1. **Stating the problem:** We are given that $y=f(x^2)$ and the derivative $\frac{dy}{dx} = 2x^3 - 4x$. We need to find $f'(4)$. 2. Since $y = f(x^2)$, by the chain rule, $$\frac{dy}{dx} = f'(x^2) \cdot \frac{d}{dx}(x^2) = f'(x^2) \cdot 2x.$$ 3. We know $$\frac{dy}{dx} = 2x^3 - 4x = 2x(x^2 - 2).$$ 4. Equating the two expressions for $\frac{dy}{dx}$: $$f'(x^2) \cdot 2x = 2x(x^2 - 2).$$ 5. For $x \neq 0$, divide both sides by $2x$: $$f'(x^2) = x^2 - 2.$$ 6. To find $f'(4)$, set $x^2 = 4$, so $x = \pm 2$. 7. Substitute $x=2$: $$f'(4) = 2^2 - 2 = 4 - 2 = 2.$$ 8. So the answer is (b) 2.