Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.7 Hyperbolic Functions - 6.7 Exercises - Page 490: 47

Answer

$\DeclareMathOperator{\sech}{sech}$$\dfrac{d}{dx}\arctan(\tanh x)=\sech{2x}$

Work Step by Step

Firstly solve the left hand side. Thus, $\DeclareMathOperator{\sech}{sech}$$\dfrac{d}{dx}\arctan(\tanh x)=\dfrac{1}{1+\tanh^2 x}\times\sech^2 x$ As we know that $\sech^2 x=1-\tanh^2 x$ Thus, $=\dfrac{1-\tanh^2 x}{1+\tanh^2 x}$ or, $=\dfrac{\cosh^2 x-\sinh^2 x}{\cosh^2x+\sinh^2 x}$ Also, $\cosh^2 x-\sinh^2 x=1$ and $\cosh(2x)=\cosh^2 x+\sinh^2 x$ Therefore, $=\dfrac{1}{\cosh {2x}}$ $=\sech{2x}$ Hence, $\dfrac{d}{dx}\arctan(\tanh x)=\sech{2x}$
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