1. Stating the problem: Find the derivatives of the functions \(y = 3x^{-4} + 2x^{-3} + x^{-2} + x^{-1}\) and \(y = \cos^3(2x)\). 2. For \(y = 3x^{-4} + 2x^{-3} + x^{-2} + x^{-1}\), use the power rule \(\frac{d}{dx} x^n = nx^{n-1}\): \[\frac{dy}{dx} = 3(-4)x^{-5} + 2(-3)x^{-4} + (-2)x^{-3} + (-1)x^{-2} = -12x^{-5} - 6x^{-4} - 2x^{-3} - x^{-2}\] 3. For \(y = \cos^3(2x) = (\cos(2x))^3\), use the chain rule: Let \(u = \cos(2x)\), so \(y = u^3\). First, \(\frac{dy}{du} = 3u^2\). Second, \(\frac{du}{dx} = \frac{d}{dx} \cos(2x) = -\sin(2x) \cdot 2 = -2\sin(2x)\). Then applying chain rule: \[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 3u^2 \cdot (-2\sin(2x)) = -6\cos^2(2x)\sin(2x)\] Final answers: \[\frac{dy}{dx} = -12x^{-5} - 6x^{-4} - 2x^{-3} - x^{-2}\] \[\frac{dy}{dx} = -6\cos^2(2x)\sin(2x)\]