How do you find the equation of the tangent line to the curve y=x^2e^-x at (1, 1/e)?

1 Answer
Apr 20, 2017

y=x/e

Explanation:

y=x^2e^{-x}

By Product Rule,

y'=2x cdot e^{-x}+x^2 cdot(-e^{-x})=(2-x)xe^{-x}

To find an equation of a line, we need two things:

  1. Point: (x_1,y_1)=(1,1/e)
  2. Slope: m=y'(1)=(2-1)cdot1cdot e^{-1}=1/e

By Point-Slope Form: y-y_1=m(x-x_1),

y-1/e=1/e(x-1)=x/e-1/e

By adding 1/eto both sides,

y=x/e

I hope that this was clear.