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: 51

Answer

\[{f^,}\,\left( x \right) = \frac{{3{e^{\sqrt {3x} }}{{\sec }^2}\,\left( {{e^{\sqrt {3x} }}} \right)}}{{2\sqrt {3x} }}\]

Work Step by Step

\[\begin{gathered} f\,\left( x \right) = \tan \,\left( {{e^{\sqrt {3x} }}} \right) \hfill \\ \hfill \\ Chain\,\,rule \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = {f^,}\,\left( {g\,\left( t \right)} \right) \cdot {g^,}\,\left( t \right) \hfill \\ \hfill \\ f'\,\left( x \right) = {\sec ^2}\,\left( {{e^{\sqrt {3x} }}} \right)\frac{d}{{dx}}\,\,\left[ {{e^{\sqrt {3x} }}} \right] \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ {f^,}\,\left( x \right) = {\sec ^2}\,\left( {{e^{\sqrt {3x} }}} \right)\,\left( {\frac{3}{{2\sqrt {3x} }}} \right){e^{\sqrt {3x} }} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ {f^,}\,\left( x \right) = \frac{{3{e^{\sqrt {3x} }}{{\sec }^2}\,\left( {{e^{\sqrt {3x} }}} \right)}}{{2\sqrt {3x} }} \hfill \\ \hfill \\ \end{gathered} \]
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