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: 52

Answer

\[{f^,}\,\left( x \right) = - \frac{{0.05{e^{ - 0.05x}}}}{{\,{{\left( {1 - {e^{ - 0.05x}}} \right)}^2}}}\]

Work Step by Step

\[\begin{gathered} f\,\left( x \right) = \,{\left( {1 - {e^{ - 0.05x}}} \right)^{ - 1}} \hfill \\ \hfill \\ Chain\,\,rule \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = {f^,}\,\left( {g\,\left( t \right)} \right) \cdot {g^,}\,\left( t \right) \hfill \\ \hfill \\ {f^,}\,\left( x \right) = - 1\,{\left( {1 - {e^{ - 0.05x}}} \right)^{ - 2}}\,{\left( {1 - {e^{ - 0.05x}}} \right)^,} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ {f^,}\,\left( x \right) = - 1\,{\left( {1 - {e^{ - 0.05x}}} \right)^{ - 2}}\,\left( {0.05{e^{ - 0.05x}}} \right) \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ {f^,}\,\left( x \right) = - \frac{{0.05{e^{ - 0.05x}}}}{{\,{{\left( {1 - {e^{ - 0.05x}}} \right)}^2}}} \hfill \\ \hfill \\ \end{gathered} \]
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