Answer
$y'=2(-1-\dfrac{csc\theta}{2}-\dfrac{\theta^2}{4})(\dfrac{csc\theta \cot \theta}{2}- \dfrac{\theta}{2})$
Work Step by Step
Given that $y=(-1-\dfrac{csc\theta}{2}-\dfrac{\theta^2}{4})^2$
Apply derivative rules of differentiation:
$f(x)=p'(x)q(x)+p(x)q'(x)$
$y'=\frac{d}{dx}[(-1-\dfrac{csc\theta}{2}-\dfrac{\theta^2}{4})^2]$
$=2(-1-\dfrac{csc\theta}{2}-\dfrac{\theta^2}{4})(-1-\dfrac{csc\theta}{2}-\dfrac{\theta^2}{4})'$
Hence, $y'=2(-1-\dfrac{csc\theta}{2}-\dfrac{\theta^2}{4})(\dfrac{csc\theta \cot \theta}{2}- \dfrac{\theta}{2})$
