Calculus with Applications (10th Edition)

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

Chapter 4 - Calculating the Derivative - 4.4 Derivatives of Exponential Functions - 4.4 Exercises: 17

Answer

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

Work Step by Step

\[\begin{gathered} y = \frac{{{x^2}}}{{{e^x}}} \hfill \\ Find\,\,the\,\,derivative\,\,using\,\,the\,\,quotient\,\,rule \hfill \\ \,\,{\left[ {\frac{u}{v}} \right]^,} = \frac{{v{u^,} - u{v^,}}}{{{v^2}}} \hfill \\ Then \hfill \\ {y^,} = \frac{{{e^x}\,{{\left( {{x^2}} \right)}^,} - {x^2}\,{{\left( {{e^x}} \right)}^,}}}{{\,{{\left( {{e^x}} \right)}^2}}} \hfill \\ {y^,} = \frac{{{e^x}\,\left( {2x} \right) - {x^2}\,\left( {{e^x}} \right)}}{{{e^{2x}}}} \hfill \\ Multiply \hfill \\ {y^,} = \frac{{2x{e^x} - {x^2}{e^x}}}{{{e^{2x}}}} \hfill \\ Factor\,\,the\,\,numerator \hfill \\ {y^,} = \,\frac{{x{e^x}\,\left( {2 - x} \right)}}{{{e^{2x}}}} \hfill \\ Cancel\,\,{e^x} \hfill \\ {y^,} = \frac{{x\,\left( {2 - x} \right)}}{{{e^x}}} \hfill \\ \end{gathered} \]
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