Answer
$$ y'= \frac{\cos \theta \cos(\sqrt{\sin\theta +1})}{2\sqrt{\sin\theta +1}}.$$
Work Step by Step
Since $ y=\sin(\sqrt{\sin\theta +1})=\sin(\sin\theta +1)^{1/2})$, by using the chain rule: $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$ and recalling that $(\sin x)'= \cos x $, the derivative $ y'$ is given by
$$ y'=\cos(\sqrt{\sin\theta +1}) ((\sin\theta +1)^{1/2})'\\
=\cos(\sqrt{\sin\theta +1}) \frac{\cos \theta}{2\sqrt{\sin\theta +1}}\\ =\frac{\cos \theta \cos(\sqrt{\sin\theta +1})}{2\sqrt{\sin\theta +1}}.$$
