## Calculus 8th Edition

Published by Cengage

# Chapter 6 - Inverse Functions - 6.4* General Logarithmic and Exponential Functions - 6.4* Exercise: 31

#### Answer

$F'(x)=\frac{3}{(3x-1)ln2}$

#### Work Step by Step

$\frac{d}{dx}F(x)=\frac{d}{dx}[log_{2}(1-3x)]$ $=\frac{1}{(1-3x)ln2}\frac{d}{dx}(1-3x)$ $=\frac{1}{(1-3x)ln2}(-3)$ $=\frac{-3}{(-(3x-1))ln2}$ Hence, $F'(x)=\frac{3}{(3x-1)ln2}$

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