1. The problem states that: Given $y = (x^3 + 2)^7$, prove that $\frac{dy}{dx} = 21x^2 y$. 2. Start by differentiating $y$ using the chain rule. Let $u = x^3 + 2$. Then $y = u^7$. 3. Differentiate $y$ with respect to $x$: $$\frac{dy}{dx} = 7u^6 \cdot \frac{du}{dx}$$ 4. Compute $\frac{du}{dx}$: $$\frac{du}{dx} = 3x^2$$ 5. Substitute back $u = x^3 + 2$: $$\frac{dy}{dx} = 7(x^3 + 2)^6 \cdot 3x^2 = 21x^2 (x^3 + 2)^6$$ 6. Since $y = (x^3 + 2)^7$, then $(x^3 + 2)^6 = \frac{y}{x^3 + 2}$. 7. Substitute this in: $$\frac{dy}{dx} = 21x^2 \cdot \frac{y}{x^3 + 2}$$ 8. Multiply both sides by $(x^3 + 2)$ to get: $$(x^3 + 2) \frac{dy}{dx} = 21x^2 y$$ This proves the statement as desired.