# How do you use the Product Rule to find the derivative of f(x) = (x^2 + 2x +3)e^-x?

Aug 9, 2015

${f}^{'} = - {e}^{- x} \cdot \left({x}^{2} + 1\right)$

#### Explanation:

Notice that you can write your function as

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

where $g \left(x\right) = \left({x}^{2} + 2 x + 3\right)$ and $h \left(x\right) = {e}^{- x}$. The product rule allows you to differentiate functions that can be written as the the product of two other functions by using the formula

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \left[\frac{d}{\mathrm{dx}} \left(g \left(x\right)\right)\right] \cdot h \left(x\right) + g \left(x\right) \cdot \frac{d}{\mathrm{dx}} \left(h \left(x\right)\right)}$

If you take into account the fact that

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

you can write

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \left[\frac{d}{\mathrm{dx}} \left({x}^{2} + 2 x + 3\right)\right] \cdot {e}^{- x} + \left({x}^{2} + 2 x + 3\right) \cdot \frac{d}{\mathrm{dx}} \left({e}^{- x}\right)$

${f}^{'} = \left(2 x + 2\right) \cdot {e}^{- x} + \left({x}^{2} + 2 x + 3\right) \cdot \left(- {e}^{- x}\right)$

${f}^{'} = {e}^{- x} \cdot \left(\textcolor{red}{\cancel{\textcolor{b l a c k}{2 x}}} + 2 - {x}^{2} - \textcolor{red}{\cancel{\textcolor{b l a c k}{2 x}}} - 3\right)$

${f}^{'} = {e}^{- x} \cdot \left(- {x}^{2} - 1\right)$

${f}^{'} = \textcolor{g r e e n}{- {e}^{- x} \cdot \left({x}^{2} + 1\right)}$