Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.4 Exercises - Page 353: 67

Answer

$(3e^{-3x})(5+6x)$

Work Step by Step

$f(x)=(3+2x)e^{-3x}$ $u=3+2x$ $u’=2$ $v=e^{-3x}$ $v’=-3e^{-3x}$ $f’(x)=2e^{-3x}+(3+2x)(-3)(e^{-3x}$ $=2e^{-3x}+(-9-6x)e^{-3x}$ $f’(x)=e^{-3x}(-7-6x)$ $u=e^{-3x}$ $y’=-3e^{-3x}$ $v=-7-6x$ $v’=-6$ $f’’(x)=(-3)(e^{-3x})(-7-6x)+(e^{-3x})(-6)$ $=(e^{-3x})(21+18x-6)$ $=(3e^{-3x})(5+6x)$
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