1. Differentiate $y = \ln(1 + \sin^2 x)$.\n\nStep 1: State the problem. We want to find $\frac{dy}{dx}$ for $y = \ln(1 + \sin^2 x)$.\n\nStep 2: Use the chain rule for differentiation. The derivative of $\ln u$ with respect to $x$ is $\frac{1}{u} \cdot \frac{du}{dx}$. Here, $u = 1 + \sin^2 x$.\n\nStep 3: Differentiate $u = 1 + \sin^2 x$. Using the chain rule again, $\frac{du}{dx} = 2 \sin x \cdot \cos x$ because $\frac{d}{dx}(\sin^2 x) = 2 \sin x \cdot \cos x$.\n\nStep 4: Combine the results:\n$$\frac{dy}{dx} = \frac{1}{1 + \sin^2 x} \cdot 2 \sin x \cos x = \frac{2 \sin x \cos x}{1 + \sin^2 x}.$$\n\nStep 5: Simplify if desired. Note that $2 \sin x \cos x = \sin 2x$, so an alternative form is:\n$$\frac{dy}{dx} = \frac{\sin 2x}{1 + \sin^2 x}.$$\n\nThis is the derivative of the given function.\n