# How do you find the derivative and double derivative of the equation f(x)=xe^(-x^2)?

Aug 31, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 {x}^{2} {e}^{- {x}^{2}} + {e}^{- {x}^{2}}$
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 2 x {e}^{- {x}^{2}} \left(2 {x}^{2} - 3 x\right)$

#### Explanation:

Given -

$y = x {e}^{- {x}^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = x . {e}^{- {x}^{2}} \left(- 2 x\right) + {e}^{- {x}^{2}} \left(1\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 {x}^{2} {e}^{- {x}^{2}} + {e}^{- {x}^{2}}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \left[- 2 {x}^{2.} {e}^{- {x}^{2}} \left(- 2 x\right) + {e}^{- {x}^{2}} \left(- 4 x\right)\right] + \left[{e}^{- {x}^{2}} \left(- 2 x\right)\right]$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 4 {x}^{3} {e}^{- {x}^{2}} - 4 x {e}^{- {x}^{2}} - 2 x {e}^{- {x}^{2}}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 4 {x}^{3} {e}^{- {x}^{2}} - 6 x {e}^{- {x}^{2}}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 2 x {e}^{- {x}^{2}} \left(2 {x}^{2} - 3 x\right)$