Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.7 - Hyperbolic Functions - Exercises 7.7 - Page 431: 76

Answer

See derivation below. The minus sign yields a negative value for $e^{y}$, so it is ommited from being a part of the solution.

Work Step by Step

$\sinh^{-1}x=y\quad$ means that $x=\sinh y$ By definition of sinh: $ x=\displaystyle \frac{e^{y}-e^{-y}}{2}\quad $... we solve this for y $2x=e^{y}-\displaystyle \frac{1}{e^{y}}\qquad $... substitute $t=e^{y} $ (we see that $t\gt 0)$ $ 0=t-\displaystyle \frac{1}{t}-2x\qquad$ ... multiply with t $t^{2}-(2x)t-1=0\qquad $... apply the quadratic formula, $a=1, b=-2x, c=-1$ $t=\displaystyle \frac{2x\pm\sqrt{4x^{2}+4}}{2}=\frac{2x\pm 2\sqrt{x^{2}+1}}{2}=x\pm\sqrt{x^{2}+1}$ $\sqrt{x^{2}+1}\gt\sqrt{x^{2}}\quad$ so we will discard the minus, because it yields a negative $t$ ($t$ needs to be positive). $t=x+\sqrt{x^{2}+1}$ $e^{y}=x+\sqrt{x^{2}+1}\qquad $... apply ln(..) to both sides $y=\ln(x+\sqrt{x^{2}+1})$
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