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First, use the logarithm property $\ln(A/B) = \ln A - \ln B$:$y = \ln(x^2) - \ln(x^2+1)$Then, use the property $\ln(x^n) = n\ln x$:$y = 2\ln x - \ln(x^2+1)$Now, differentiate $y$ with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(2\ln x) - \frac{d}{dx}(\ln(x^2+1))$$\frac{d}{dx}(2\ln x) = 2 \cdot \frac{1}{x} = \frac{2}{x}$For the second term, use the chain rule $\frac{d}{dx}(\ln u) = \frac{1}{u} \frac{du}{dx}$ where $u = x^2+1$:$\frac{d}{dx}(\ln(x^2+1)) = \frac{1}{x^2+1} \cdot \frac{d}{dx}(x^2+1) = \frac{1}{x^2+1} \cdot 2x = \frac{2x}{x^2+1}$Combine the derivatives:$\frac{dy}{dx} = \frac{2}{x} - \frac{2x}{x^2+1}$To simplify, find a common denominator, which is $x(x^2+1)$:$\frac{dy}{dx} = \frac{2(x^2+1)}{x(x^2+1)} - \frac{2x \cdot x}{x(x^2+1)} = \frac{2x^2 + 2 - 2x^2}{x(x^2+1)} = \frac{2}{x(x^2+1)}$