## Calculus 8th Edition

$(2x+1)e^{-1/x}$
$\frac{d}{dx}x^2e^{-1/x}$ Use chain rule: $=x^2\frac{d}{dx}e^{-1/x}+e^{-1/x}\frac{d}{dx}x^2$ Use chain rule again to evaluate $\frac{d}{dx}e^{-1/x}$: $=x^2e^{-1/x}\frac{d}{dx}(-\frac{1}{x})+e^{-1/x}*2x$ $=x^2e^{-1/x}\frac{d}{dx}(-x^{-1})+e^{-1/x}*2x$ $=x^2e^{-1/x}x^{-2}+e^{-1/x}*2x$ $=e^{-1/x}+e^{-1/x}*2x$ $=(2x+1)e^{-1/x}$