Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 374: 34

Answer

$$ y'= \frac{1}{3(1+(x/3)^2)}.$$

Work Step by Step

Since $ y=arctan(x/3)=\tan^{-1}(x/3)$, then $$ y'= \frac{1}{1+(x/3)^2}(x/3)'=\frac{1}{3(1+(x/3)^2)}.$$
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