Chapter 7 - Exponential Functions - 7.1 Derivative of f(x)=bx and the Number e - Exercises - Page 327: 18

$$ y=2ex-e=e(2x-1).$$

We have $$ y'=2xe^{x^2}\Longrightarrow y'(1)=2e.$$ Then the slope of the tangent line at $ x=1$ is $ m= 2e $. Hence the equation of the tangent line is given by $$ y=mx+c=2ex+c.$$ Since $ y(1)=e $, then $ c=-e $, hence the tangent line at $ x=1$ is given by $$ y=2ex-e=e(2x-1).$$