1. **Problem:** Find the derivative of $y = \arcsin(4x^4)$. 2. **Formula:** The derivative of $y = \arcsin(u)$ is $\frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$. 3. **Step 1:** Identify $u = 4x^4$. Then $\frac{du}{dx} = 16x^3$. 4. **Step 2:** Apply the chain rule: $$\frac{dy}{dx} = \frac{1}{\sqrt{1-(4x^4)^2}} \cdot 16x^3 = \frac{16x^3}{\sqrt{1-16x^8}}.$$ 5. **Answer:** $$\boxed{\frac{dy}{dx} = \frac{16x^3}{\sqrt{1-16x^8}}}.$$