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: 41

Answer

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

Work Step by Step

\[\begin{gathered} y = \frac{{x{e^x}}}{{\ln \,\left( {{x^2} - 1} \right)}} \hfill \\ Find\,\,the\,\,derivative\,\,of\,\,the\,\,\,function \hfill \\ {y^,} = \,\,\left[ {\frac{{x{e^x}}}{{\ln \,\left( {{x^2} - 1} \right)}}} \right] \hfill \\ Use\,\,the\,\,quotient\,\,rule \hfill \\ {y^,} = \frac{{\,\left( {\ln \,\left( {{x^2} - 1} \right)} \right)\,{{\left( {x{e^x}} \right)}^,} - x{e^x}\,{{\left( {\ln \,\left( {{x^2} - 1} \right)} \right)}^,}}}{{\,{{\left( {\ln \,\left( {{x^2} - 1} \right)} \right)}^2}}} \hfill \\ Use\,\,the\,\,product\,\,rule\,\,for\,\,\,{\left( {x{e^x}} \right)^,} \hfill \\ {y^,} = \frac{{\ln \,\left( {{x^2} - 1} \right)\,\left( {x{e^x} + {e^x}} \right) - x{e^x}\,\left( {\frac{{2x}}{{{x^2} - 1}}} \right)}}{{\,\,{{\left[ {\ln \,\left( {{x^2} - 1} \right)} \right]}^{^2}}}} \hfill \\ Simplifying \hfill \\ {y^,} = \frac{{\,\,\left( {{x^2} - 1} \right)\ln \,\left( {{x^2} - 1} \right)\,\left( {x{e^x} + {e^x}} \right) - 2{x^2}{e^x}}}{{\,\,\,\,\left( {{x^2} - 1} \right){{\left[ {\ln \,\left( {{x^2} - 1} \right)} \right]}^2}}} \hfill \\ \hfill \\ \end{gathered} \]
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