Answer
$h'(t)=\frac{1}{t^2(1+\frac{1}{t^2})}-\frac{1}{1+t^2}=0$
Work Step by Step
$h(t)=cot^{-1}(t)+cot^{-1}(\frac{1}{t})\\
h'(t)=-\frac{1}{1+t^2}+(-\frac{1}{1+(\frac{1}{t})^2})\times(-{\frac{1}{t^2}})\\
h'(t)=-\frac{1}{1+t^2}+\frac{1}{t^2(1+\frac{1}{t^2})}\\
h'(t)=\frac{1}{t^2(1+\frac{1}{t^2})}-\frac{1}{1+t^2}=0$