Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.6 Inverse Trigonometric Functions - 6.6 Exercises - Page 481: 33

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 \[h'(t)=0\]
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