## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 3 - Derivatives - 3.7 The Chain Rule - 3.7 Exercises - Page 192: 61

#### Answer

$= 2\theta \sec 5\theta + 5{\theta ^2}\sec 5\theta \tan 5\theta$

#### Work Step by Step

$\begin{gathered} y = {\theta ^2}\sec 5\theta \hfill \\ \hfill \\ {y^,} = \frac{d}{{d\theta }}\,\left( {{\theta ^2}\sec 5\theta } \right) \hfill \\ \hfill \\ use\,\,the\,\,Product\,\,\,rule \hfill \\ \hfill \\ = \,{\left( {{\theta ^2}} \right)^,} \cdot \sec 5\theta + {\theta ^2} \cdot \,{\left( {\sec 5\theta } \right)^,} \hfill \\ \hfill \\ use\,\,the\,\,Chain\,\,rule \hfill \\ \hfill \\ = 2\theta \sec 5\theta + {\theta ^2} \cdot \,\left( {\sec 5\theta \tan 5\theta } \right) \cdot \,{\left( {5\theta } \right)^,} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = 2\theta \sec 5\theta + 5{\theta ^2}\sec 5\theta \tan 5\theta \hfill \\ \hfill \\ \end{gathered}$

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