# What is the derivative of xe^(-kx)?

Assuming $k$ as a constant, we can derivate this expression using product rule, which states that, be $y = f \left(x\right) g \left(x\right)$, then $y ' = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$.
Naming $f \left(x\right) = x$ and $g \left(x\right) = {e}^{- k x}$, we have that the derivates of these are given by: $f ' \left(x\right) = 1$ and $g ' \left(x\right) = - k {e}^{- k x}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 1 \cdot {e}^{- k x} + x \left(- k {e}^{- k x}\right) = \textcolor{g r e e n}{{e}^{- k x} \left(1 - k x\right)}$