## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 7: Transcendental Functions - Section 7.2 - Natural Logarithms - Exercises 7.2 - Page 381: 19

#### Answer

$$\frac{1}{{{t^2}}}\left( {1 - \ln t} \right)$$

#### Work Step by Step

\eqalign{ & y = \frac{{\ln t}}{t} \cr & {\text{use }}\frac{1}{t} = {t^{ - 1}} \cr & {\text{Find the derivative of }}y{\text{ with respect to }}t \cr & y = {t^{ - 1}}\ln t \cr & {\text{use the product rule for derivatives}} \cr & \frac{{dy}}{{dt}} = {t^{ - 1}}\frac{d}{{dt}}\left[ {\ln t} \right] + \ln t\frac{d}{{dt}}\left[ {{t^{ - 1}}} \right] \cr & {\text{solve the derivatives}} \cr & \frac{{dy}}{{dt}} = {t^{ - 1}}\left( {\frac{1}{t}} \right) + \ln t\left( { - {t^{ - 2}}} \right) \cr & {\text{simplify}} \cr & \frac{{dy}}{{dt}} = {t^{ - 2}} - {t^{ - 2}}\ln t \cr & \frac{{dy}}{{dt}} = {t^{ - 2}}\left( {1 - \ln t} \right) \cr & \frac{{dy}}{{dt}} = \frac{1}{{{t^2}}}\left( {1 - \ln t} \right) \cr}

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