Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.11 - Hyperbolic Functions - 3.11 Exercises: 31

Answer

$f'(x)=\frac{\sec^2 {h}{\sqrt{x}}}{2\sqrt{x}}$

Work Step by Step

$f'(x)=\frac{d}{dx}\tanh {\sqrt{x}}$ Using the chain rule: $f'(x)=\frac{d\tanh{\sqrt{x}}}{d(\sqrt{x})} \times\frac{d(\sqrt{x})}{dx}$ $\DeclareMathOperator{\sech}{sech}$ $=(\sech)^2 {\sqrt{x}} \times\frac{1}{2\sqrt{x}}$ $=\frac{(\sech)^2{\sqrt{x}}}{2\sqrt{x}}$
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