## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 4 - Calculating the Derivative - 4.4 Derivatives of Exponential Functions - 4.4 Exercises - Page 232: 14

#### Answer

${y^{^,}} = - 2{x^2}{e^{ - 2x}} + 2x{e^{ - 2x}}$

#### Work Step by Step

$\begin{gathered} y = {x^2}{e^{ - 2x}} \hfill \\ Use\,\,the\,\,product\,\,rule\,\,to\,\,find\,\,the\,\,derivative \hfill \\ {y^,} = {x^2}\,{\left( {{e^{ - 2x}}} \right)^,} + \,\left( {{e^{ - 2x}}} \right)\,\left( {{x^2}} \right) \hfill \\ and \hfill \\ \frac{d}{{dx}}\,\,\left[ {{e^{g\,\left( x \right)}}} \right] = {e^{g\,\left( x \right)}}{g^,}\,\left( x \right) \hfill \\ Then \hfill \\ {y^,} = {x^2}\,\left( {{e^{ - 2x}}} \right)\,\left( { - 2} \right) + \,\,\left( {{e^{ - 2x}}} \right)\left( {2x} \right) \hfill \\ Multiplying\, \hfill \\ {y^{^,}} = - 2{x^2}{e^{ - 2x}} + 2x{e^{ - 2x}} \hfill \\ \end{gathered}$

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