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