Just in case a picture helps...
... where (key in spoiler) ...
Spoiler:
__________________________________________________ __________
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
You're going to have to use the product rule, chain rule, and also the derivative of an exponential function.
Assuming you're differentiating with respect to x, and k is a constant: y' = (x)' (e^-kx) + x(e^-kx)' . Think you can get it now?
$\displaystyle \displaystyle \begin{align*} y &= x\,e^{-k\,x} \\ \frac{d}{dx} \left( y \right) &= \frac{d}{dx} \left( x\, e^{-k\,x} \right) \\ \frac{dy}{dx} &= x \, \frac{d}{dx} \left( e^{-k\, x} \right) + e^{-k\, x}\,\frac{d}{dx}\left( x \right) \\ \frac{dy}{dx} &= x\left( -k \, e^{-k\, x} \right) + e^{-k\, x} \left( 1 \right) \\ \frac{dy}{dx} &= -k\, x \, e^{-k\, x} + e^{-k \, x} \\ \frac{dy}{dx} &= e^{-k\, x} \left( -k\, x + 1 \right) \end{align*}$