1. **Problem Statement:** Find the derivative of the vector function $$F(t)=e^{9t} \mathbf{i} + \sin^8(t) \mathbf{j} - \cos(4t) \mathbf{k}$$ with respect to $t$, and then calculate the magnitude of the derivative vector $F'(t)$. 2. **Derivative of each component:** - For the $\mathbf{i}$ component: $$\frac{d}{dt} e^{9t} = 9e^{9t}$$ - For the $\mathbf{j}$ component: Use the chain rule for $\sin^8(t)$: $$\frac{d}{dt} \sin^8(t) = 8\sin^7(t) \cdot \cos(t)$$ - For the $\mathbf{k}$ component: $$\frac{d}{dt} \left(-\cos(4t)\right) = -(-\sin(4t) \cdot 4) = 4\sin(4t)$$ 3. **Derivative vector $F'(t)$:** $$F'(t) = 9e^{9t} \mathbf{i} + 8\sin^7(t) \cos(t) \mathbf{j} + 4\sin(4t) \mathbf{k}$$ 4. **Calculate magnitude $||F'(t)||$:** $$||F'(t)|| = \sqrt{(9e^{9t})^2 + (8\sin^7(t) \cos(t))^2 + (4\sin(4t))^2}$$ Simplify inside the square root: $$= \sqrt{81e^{18t} + 64 \sin^{14}(t) \cos^2(t) + 16 \sin^2(4t)}$$ This is the simplified expression for the magnitude. **Final answers:** $$\boxed{F'(t) = 9e^{9t} \mathbf{i} + 8\sin^7(t) \cos(t) \mathbf{j} + 4\sin(4t) \mathbf{k}}$$ $$\boxed{||F'(t)|| = \sqrt{81e^{18t} + 64 \sin^{14}(t) \cos^2(t) + 16 \sin^2(4t)}}$$