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: 59

Answer

$y=\frac{x}{e}-\frac{1}{e}$

Work Step by Step

$f(x)=e^{-x}ln(x)$ $(1,0)$ $u=e^{-x}$ $u’=-e^{-x}$ $v=ln(x)$ $v’=\frac{1}{x}$ $f’(x)=(-e^{-x})(ln(x))+(e^{-x})(\frac{1}{x})$ $=e^{-x}(-ln(x)+\frac{1}{x})$ $f’(1)=M=e^{-1}(-ln(1)+1)$ $=\frac{1}{e}$ Equation of Tangent: $y-0=\frac{1}{e}(x-1)$ $y=\frac{x}{e}-\frac{1}{e}$
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