Answer
$y=ex-e$
Work Step by Step
$y=xe^x-e^x$ $(1,0)$
$=e^x(x-1)$
$u=e^x$
$u’=e^x$
$v=x-1$
$v’=1$
$y’=(e^x)(x-1)+(e^)(1)$
$=(e^x)(x-1+1)$
$=(e^x)(x)$
$y’(1)=e^1(1)$
$=e$
Equation of tangent:
$y-0=e(x-1)$
$y=ex-e$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
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: 62
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