1. **State the problem:** We are given the function $y = x^3 \ln \sqrt{x^2 + 1}$ and need to find its derivative $\frac{dy}{dx}$. 2. **Rewrite the function:** Recall that $\sqrt{x^2 + 1} = (x^2 + 1)^{\frac{1}{2}}$, so $$y = x^3 \ln \left( (x^2 + 1)^{\frac{1}{2}} \right) = x^3 \cdot \frac{1}{2} \ln (x^2 + 1) = \frac{x^3}{2} \ln (x^2 + 1).$$ 3. **Use the product rule:** For $y = u \cdot v$ where $u = \frac{x^3}{2}$ and $v = \ln (x^2 + 1)$, $$\frac{dy}{dx} = u'v + uv'.$$ 4. **Find derivatives of $u$ and $v$:** - $u = \frac{x^3}{2} \implies u' = \frac{3x^2}{2}$. - $v = \ln (x^2 + 1) \implies v' = \frac{1}{x^2 + 1} \cdot 2x = \frac{2x}{x^2 + 1}$ by chain rule. 5. **Substitute back:** $$\frac{dy}{dx} = \frac{3x^2}{2} \ln (x^2 + 1) + \frac{x^3}{2} \cdot \frac{2x}{x^2 + 1} = \frac{3x^2}{2} \ln (x^2 + 1) + \frac{x^4}{x^2 + 1}.$$ 6. **Final answer:** $$\boxed{\frac{dy}{dx} = \frac{3x^2}{2} \ln (x^2 + 1) + \frac{x^4}{x^2 + 1}}.$$