Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 4 - Calculating the Derivative - 4.5 Derivatives of Logarithmic Functions - 4.5 Exercises - Page 240: 39

Answer

$$\frac{{dw}}{{dp}} = \frac{{{2^p}}}{{3\left( {{2^p} - 1} \right)}}$$

Work Step by Step

$$\eqalign{ & w = {\log _8}\left( {{2^p} - 1} \right) \cr & {\text{differentiate with respect to }}p \cr & \frac{{dw}}{{dp}} = \frac{d}{{dp}}\left[ {{{\log }_8}\left( {{2^p} - 1} \right)} \right] \cr & {\text{use derivative}}{\log _a}x,\,\,\,\,\frac{d}{{dx}}\left[ {{{\log }_a}g\left( x \right)} \right] = \frac{1}{{\ln a}}\left( {\frac{{g'\left( x \right)}}{{g\left( x \right)}}} \right).{\text{ set }}x = p,{\text{ }}a = 8 \cr & \frac{{dw}}{{dp}} = \frac{1}{{\ln 8}}\left( {\frac{{\left( {{2^p} - 1} \right)'}}{{{2^p} - 1}}} \right) \cr & \frac{{dw}}{{dp}} = \frac{1}{{\ln 8}}\left( {\frac{{{2^p}\ln 2}}{{{2^p} - 1}}} \right) \cr & {\text{Simplifying}} \cr & \frac{{dw}}{{dp}} = \frac{{\ln 2}}{{3\ln 2}}\left( {\frac{{{2^p}}}{{{2^p} - 1}}} \right) \cr & \frac{{dw}}{{dp}} = \frac{{{2^p}}}{{3\left( {{2^p} - 1} \right)}} \cr} $$
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