1. Problem 1: Find the derivative of $$f(x) = (2x^3 - 8x^2 + 5)^4$$. 2. We apply the chain rule: If $$f(x) = [u(x)]^4$$, then $$f'(x) = 4[u(x)]^3 \cdot u'(x)$$. 3. Here, $$u(x) = 2x^3 - 8x^2 + 5$$. 4. Compute $$u'(x) = \frac{d}{dx}(2x^3 - 8x^2 + 5) = 6x^2 - 16x$$. 5. Substitute back: $$f'(x) = 4(2x^3 - 8x^2 + 5)^3 \cdot (6x^2 - 16x)$$. --- 6. Problem 2: Find the derivative of $$f(x) = \sqrt[3]{\frac{1}{4x^2} - 5}$$. 7. Rewrite as: $$f(x) = \left(\frac{1}{4x^2} - 5\right)^{\frac{1}{3}}$$. 8. Let $$v(x) = \frac{1}{4x^2} - 5$$, so $$f(x) = [v(x)]^{1/3}$$. 9. Using the chain rule: $$f'(x) = \frac{1}{3} [v(x)]^{-\frac{2}{3}} \cdot v'(x)$$. 10. Compute $$v'(x)$$: $$v'(x) = \frac{d}{dx} \left(\frac{1}{4x^2} - 5\right) = \frac{d}{dx} \left(\frac{1}{4x^2}\right) - 0$$. 11. Since $$\frac{1}{4x^2} = \frac{1}{4} x^{-2}$$, $$v'(x) = \frac{1}{4} \cdot (-2) x^{-3} = -\frac{1}{2x^3}$$. 12. Substitute back: $$f'(x) = \frac{1}{3} \left(\frac{1}{4x^2} - 5\right)^{-\frac{2}{3}} \cdot\left(-\frac{1}{2x^3}\right) = -\frac{1}{6x^{3}} \left(\frac{1}{4x^2} - 5\right)^{-\frac{2}{3}}$$. Final answers: $$f'(x) = 4(2x^3 - 8x^2 + 5)^3 (6x^2 - 16x)$$ and $$f'(x) = -\frac{1}{6x^{3}} \left(\frac{1}{4x^2} - 5\right)^{-\frac{2}{3}}$$