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: 7

Answer

\[\frac{{dy}}{{dx}} = 30\,{\left( {3x + 7} \right)^9}\]

Work Step by Step

\[\begin{gathered} y = \,{\left( {3x + 7} \right)^{10}} \hfill \\ \hfill \\ Use\,\,the\,\,version\,\,1\,\,of\,\,the\,\,chain\,\,rule \hfill \\ \hfill \\ {\text{ }}\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}} \hfill \\ \hfill \\ set{\text{ }}u = 3x + 7 \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = 10\,{\left( {3x + 7} \right)^{10 - 1}}\,{\left( {3x + 7} \right)^,} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = 10\,{\left( {3x + 7} \right)^9}\,\left( 3 \right) \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = 30\,{\left( {3x + 7} \right)^9} \hfill \\ \end{gathered} \]
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