## Calculus (3rd Edition)

$$h'(t)=-9\csc t \cot t+ \cot t -t \csc^2 t.$$
Since $h(t)=9\csc t+t \cot t$, then, using the product rule: $(uv)'=u'v+uv’$ and also, using the facts that $(\csc x)'=-\csc x \cot x$ and $(\cot x)'=-\csc^2 x$, we have $$h'(t)=(9\csc t)' +(t \cot t)'\\ =-9\csc t \cot t+ \cot t -t \csc^2 t.$$