Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.4 Exercises - Page 352: 65

Answer

$y=(-e-1)x+1$

Work Step by Step

$xe^y+ye^x=1$ $(0,1)$ $e^y+xe^y\frac{dy}{dx}+\frac{dy}{dx}e^x+ye^x=0$ $e^y+ye^x+\frac{dy}{dx}(xe^y+e^x)=0$ $\frac{dy}{dx}=\frac{-(e^y+ye^x)}{xe^y+e^x}$ $\frac{dy}{dx}(0)=-(e+1)$ Equation of Tangent: $y-1=-(e+1)(x-0)$ $y-1=(-e-1)x$ $y=(-e-1)x+1$
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