Answer
$$y=x$$
Work Step by Step
Finding the derivative of the curve:
$y'=\frac{d[xe^{-x^2}]}{dx}$
Using the product rule:
$y'=e^{-x^2}+x\frac{d}{dx}[e^{-x^2}]$
Using the chain rule:
$y'=e^{-x^2}+x[\frac{d[e^{-x^2}]}{d[-x^2]}\times\frac{d[-x^2]}{dx}]$
$y'=e^{-x^2}+xe^{-x^2}\times-2x=e^{-x^2}(1-2x^2)$
Thus, the slope ($m$) of the tangent line at (0,0) is:
$m=y'(0)$
$=e^{-(0)^2}(1-2(0)^2)=1\times(1-0)=1$
From the point-slope form of a linear equation:
$y-0=1(x-0)$
Thus, an equation of the tangent line would be:
$y=x$
