Calculus (3rd Edition)

$$y'= \csc(9z+1)-9z\csc (9z+1) \cot (9z+1).$$
Recall the product rule: $(uv)'=u'v+uv'$ Recall that $(\csc x)'=-\csc x \cot x$. Since $y=z\csc(9z+1)$, then by the product rule, the derivative $y'$ is given by $$y'= \csc(9z+1)+9z(-\csc (9z+1) \cot (9z+1))\\= \csc(9z+1)-9z\csc (9z+1) \cot (9z+1).$$