Answer
$$y'=\frac{2\sin\theta+2\theta\cos\theta-\theta\sin\theta\tan\theta}{2\sec\theta}$$
Work Step by Step
$$y=\frac{\theta\sin\theta}{\sqrt{\sec\theta}}$$
Logarithmic differentiation helps to simplify the differentiation of otherwise complex functions by using the algebraic properties of logarithm. In detail,
1) Take the natural logarithm of both sides and simplify:
$$\ln y=\ln\Big(\frac{\theta\sin\theta}{\sqrt{\sec\theta}}\Big)$$
Recall that $\ln(A\times B)=\ln A+\ln B$ and $\ln(A/B)=\ln A-\ln B$: $$\ln y=\ln(\theta\sin\theta)-\ln(\sqrt{\sec\theta})$$ $$\ln y=\ln\theta+\ln\sin\theta-\ln\sqrt{\sec\theta}$$
2) Then we take the derivative of both sides with respect to $\theta$ and remember that $(\ln \theta)'=1/\theta$:
$$\frac{1}{y}\times y'=\frac{1}{\theta}+\frac{1}{\sin\theta}(\sin\theta)'-\frac{1}{\sqrt{\sec\theta}}(\sqrt{\sec\theta})'$$ $$\frac{y'}{y}=\frac{1}{\theta}+\frac{\cos\theta}{\sin\theta}-\frac{1}{\sqrt{\sec\theta}}\times\frac{1}{2\sqrt{\sec\theta}}(\sec\theta)'$$ $$\frac{y'}{y}=\frac{1}{\theta}+\frac{\cos\theta}{\sin\theta}-\frac{(\sec\theta)'}{2\sec\theta}$$ $$\frac{y'}{y}=\frac{1}{\theta}+\frac{\cos\theta}{\sin\theta}-\frac{\sec\theta\tan\theta}{2\sec\theta}=\frac{1}{\theta}+\frac{\cos\theta}{\sin\theta}-\frac{\tan\theta}{2}$$ $$\frac{y'}{y}=\frac{2\sin\theta+2\theta\cos\theta-\theta\sin\theta\tan\theta}{2\theta\sin\theta}$$
3) Solve for $y'$: $$y'=\frac{2\sin\theta+2\theta\cos\theta-\theta\sin\theta\tan\theta}{2\theta\sin\theta}\times y$$
4) Finally, substitute for $y=\frac{\theta\sin\theta}{\sqrt{\sec\theta}}$
$$y'=\frac{2\sin\theta+2\theta\cos\theta-\theta\sin\theta\tan\theta}{2\theta\sin\theta}\times\frac{\theta\sin\theta}{\sqrt{\sec\theta}}$$ $$y'=\frac{2\sin\theta+2\theta\cos\theta-\theta\sin\theta\tan\theta}{2\sec\theta}$$
