May be the first derivative at stationary point is zero.
Product rule? Just in case a picture helps...
Or, zooming in on the chain rule...
... where (key in spoiler) ...
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Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
From Tom's differentiation we have [tex]f'(x) = 2xe^{-kx}-kx^2e^{-kx} = 0 \Rightarrow e^{-kx}\left(2x-kx^2\right) = 0[/tex].
Since for any real number , we have . We know that ,
so . You can find easily now as it's just .