Calculus: Early Transcendentals 8th Edition

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

Chapter 10 - Section 10.1 - Curves Defined by Parametric Equations - 10.1 Exercises - Page 645: 14

Answer

$y=e^{-2\ln\left(x\right)}=e^{\ln\left(x^{-2}\right)}=x^{-2}={\frac{1}{x^{2}}}$

Work Step by Step

a) $\ln(x)=t$ $y=e^{-2\ln\left(x\right)}=e^{\ln\left(x^{-2}\right)}=x^{-2}={\frac{1}{x^{2}}}$
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