1. **State the problem:** Find $\frac{dy}{dx}$ for $y = u(u^2 + 3)^3$ where $u = (x + 3)^2$ at $x = -2$ using the chain rule in Leibniz notation. 2. **Recall the chain rule:** If $y$ is a function of $u$ and $u$ is a function of $x$, then $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$ 3. **Find $\frac{dy}{du}$:** Given $y = u(u^2 + 3)^3$, use the product rule: $$\frac{dy}{du} = (1) \cdot (u^2 + 3)^3 + u \cdot 3(u^2 + 3)^2 \cdot 2u$$ Simplify: $$\frac{dy}{du} = (u^2 + 3)^3 + 6u^2 (u^2 + 3)^2$$ 4. **Find $\frac{du}{dx}$:** Given $u = (x + 3)^2$, differentiate: $$\frac{du}{dx} = 2(x + 3)$$ 5. **Evaluate at $x = -2$:** Calculate $u$: $$u = (-2 + 3)^2 = 1^2 = 1$$ Calculate $\frac{dy}{du}$ at $u=1$: $$\frac{dy}{du} = (1^2 + 3)^3 + 6 \cdot 1^2 \cdot (1^2 + 3)^2 = 4^3 + 6 \cdot 1 \cdot 4^2 = 64 + 6 \cdot 16 = 64 + 96 = 160$$ Calculate $\frac{du}{dx}$ at $x=-2$: $$\frac{du}{dx} = 2(-2 + 3) = 2(1) = 2$$ 6. **Calculate $\frac{dy}{dx}$ at $x = -2$:** $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 160 \cdot 2 = 320$$ **Final answer:** $$\boxed{\frac{dy}{dx} = 320 \text{ at } x = -2}$$