University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Practice Exercises - Page 202: 48



Work Step by Step

$y=\log_{5}{({3x-7})}=\frac{\ln{({3x-7})}}{ln5}$ On differentiating both sides: $\frac{dy}{dx}=\frac{1}{ln{5}}\frac{d(\ln{({3x-7})})}{dx}$ $\frac{dy}{dx}=\frac{1}{\ln{5}}\frac{1}{(3x-7)}\frac{d(3x-7)} {dx}$ $\frac{dy}{dx}=\frac{1}{\ln{5}}\frac{3}{3x-7}$
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