## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 7: Transcendental Functions - Section 7.2 - Natural Logarithms - Exercises 7.2 - Page 381: 6

#### Answer

$$\frac{1}{x}$$

#### Work Step by Step

\eqalign{ & y = \ln kx,{\text{ where }}k{\text{ is a constant}} \cr & {\text{Find the derivative of }}y{\text{ with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\ln kx} \right] \cr & {\text{use the formula }}\frac{d}{{dx}}\ln u = \frac{1}{u}\frac{{du}}{{dx}}{\text{, }}u > 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}{{kx}}\frac{d}{{dx}}\left[ {kx} \right] \cr & {\text{solve the derivative and simplify}} \cr & \frac{{dy}}{{dx}} = \frac{1}{{kx}}\left( k \right) \cr & \frac{{dy}}{{dx}} = \frac{1}{x} \cr}

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