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 - Preliminary Questions - Page 374: 5



Work Step by Step

If we assume that $x=4u$, then we have $$\sqrt{16+x^2}=\sqrt{16+16u^2}=4\sqrt{1+u^2}.$$ Hence, the relation between $x$ and $u$ is $x=4u$.
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