Answer
$$ h'(t)=-\frac{2t(\cot t+1)\csc^2t}{t^2}.$$
Work Step by Step
Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$
Recall that $(\csc x)'=-\csc x \cot x$.
Since $ h(t)=\frac{\csc^2t}{t}$, then using the quotient rule, we have
$$ h'(t)=\frac{t(t(\csc^2t)'-(t^2)'\csc^2t}{t^2}\\
=\frac{t(-2\csc t \csc t\cot t)-2t\csc^2t}{t^2}
\\=-\frac{2t(\cot t+1)\csc^2t}{t^2}.$$
