Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Practice Exercises - Page 177: 14


$$\frac{{ds}}{{dt}} = \frac{6}{{{t^2}}}{\cot ^2}\left( {\frac{2}{t}} \right){\csc ^2}\left( {\frac{2}{t}} \right)$$

Work Step by Step

$$\eqalign{ & s = {\cot ^3}\left( {\frac{2}{t}} \right) \cr & {\text{differentiate with respect to }}t \cr & \frac{{ds}}{{dt}} = \frac{d}{{dt}}\left[ {{{\cot }^3}\left( {\frac{2}{t}} \right)} \right] \cr & {\text{use the chain rule}} \cr & \frac{{ds}}{{dt}} = 3{\cot ^2}\left( {\frac{2}{t}} \right)\frac{d}{{dt}}\left[ {\cot \left( {\frac{2}{t}} \right)} \right] \cr & {\text{use }}\frac{d}{{dt}}\left[ {\cot u} \right] = - {\csc ^2}u\frac{{du}}{{dx}} \cr & \frac{{ds}}{{dt}} = 3{\cot ^2}\left( {\frac{2}{t}} \right)\left( { - {{\csc }^2}\left( {\frac{2}{t}} \right)} \right)\frac{d}{{dt}}\left[ {\frac{2}{t}} \right] \cr & {\text{find the derivative and simplify}} \cr & \frac{{ds}}{{dt}} = 3{\cot ^2}\left( {\frac{2}{t}} \right)\left( { - {{\csc }^2}\left( {\frac{2}{t}} \right)} \right)\left( { - \frac{2}{{{t^2}}}} \right) \cr & \frac{{ds}}{{dt}} = \frac{6}{{{t^2}}}{\cot ^2}\left( {\frac{2}{t}} \right){\csc ^2}\left( {\frac{2}{t}} \right) \cr} $$
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