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 - Page 265: 47


LHS $=\frac{d}{dx}\arctan(\tanh x)$ $=\sech {2x}$ $=$RHS (as needed)

Work Step by Step

LHS $=\frac{d}{dx}\arctan(\tanh x)$ Using the chain rule: LHS $=\frac{d\arctan(\tanh x)}{d\tanh x} \times\frac{d\tanh x}{dx}$ $\DeclareMathOperator{\sech}{sech}$ $=\frac{1}{1+\tanh^2 x}\times\sech^2 x$ Recall that: $$\sech^2 x=1-\tanh^2 x$$ LHS $=\frac{1-\tanh^2 x}{1+\tanh^2 x}$ Multiply the numerator and denominator by $\cosh^2x$: LHS $=\frac{\cosh^2 x-\sinh^2 x}{\cosh^2x+\sinh^2 x}$ Recall that: 1) $$\cosh^2 x-\sinh^2 x=1$$ 2) $$\cosh(2x)=\cosh^2 x+\sinh^2 x$$ (derived from the compound formula for $\cosh$) Thus, LHS $=\frac{1}{\cosh {2x}}$ $=\sech{2x}$ $=$RHS (as needed)
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.