Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.4 Exercises - Page 352: 61

$y=ex$

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