Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.9 Derivatives of Logarithmic and Exponential Functions - 3.9 Exercises - Page 210: 3

Answer

\[\frac{d}{dx}\left[ \ln kx \right]=\frac{d}{dx}\left[ \ln x \right]\]

Work Step by Step

\[\begin{align} & \frac{d}{dx}\left[ \ln kx \right],\text{ where }x>0 \\ & \text{Apply the product property for logarithms ln}\left( ab \right)=\ln a+\ln b \\ & \frac{d}{dx}\left[ \ln kx \right]=\frac{d}{dx}\left[ \ln k+\ln x \right] \\ & \text{The sum property for derivatives} \\ & \frac{d}{dx}\left[ \ln kx \right]=\frac{d}{dx}\left[ \ln k \right]+\frac{d}{dx}\left[ \ln x \right] \\ & \text{Where the derivative of a constant is 0, then} \\ & \frac{d}{dx}\left[ \ln kx \right]=0+\frac{d}{dx}\left[ \ln x \right] \\ & \frac{d}{dx}\left[ \ln kx \right]=\frac{d}{dx}\left[ \ln x \right] \\ \end{align}\]
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