# How do you differentiate y = e^(-2x + x^2)?

Jan 7, 2017

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

#### Explanation:

Using the chain rule, along with the derivatives $\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$ and $\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$, we have

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

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

$= {e}^{- 2 x + {x}^{2}} \left(- 2 + 2 x\right)$