Calculus with Applications (10th Edition)

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

Chapter 4 - Calculating the Derivative - Chapter Review - Review Exercises: 49

Answer

\[{f^,}\,\left( x \right) = \frac{{x{e^{3x}} + {e^{3x}}}}{{x{e^x} + 1}} + 2{e^{2x}}\ln \,\left( {x{e^x} + 1} \right)\]

Work Step by Step

\[\begin{gathered} f\,\left( x \right) = {e^{2x}}\ln \,\left( {x{e^x} + 1} \right) \hfill \\ Find\,\,the\,\,derivative\,\,of\,\,the\,\,\,function \hfill \\ {f^,}\,\left( x \right) = \,\,{\left[ {{e^{2x}}\ln \,\left( {x{e^x} + 1} \right)} \right]^,} \hfill \\ Use\,\,the\,\,product\,\,rule \hfill \\ {f^,}\,\left( x \right) = {e^{2x}}\,\,{\left[ {\ln \,\left( {x{e^x} + 1} \right)} \right]^,} + \ln \,\left( {x{e^x} + 1} \right)\,{\left( {{e^{2x}}} \right)^,} \hfill \\ Where\,\,\frac{d}{{dx}}\,\,\left[ {\ln g\,\left( x \right)} \right] = \frac{{{g^,}\,\left( x \right)}}{{g\,\left( x \right)}} \hfill \\ {f^,}\,\left( x \right) = {e^{2x}}\,\left( {\frac{{x{e^x} + {e^x}}}{{x{e^x} + 1}}} \right) + 2{e^{2x}}\ln \,\left( {x{e^x} + 1} \right) \hfill \\ Multiplying \hfill \\ {f^,}\,\left( x \right) = \frac{{x{e^{3x}} + {e^{3x}}}}{{x{e^x} + 1}} + 2{e^{2x}}\ln \,\left( {x{e^x} + 1} \right) \hfill \\ \end{gathered} \]
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