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: 68

Answer

\[ = \,{\left( {z + 4} \right)^2}\,\left( {3\tan z + \,\left( {z + 4} \right){{\sec }^2}z} \right)\]

Work Step by Step

\[\begin{gathered} y = \,{\left( {z + 4} \right)^3}\tan z \hfill \\ \hfill \\ differentiate \hfill \\ \hfill \\ {y^,} = \frac{d}{{dz}}\,\left( {\,{{\left( {z + 4} \right)}^3}\tan z} \right) \hfill \\ \hfill \\ use\,\,the\,\,Product\,\,rule \hfill \\ \hfill \\ = \,{\left( {\,{{\left( {z + 4} \right)}^3}} \right)^,}\tan z + \,{\left( {z + 4} \right)^3}\,{\left( {\tan z} \right)^,} \hfill \\ \hfill \\ use\,\,the\,\,Chain\,\,rule \hfill \\ \hfill \\ = 3\,{\left( {z + 4} \right)^2}\tan z + \,{\left( {z + 4} \right)^3}{\sec ^2}z \hfill \\ \hfill \\ factor \hfill \\ \hfill \\ = \,{\left( {z + 4} \right)^2}\,\left( {3\tan z + \,\left( {z + 4} \right){{\sec }^2}z} \right) \hfill \\ \hfill \\ \end{gathered} \]
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