91Ó°ÊÓ

To find the derivative of the given vector function, we will differentiate each component with respect to $$t$$. We have: \begin{aligned} \frac{d\mathbf{u}}{dt} &= \frac{d}{dt}(2t^3\mathbf{i}+(t^2-1)\mathbf{j}-8\mathbf{k}) \\ &= \left(\frac{d}{dt} 2t^3\right)\mathbf{i} + \left(\frac{d}{dt} (t^2-1)\right)\mathbf{j} + \left(\frac{d}{dt}(-8)\right)\mathbf{k} \\ &= (6t^2\mathbf{i} + 2t\mathbf{j}) \\ \end{aligned}

Now that we have found the derivative of $$\mathbf{u}(t)$$, we can use the chain rule to find the derivative of the function $$\left(t^{12}+3t\right)\mathbf{u}(t)$$: \begin{aligned} \frac{d}{dt}\left((t^{12}+3t)\mathbf{u}(t)\right) &= \frac{d}{dt}(t^{12}+3t) \cdot \mathbf{u}(t) + (t^{12}+3t) \cdot \frac{d\mathbf{u}}{dt} \\ &= \left(12t^{11}+3\right) \left( 2t^3\mathbf{i}+(t^2-1)\mathbf{j}-8\mathbf{k}\right) \\ & \qquad + \left(t^{12}+3t\right) \left(6t^2\mathbf{i} + 2t\mathbf{j}\right) \\ \end{aligned} Now we distribute and simplify the expression. \begin{aligned} &= \left(12t^{11}+3\right) (2t^3\mathbf{i}+(t^2-1)\mathbf{j}-8\mathbf{k}) + (t^{12}+3t)(6t^2\mathbf{i} + 2t\mathbf{j})\\ &= 24t^{14}\mathbf{i} + 12t^{11}(t^2-1)\mathbf{j} - 12t^{11}\cdot8\mathbf{k} + 6t^{14}\mathbf{i} + 18t^3\mathbf{i} + 2t^{13}\mathbf{j} + 6t^2\mathbf{j} \\ &= (24t^{14} + 6t^{14} + 18t^3)\mathbf{i} + (12t^{11}(t^2-1) + 2t^{13} + 6t^2)\mathbf{j} - 96t^{11}\mathbf{k} \end{aligned} The final derivative is: $$\frac{d}{dt}\left((t^{12}+3t)\mathbf{u}(t)\right) = \left(30t^{14} + 18t^3\right)\mathbf{i} + \left(12t^{11}(t^2-1) + 2t^{13} + 6t^2\right)\mathbf{j} - 96t^{11}\mathbf{k}$$