(x^2) - \ln(x^2+1) = 2\ln(x) - \ln(x^2+1)$Now, differentiate term by term:$\frac{dy}{dx} = \frac{d}{dx}(2\ln x) - \frac{d}{dx}(\ln(x^2+1))$Using the chain rule for $\ln(u)$, which is $\frac{1}{u}\frac{du}{dx}$:$\frac{dy}{dx} = 2 \cdot \frac{1}{x} - \frac{1}{x^2+1} \cdot \frac{d}{dx}(x^2+1)$$\frac{dy}{dx} = \frac{2}{x} - \frac{1}{x^2+1} \cdot (2x)$$\frac{dy}{dx} = \frac{2}{x} - \frac{2x}{x^2+1}$To combine these fractions, find a common denominator:$\frac{dy}{dx} = \frac{2(x^2+1) - 2x(x)}{x(x^2+1)}$$\frac{dy}{dx} = \frac{2x^2+2 - 2x^2}{x(x^2+1)}$$\frac{dy}{dx} = \frac{2}{x(x^2+1)}$