1. The problem states: Given $y=(x^3+2)^7$, prove that $\frac{dy}{dx} = 21x^2y/(x^3+2)$. 2. Start by differentiating $y$ using the chain rule. Let $u = x^3 + 2$, so $y = u^7$. 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: $$\frac{dy}{dx} = 7 (x^3 + 2)^6 \cdot 3x^2 = 21x^2 (x^3 + 2)^6.$$ 6. Recall that $y = (x^3 + 2)^7$, so $y = (x^3 + 2) \cdot (x^3 + 2)^6$. 7. Therefore, rewrite $\frac{dy}{dx}$ as: $$\frac{dy}{dx} = 21x^2 \frac{y}{x^3 + 2}.$$ 8. Hence, it is proved that: $$\frac{dy}{dx} = \frac{21x^2 y}{x^3 + 2}.$$