Question

Find an equation of the tangent line to the curve
`y=xe^{-x^(2) }` at the point (0,0).

Answer 1

y=x

Explanation:

We must find the derivative (the formula for the slopes of the tangent lines at any x).

We will need product rule:

y'=(1)e^(-x^2)+x(-2x)e^(-x^2)

y'=e^(-x^2)-2x^2e^(-x^2)

At x=0 we get the slope is e^(-0^2)-2(0)^2e^(-0^2)

=1-0

=1.

So the line going through (0,0) with slope 1 is

y-0=1(x-0)

Or

y=x