Answer
$f'(x)=4^{-3x}(1-3\ln(4)x)$
Work Step by Step
$f(x)=x(4^{-3x})$
$f'(x)=x'(4^{-3x})+(4^{-3x})'x$
$f'(x)=4^{-3x}+4^{-3x}(-3x)'(ln(4))x$
$f'(x)=4^{-3x}+4^{-3x}(-3)(ln(4))x$
Or, $f'(x)=4^{-3x}(1-3\ln(4)x)$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
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 - Review Exercises - Page 394: 62
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