Chapter 3 - Section 3.5 - Implicit Differentiation - 3.5 Exercises - Page 216: 55

$h'(t)=\frac{1}{t^2(1+\frac{1}{t^2})}-\frac{1}{1+t^2}=0$

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