Calculus (3rd Edition)

$$y'= \frac{8\csc^2\theta}{(1+\cot \theta)^2}.$$
Recall that $(\cot x)'=-\csc^2 x$. Since $y=\frac{8}{1+\cot \theta }$, then by the quotient rule $(u/v)'=\frac{vu'-uv'}{v^2}$, the derivative $y'$ is given by $$y'= \frac{(1+\cot \theta) (0)-8(-\csc^2\theta)}{(1+\cot \theta)^2}\\ =\frac{8\csc^2\theta}{(1+\cot \theta)^2}.$$