Respuesta :

[tex]y=11xe^x\implies y'=11e^x(x+1)[/tex]

When [tex]x=0[/tex], the slope of the tangent line is [tex]11e^0(0+1)=11[/tex]. The equation of the tangent line in point-slope form is then

[tex]y-0=11(x-0)\implies y=11x[/tex]