Determine the equation of the tangent line to y=x^2 (1+In x) at the point where x = e.

Answer 1

y(e) = e^2 (1+1) = 2e^2
y'(x) = x(3+2lnx)
y'(e) = e(3+2) = 5e
so the line is
y - 2e^2 = 5e(x-e)

see the graphs at

https://www.wolframalpha.com/input/?i=plot+y%3Dx%5E2%281%2Blnx%29%2C+y%3D5e%28x-e%29%2B2e%5E2+for+-2+%3C%3D+x+%3C%3D+4