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

##### 1 Answer
Dec 22, 2014

Answer :

$y ' = {e}^{- k x} \left(1 - k \cdot x\right)$

Solution :

Suppose :

$y = f \left(x\right) \cdot g \left(x\right)$

Using Product Rule which is,

$y ' = f \left(x\right) \cdot g ' \left(x\right) + f ' \left(x\right) \cdot g \left(x\right)$

Similarly following for the given problem,

$y = x \cdot {e}^{- k x}$

Differentiating with respect to $x$,

$y ' = x \cdot \left({e}^{- k x}\right) ' + {e}^{- k x} \cdot \left(x\right) '$

$y ' = x \cdot \left(- k \cdot {e}^{- k x}\right) + {e}^{- k x}$

$y ' = {e}^{- k x} \left(1 - k \cdot x\right)$