Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Practice Exercises - Page 439: 30


$$\frac{{dy}}{{dx}} = \frac{{{{\left( {\ln x} \right)}^{1/\ln x}}\left( {1 - \ln \left( {\ln x} \right)} \right)}}{{x{{\left( {\ln x} \right)}^2}}}$$

Work Step by Step

$$\eqalign{ & y = {\left( {\ln x} \right)^{1/\ln x}} \cr & {\text{Take the natural logarithm on both sides}} \cr & \ln y = \ln {\left( {\ln x} \right)^{1/\ln x}} \cr & {\text{use power property for logarithms}} \cr & \ln y = \frac{1}{{\ln x}}\ln \left( {\ln x} \right) \cr & {\text{differentiate both sides with respect to }}x \cr & \frac{1}{y}\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {\frac{{\ln \left( {\ln x} \right)}}{{\ln x}}} \right) \cr & {\text{use quotient rule for derivatives}} \cr & \frac{1}{y}\frac{{dy}}{{dx}} = \frac{{\ln x\left( {\frac{1}{{x\ln x}}} \right) - \ln \left( {\ln x} \right)\left( {\frac{1}{x}} \right)}}{{{{\left( {\ln x} \right)}^2}}} \cr & \frac{1}{y}\frac{{dy}}{{dx}} = \frac{{\frac{1}{x} - \frac{1}{x}\ln \left( {\ln x} \right)}}{{{{\left( {\ln x} \right)}^2}}} \cr & \frac{1}{y}\frac{{dy}}{{dx}} = \frac{{1 - \ln \left( {\ln x} \right)}}{{x{{\left( {\ln x} \right)}^2}}} \cr & {\text{solve the equation for }}\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}} = \frac{{y\left( {1 - \ln \left( {\ln x} \right)} \right)}}{{x{{\left( {\ln x} \right)}^2}}} \cr & {\text{replace }}y = {\left( {\ln x} \right)^{1/\ln x}} \cr & \frac{{dy}}{{dx}} = \frac{{{{\left( {\ln x} \right)}^{1/\ln x}}\left( {1 - \ln \left( {\ln x} \right)} \right)}}{{x{{\left( {\ln x} \right)}^2}}} \cr} $$
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