Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 6 - Applications of Integration - 6.8 Logarithmic and Exponential - 6.8 Exercises: 11

Answer

$$ - \frac{5}{{x\ln {{\left( {2x} \right)}^6}}}$$

Work Step by Step

$$\eqalign{ & \frac{d}{{dx}}\left( {\ln {{\left( {2x} \right)}^{ - 5}}} \right) \cr & {\text{use the chain rule}} \cr & = - 5\ln {\left( {2x} \right)^{ - 6}}\frac{d}{{dx}}\left( {\ln \left( {2x} \right)} \right) \cr & {\text{use the rule }}\frac{d}{{dx}}\left( {\ln u} \right) = \frac{1}{u}\frac{{du}}{{dx}},{\text{ letting }}u = 2x \cr & = - 5\ln {\left( {2x} \right)^{ - 6}}\left( {\frac{2}{{2x}}} \right) \cr & {\text{simplify}} \cr & = - \frac{5}{{x\ln {{\left( {2x} \right)}^6}}} \cr} $$
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