Question

Find an equation for a line that is normal to the graph of y=xe^x and goes through the origin

Answer 1

Y = xe^x
dy/dx(e^x x)=>use the product rule, d/dx(u v) = v*(du)/(dx)+u*(dv)/(dx), where u = e^x and v = x:
= e^x (d/dx(x))+x (d/dx(e^x))
y' = e^x x+ e^x
y'(0) = 1 => slope of the tangent
slope of the normal = -1
y - 0 = -1(x - 0)
y = -x => normal at origin