Answer
\[h'(t)=0\]
Work Step by Step
\[h(t)=\cot^{-1}(t)+\cot^{-1}\left(\frac{1}{t}\right)\;\;\;...(1)\]
We know that \[\frac{d}{dx}(\cot^{-1}x)=\frac{-1}{1+x^2}\;\;\;...(2)\]
Differentiate (1) with respect to $t$ using (2)
\[h'(t)=\frac{-1}{1+t^2}+\frac{-1}{1+\frac{1}{t^2}}\cdot \left(\frac{1}{t}\right)'\]
\[h'(t)=\frac{-1}{1+t^2}+\frac{-1}{1+\frac{1}{t^2}}\cdot \left(\frac{-1}{t^2}\right)\]
\[h'(t)=\frac{-1}{1+t^2}+\frac{-t^2}{1+t^2}\cdot \left(\frac{-1}{t^2}\right)\]
\[h'(t)=\frac{-1}{1+t^2}+\frac{1}{1+t^2}\]
\[h'(t)=0\]
Hence \[h'(t)=0\]