## Calculus: Early Transcendentals 8th Edition

$$y=x$$
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: $y=x$