# Chapter 11 - Vectors and Vector-Valued Functions - 11.6 Calculus of Vector-Valued Functions - 11.6 Exercises: 43

$\textbf{r}''(t) = \langle -9cos(3t), -16sin(4t), -36cos(6t)\rangle$ $\textbf{r}'''(t) = \langle 27sin(3t), -64cos(4t), 216sin(6t)\rangle$

#### Work Step by Step

To find the derivative, simply find the derivative of each component. $\textbf{r}(t) = \langle cos(3t), sin(4t), cos(6t)\rangle$ $\textbf{r}'(t) = \langle -3sin(3t), 4cos(4t), -6sin(6t)\rangle$ $\textbf{r}''(t) = \langle -9cos(3t), -16sin(4t), -36cos(6t)\rangle$ $\textbf{r}'''(t) = \langle 27sin(3t), -64cos(4t), 216sin(6t)\rangle$

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