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: 39

Answer

$\DeclareMathOperator{\csch}{csch}$ $$g'(t)=\coth{\sqrt{t^2+1}}-\frac{t^2\csch^2{\sqrt{t^2+1}}}{\sqrt{t^2+1}}$$

Work Step by Step

$g'(t)=\frac{d}{dt}t\coth{\sqrt{t^2+1}}$ Using the product rule: $g'(t)=1\times\coth{\sqrt{t^2+1}}+t\times\frac{d}{dt}\coth{\sqrt{t^2+1}}$ Using the chain rule: $g'(t)=\coth{\sqrt{t^2+1}}+t(\frac{d\coth{\sqrt{t^2+1}}}{d\sqrt{t^2+1}} \times \frac{d\sqrt{t^2+1}}{dt^2+1} \times \frac{dt^2+1}{dt})$ $\DeclareMathOperator{\csch}{csch}$ $=\coth{\sqrt{t^2+1}}+t(-\csch^2{\sqrt{t^2+1}} \times\frac{1}{2\sqrt{t^2+1}} \times (2t)$ $=\coth{\sqrt{t^2+1}}-\frac{t^2\csch^2{\sqrt{t^2+1}}}{\sqrt{t^2+1}}$
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