1. **State the problem:** Find the derivative $y'$ of the function $$y = \ln(\cos(x^2 + 1)).$$ 2. **Recall the formula:** The derivative of $\ln(u)$ with respect to $x$ is $$\frac{d}{dx} \ln(u) = \frac{1}{u} \cdot \frac{du}{dx}.$$ Also, the derivative of $\cos(v)$ is $$\frac{d}{dx} \cos(v) = -\sin(v) \cdot \frac{dv}{dx}.$$ 3. **Apply the chain rule:** Let $$u = \cos(x^2 + 1),$$ so $$y = \ln(u).$$ Then, $$y' = \frac{1}{u} \cdot u' = \frac{1}{\cos(x^2 + 1)} \cdot \frac{d}{dx} \cos(x^2 + 1).$$ 4. **Differentiate $u = \cos(x^2 + 1)$:** $$u' = -\sin(x^2 + 1) \cdot \frac{d}{dx} (x^2 + 1) = -\sin(x^2 + 1) \cdot 2x.$$ 5. **Substitute back:** $$y' = \frac{1}{\cos(x^2 + 1)} \cdot (-\sin(x^2 + 1) \cdot 2x) = -2x \frac{\sin(x^2 + 1)}{\cos(x^2 + 1)}.$$ 6. **Simplify using tangent:** $$y' = -2x \tan(x^2 + 1).$$ **Final answer:** $$y' = -2x \tan(x^2 + 1).$$