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: 8

Answer

\[\frac{{dy}}{{dx}} = \,\left( {200x + 220} \right)\,{\left( {5{x^2} + 11x} \right)^{19}}\]

Work Step by Step

\[\begin{gathered} y = \,{\left( {5{x^2} + 11x} \right)^{20}} \hfill \\ \hfill \\ Use\,\,the\,\,version\,\,1\,\,of\,\,the\,\,chain\,\,rule \hfill \\ \hfill \\ {\text{ }}\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}} \hfill \\ \hfill \\ set\,\,u = 5{x^2} + 11x \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = 20\,{\left( {5{x^2} + 11x} \right)^{19}}\frac{d}{{dx}}\,\,\left[ {5{x^2} + 11x} \right] \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = 20\,{\left( {5{x^2} + 11x} \right)^{19}}\,\left( {10x + 11} \right) \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \,\left( {200x + 220} \right)\,{\left( {5{x^2} + 11x} \right)^{19}} \hfill \\ \end{gathered} \]
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