1. We start with the function $f(x) = (x^2 - 4)^2$. 2. To find the derivative $f'(x)$, we use the chain rule: if $f(x) = [g(x)]^2$, then $f'(x) = 2g(x) \cdot g'(x)$. 3. Here, $g(x) = x^2 - 4$, so $g'(x) = 2x$. 4. Applying the chain rule: $$ f'(x) = 2(x^2 - 4) \cdot 2x $$ 5. Simplify the expression: $$ f'(x) = 4x(x^2 - 4) $$ 6. We can expand the product: $$ f'(x) = 4x^3 - 16x $$ 7. This is the derivative of the function $f(x)$. **Final answer:** $$ f'(x) = 4x^3 - 16x $$