Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.7 The Chain Rule - 3.7 Exercises: 30

Answer

\[\frac{{dy}}{{dt}} = - \,\left( {2t + 1} \right)\csc \,\left( {{t^2} + t} \right)\cot \,\left( {{t^2} + t} \right)\]

Work Step by Step

\[\begin{gathered} y = \csc \,\left( {{t^2} + t} \right) \hfill \\ \hfill \\ Use\,\,the\,\,version\,\,2\,\,of\,\,the\,\,chain\,\,rule\, \hfill \\ \hfill \\ \,\,\frac{d}{{dx}}\,\,\left[ {f\,\left( {g\,\left( x \right)} \right)} \right] = {f^,}\,\left( {g\,\left( x \right)} \right) \cdot {g^,}\,\left( x \right) \hfill \\ \hfill \\ set\,\,\,g\,\left( t \right) = {t^2} + t \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \,\,\,\,\,\,\,\,\,\,\,{g^,}\,\left( t \right) = 2t + 1 \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = - \csc \,\left( {{t^2} + t} \right)\cot \,\left( {{t^2} + t} \right)\,\left( {2t + 1} \right) \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ \frac{{dy}}{{dt}} = - \,\left( {2t + 1} \right)\csc \,\left( {{t^2} + t} \right)\cot \,\left( {{t^2} + t} \right) \hfill \\ \hfill \\ \end{gathered} \]
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