Chapter 14 - Calculus of Vector-Valued Functions - 14.2 Calculus of Vector-Valued Functions - Exercises - Page 720: 12

$$ a'(\theta) =(-3\sin \theta) i+( \sin2\theta ) j+(\sec^2 \theta) k.$$

Since $ a(\theta)=(\cos3 \theta) i+(\sin^2 \theta) j+(\tan \theta) k $, then by using the chain rule, the derivative $ a'(\theta)$ is given by $$ a'(\theta)=(-3\sin \theta) i+(2\sin \theta \cos \theta) j+(\sec^2 \theta) k\\ =(-3\sin \theta) i+( \sin2\theta ) j+(\sec^2 \theta) k.$$