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: 16

Answer

\[{y^,} = - 15{x^3}{e^{ - 5x}} + 20x{e^{ - 5x}} + 9{x^2}{e^{ - 5x}} - 4{e^{ - 5x}}\]

Work Step by Step

\[\begin{gathered} y = \,\left( {3{x^3} - 4x} \right){e^{ - 5x}} \hfill \\ Find\,\,the\,\,derivative\,\,using\,\,the\,\,product\,\,rule \hfill \\ {y^,} = \,\left( {3{x^3} - 4x} \right)\,{\left( {{e^{ - 5x}}} \right)^,} + \left( {{e^{ - 5x}}} \right)\left( {3{x^3} - 4x} \right){\,^,} \hfill \\ Use\,\,\,\,\frac{d}{{dx}}\,\,\left[ {{e^{g\,\left( x \right)}}} \right] = {e^{g\,\left( x \right)}}{g^,}\,\left( x \right) \hfill \\ Then \hfill \\ {y^,} = \left( {3{x^3} - 4x} \right)\,\left( { - 5{e^{ - 5x}}} \right) + {e^{ - 5x}}\,\left( {9{x^2} - 4} \right) \hfill \\ Simplify\,\,by\,\,multiplying\, \hfill \\ {y^,} = - 15{x^3}{e^{ - 5x}} + 20x{e^{ - 5x}} + 9{x^2}{e^{ - 5x}} - 4{e^{ - 5x}} \hfill \\ \end{gathered} \]
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