Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 4 - Calculating the Derivative - 4.5 Derivatives of Logarithmic Functions - 4.5 Exercises: 26

Answer

\[{y^,} = \frac{{2{e^{2x - 1}}}}{{2x - 1}} + 2{e^{2x - 1}}\ln \,\left( {2x - 1} \right)\]

Work Step by Step

\[\begin{gathered} y = {e^{2x - 1}}\ln \,\left( {2x - 1} \right) \hfill \\ Differentiate \hfill \\ {y^,} = \,\,{\left[ {{e^{2x - 1}}\ln \,\left( {2x - 1} \right)} \right]^,} \hfill \\ Use\,\,the\,\,product\,\,rule\, \hfill \\ {y^,} = {e^{2x - 1}}\,{\left( {\ln \,\left( {2x - 1} \right)} \right)^,} + \ln \,\left( {2x - 1} \right)\,{\left( {{e^{2x - 1}}} \right)^,} \hfill \\ Use\,\,the\,\,\,formula \hfill \\ \frac{d}{{dx}}\,\,\left[ {\ln g\,\left( x \right)} \right] = \frac{{{g^,}\,\left( x \right)}}{{g\,\left( x \right)}},\frac{d}{{dx}}\,\,\left[ {{e^{g\,\left( x \right)}}} \right] = {e^{g\,\left( x \right)}}{g^,}\,\left( x \right) \hfill \\ Then \hfill \\ {y^,} = {e^{2x - 1}}\,\left( {\frac{2}{{2x - 1}}} \right) + \ln \,\left( {2x - 1} \right)\,\left( {2{e^{2x - 1}}} \right) \hfill \\ Multipliying \hfill \\ {y^,} = \frac{{2{e^{2x - 1}}}}{{2x - 1}} + 2{e^{2x - 1}}\ln \,\left( {2x - 1} \right) \hfill \\ \end{gathered} \]
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