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: 11

Answer

\[ = 5{e^{5x - 7}}\]

Work Step by Step

\[\begin{gathered} y = {e^{5x - 7}} \hfill \\ \hfill \\ y = f\,\left( u \right) = {e^u} \hfill \\ \hfill \\ set\,\,u = g\,\left( x \right) = 5x - 7 \hfill \\ \hfill \\ Use\,\,the\,\,version\,\,1\,\,of\,\,the\,\,chain\,\,rule \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}} \hfill \\ \hfill \\ Therefore \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{d}{{du}}\,\left( {{e^u}} \right) \cdot \frac{d}{{dx}}\,\left( {5x - 7} \right) \hfill \\ \hfill \\ = {e^u} \cdot 5 \hfill \\ \hfill \\ substitute\,\,\,back\,\,u = 5x - 7 \hfill \\ \hfill \\ = 5{e^{5x - 7}} \hfill \\ \hfill \\ \end{gathered} \]
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