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: 12

Answer

\[ = \frac{7}{{2\sqrt {7x - 1} }}\]

Work Step by Step

\[\begin{gathered} y = \sqrt {7x - 1} \hfill \\ \hfill \\ y = f\,\left( u \right) = \sqrt u \hfill \\ \hfill \\ set\,\,u = g\,\left( x \right) = 7x - 1 \hfill \\ \hfill \\ Use\,\,the\,\,version\,\,1\,\,of\,\,the\,\,chain\,\,rule \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}} \hfill \\ \hfill \\ Therefore \hfill \\ \hfill \\ \,\,\frac{{dy}}{{dx}} = \frac{d}{{du}}\,\left( {\sqrt u } \right) \cdot \frac{d}{{dx}}\,\left( {7x - 1} \right) \hfill \\ \hfill \\ then \hfill \\ \hfill \\ = \frac{1}{{2\sqrt u }} \cdot 7 \hfill \\ \hfill \\ substitute\,\,\,back\,\,u = 7x - 1 \hfill \\ \hfill \\ = \frac{7}{{2\sqrt {7x - 1} }} \hfill \\ \hfill \\ \end{gathered} \]
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