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: 66

Answer

\[ = \frac{{{e^t}\,\left( {{t^2} + t + 1} \right)}}{{\,{{\left( {t + 1} \right)}^2}}}\]

Work Step by Step

\[\begin{gathered} y = \frac{{t{e^t}}}{{t + 1}} \hfill \\ \hfill \\ {y^,} = \frac{d}{{dt}}\,\left( {\frac{{t{e^t}}}{{t + 1}}} \right) \hfill \\ \hfill \\ use\,\,quotient\,\,rule \hfill \\ \hfill \\ {y^,} = \frac{{\,{{\left( {t{e^t}} \right)}^,}\,\left( {t + 1} \right) - t{e^t}\,{{\left( {t + 1} \right)}^,}}}{{\,{{\left( {t + 1} \right)}^2}}} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ {y^,} = \frac{{\,\left( {{t^,}{e^t} + t\,{{\left( {{e^t}} \right)}^,}} \right)\,\left( {t + 1} \right) - t{e^t}}}{{\,{{\left( {t + 1} \right)}^2}}} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ {y^,} = \frac{{\,\left( {{e^t} + t{e^t}} \right)\,\left( {t + 1} \right) - t{e^t}}}{{\,{{\left( {t + 1} \right)}^2}}} \hfill \\ \hfill \\ multiply\,\,and\,\,simplify \hfill \\ \hfill \\ \hfill \\ {y^,} = \frac{{t{e^t} + {e^t} + {t^2}{e^t} + t{e^t} - t{e^t}}}{{\,{{\left( {t + 1} \right)}^2}}} \hfill \\ \hfill \\ = \frac{{{e^t}\,\left( {{t^2} + t + 1} \right)}}{{\,{{\left( {t + 1} \right)}^2}}} \hfill \\ \end{gathered} \]
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