Answer:
y=x
Step-by-step 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
