1. **Problem Statement:** Find the derivative $\frac{dy}{dx}$ of the function $$y = x^{2} \cos \left(\sqrt{x^{3} - 1} + 2 \right)$$ 2. **Formula and Rules:** We will use the product rule for derivatives since $y$ is a product of two functions: $$u = x^{2} \quad \text{and} \quad v = \cos \left(\sqrt{x^{3} - 1} + 2 \right)$$ The product rule states: $$\frac{d}{dx}(uv) = u'v + uv'$$ We also need the chain rule to differentiate $v$ because it is a composition of functions. 3. **Step 1: Differentiate $u = x^{2}$** $$u' = 2x$$ 4. **Step 2: Differentiate $v = \cos \left(\sqrt{x^{3} - 1} + 2 \right)$** Let $$w = \sqrt{x^{3} - 1} + 2$$ Then $$v = \cos(w)$$ Using the chain rule: $$v' = -\sin(w) \cdot \frac{dw}{dx}$$ 5. **Step 3: Differentiate $w = \sqrt{x^{3} - 1} + 2$** Rewrite $w$ as: $$w = (x^{3} - 1)^{\frac{1}{2}} + 2$$ Derivative of constant 2 is 0, so: $$\frac{dw}{dx} = \frac{1}{2}(x^{3} - 1)^{-\frac{1}{2}} \cdot 3x^{2} = \frac{3x^{2}}{2\sqrt{x^{3} - 1}}$$ 6. **Step 4: Substitute back to find $v'$** $$v' = -\sin \left(\sqrt{x^{3} - 1} + 2 \right) \cdot \frac{3x^{2}}{2\sqrt{x^{3} - 1}}$$ 7. **Step 5: Apply the product rule** $$\frac{dy}{dx} = u'v + uv' = 2x \cdot \cos \left(\sqrt{x^{3} - 1} + 2 \right) + x^{2} \cdot \left(-\sin \left(\sqrt{x^{3} - 1} + 2 \right) \cdot \frac{3x^{2}}{2\sqrt{x^{3} - 1}} \right)$$ 8. **Step 6: Simplify the expression** $$\frac{dy}{dx} = 2x \cos \left(\sqrt{x^{3} - 1} + 2 \right) - \frac{3x^{4}}{2\sqrt{x^{3} - 1}} \sin \left(\sqrt{x^{3} - 1} + 2 \right)$$ **Final answer:** $$\boxed{\frac{dy}{dx} = 2x \cos \left(\sqrt{x^{3} - 1} + 2 \right) - \frac{3x^{4}}{2\sqrt{x^{3} - 1}} \sin \left(\sqrt{x^{3} - 1} + 2 \right)}$$