## Calculus (3rd Edition)

$$A_\theta = 4\cos(4\theta-9t), A_t=-9\cos(4\theta-9t) .$$
Recall that $(\sin x)'=\cos x$. Since $A=\sin(4\theta-9t)$, then by using the chain rule, we have $$A_\theta = \cos(4\theta-9t) (4)=4\cos(4\theta-9t), \quad \\ A_t= cos(4\theta-9t) (-9)=-9\cos(4\theta-9t) .$$