Derivatives of exponential functions and a stationary point

Statement

• (a) Differentiate $f(x)=e^{2x}(x-1)$.
• (b) Differentiate $g(x)=x\,e^{-x}$ and find its stationary point, stating its nature.

Solution

(a) Product rule: $f'(x)=2e^{2x}(x-1)+e^{2x}=e^{2x}(2x-1)$.

(b) $g'(x)=e^{-x}-x\,e^{-x}=e^{-x}(1-x)$. It vanishes at $x=1$. Since $g'>0$ for $x<1$ and $g'<0$ for $x>1$, it is a maximum; $g(1)=e^{-1}\approx 0{.}368$.

$\boxed{f'(x)=e^{2x}(2x-1);\quad \text{max at }(1,\ e^{-1})\approx(1,\ 0{.}368)}$