Answer
$$ y'= \frac{1}{3(1+(x/3)^2)}.$$
Work Step by Step
Since $ y=arctan(x/3)=\tan^{-1}(x/3)$, then
$$ y'= \frac{1}{1+(x/3)^2}(x/3)'=\frac{1}{3(1+(x/3)^2)}.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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