Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 386: 32


$$ G'(s)=\frac{1}{2\sqrt s(1+ s)}.$$

Work Step by Step

Recall that $(\tan^{-1} x)'=\dfrac{1}{1+x^2}$ Since $ G(s)=\tan^{-1}(\sqrt s)$, then the derivative, using the chain rule, is given by $$ G'(s)=\frac{1}{1+(\sqrt s)^2}(\sqrt s)'=\frac{1}{2\sqrt s(1+ s)}.$$
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