Answer
$$ y'= -\frac{4}{\theta^2}\cos\frac{4}{\theta}.$$
Work Step by Step
Recall that $(\sin x)'=\cos x$.
Recall that $(x^n)'=nx^{n-1}$
Since $ y=\sin \frac{4}{\theta}$, then by the chain rule, the derivative $ y'$ is given by
$$ y'= \cos\frac{4}{\theta} \left(-\frac{4}{\theta^2}\right)=-\frac{4}{\theta^2}\cos\frac{4}{\theta}.$$
