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, $F'(x)=\frac{3}{(3x-1)ln2}$
Calculus 8th Edition
by Stewart, James
Published by
Cengage
ISBN 10:
1285740629
ISBN 13:
978-1-28574-062-1
Chapter 6 - Inverse Functions - 6.4* General Logarithmic and Exponential Functions - 6.4* Exercise - Page 464: 31
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