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: 41

Answer

\[\frac{{dy}}{{dx}} = \,\left( {300{x^5} - 225{x^2}} \right)\,{\left( {2{x^6} - 3{x^3} + 3} \right)^{24}}\]

Work Step by Step

\[\begin{gathered} y = \,{\left( {2{x^6} - 3{x^3} + 3} \right)^{25}} \hfill \\ \hfill \\ Use\,\,the\,\,version\,\,1\,\,of\,\,the\,\,chain\,\,rule \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}} \hfill \\ \hfill \\ set\,\,u = 2{x^6} - 3{x^3} + 3 \hfill \\ \hfill \\ \frac{{du}}{{dx}} = 12{x^5} - 9{x^2} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = 25\,{\left( {2{x^6} - 3{x^3} + 3} \right)^{24}}\,\left( {12{x^5} - 9{x^2}} \right) \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \,\left( {300{x^5} - 225{x^2}} \right)\,{\left( {2{x^6} - 3{x^3} + 3} \right)^{24}} \hfill \\ \end{gathered} \]
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