Calculus (3rd Edition)

$$-e^{2}$$
Given $$f(x) =xe^{-x}$$ Since \begin{align*} f'(x) &= \frac{d}{dx}\left(x\right)e^{-x}+\frac{d}{dx}\left(e^{-x}\right)x\\ &= e^{-x}-e^{-x}x\\ &=e^{-x}(1-x) \end{align*} Since $f^{\prime}(x)<0$ for $x>1$, then the function is strictly decreasing and one-to-one. Hence, the inverse exists for $f(x)$ on $[1, \infty)$ Consider that $g$ is the inverse of $f$. Then $$f(2)=2e^{-2}\ \ \ \ \to\ \ g( 2e^{-2}) =2$$ and \begin{align*} g^{\prime}\left(2 e^{-2}\right)&=\frac{1}{f^{\prime}\left(g\left(2 e^{-2}\right)\right)}\\ &=\frac{1}{f^{\prime}(2)}\\ &=\frac{1}{e^{-2}(1-2)}=-e^{2} \end{align*}