We use the product rule to find the derivative.
f ( x ) = ( x + 2 ) ( x + 1 ) e − x {\displaystyle f(x)=(x+2)(x+1)e^{-x}}
then
f ′ ( x ) = ( x + 2 ) ′ ( x + 1 ) e − x + ( x + 2 ) ( x + 1 ) ′ e − x + ( x + 2 ) ( x + 1 ) ( e − x ) ′ = 1 ⋅ ( x + 1 ) e − x + ( x + 2 ) ⋅ 1 ⋅ e − x + ( x + 2 ) ( x + 1 ) ( − e − x ) = ( x + 1 ) e − x + ( x + 2 ) e − x − ( x + 2 ) ( x + 1 ) e − x = e − x ( ( x + 1 ) + ( x + 2 ) − ( x 2 + 3 x + 2 ) ) = e − x ( x + 1 + x + 2 − x 2 − 3 x − 2 ) = e − x ( − x 2 − x + 1 ) = − e − x ( x 2 + x − 1 ) {\displaystyle {\begin{aligned}f'(x)&=(x+2)'(x+1)e^{-x}+(x+2)(x+1)'e^{-x}+(x+2)(x+1)(e^{-x})'\\&=1\cdot (x+1)e^{-x}+(x+2)\cdot 1\cdot e^{-x}+(x+2)(x+1)(-e^{-x})\\&=(x+1)e^{-x}+(x+2)e^{-x}-(x+2)(x+1)e^{-x}\\&=e^{-x}\left((x+1)+(x+2)-(x^{2}+3x+2)\right)\\&=e^{-x}\left(x+1+x+2-x^{2}-3x-2\right)\\&=e^{-x}\left(-x^{2}-x+1\right)\\&=-e^{-x}\left(x^{2}+x-1\right)\\\end{aligned}}}
f ′ ( x ) = − e − x ( x 2 + x − 1 ) {\displaystyle \color {blue}f'(x)=-e^{-x}\left(x^{2}+x-1\right)}
