Thomas' Calculus 13th Edition

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

Chapter 7: Transcendental Functions - Section 7.2 - Natural Logarithms - Exercises 7.2 - Page 381: 5



Work Step by Step

$$\eqalign{ & y = \ln 3x \cr & {\text{Find the derivative of }}y{\text{ with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\ln 3x} \right] \cr & {\text{use the formula }}\cr & \frac{d}{{dx}}\ln u = \frac{1}{u}\frac{{du}}{{dx}}{\text{, }}u \gt 0\cr & {\text{ where }}u{\text{ is any differentiable function of }}x \cr & {\text{For this exercise you can note that }}u = 3x;{\text{ then}} \cr & \frac{{dy}}{{dx}} = \frac{1}{{3x}}\frac{d}{{dx}}\left[ {3x} \right] \cr & {\text{solve the derivative and simplify}} \cr & \frac{{dy}}{{dx}} = \frac{1}{{3x}}\left( 3 \right) \cr & \frac{{dy}}{{dx}} = \frac{1}{x} \cr} $$
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