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: 61

Answer

$y=ex$

Work Step by Step

$y=x^2e^x-2xe^x+2e^x$ $(1,e)$ $=e^x(x^2-2x+2)$ $u=e^x$ $u’=e^x$ $v=x^2-2x+2$ $v’=2x-2$ $y’=(e^x)(x^2-2x+2)+(e^x)(2x-2)$ $=(e^x)(x^2)$ $y’(1)=e^1(1^2)=e$ Equation of Tangent: $y-e=e(x-1)$ $y=ex-e+e$ $y=ex$
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