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how do you find the derivative of xe^-2x?
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use the product rule and the chain rule
\[(fg)'=f'g+g'f\] with \[f(x)=x, f'(x)=1,g(x)=e^{-2x}, g'(x)=-2e^{-2x}\]
\[g'\] was found by the chain rule
so the answer is e^-2x + -2e^-2x?
no
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\[\huge e^{-2x} -2\color{red}xe^{-2x}\]
great thanks!
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