# How do you differentiate f(x) = xe^(-2x^2) using the product rule?

Nov 14, 2015

$f ' \left(x\right) = \frac{1 - 4 {x}^{2}}{{e}^{2 {x}^{2}}}$

#### Explanation:

We have two terms: $x$ and ${e}^{- 2 {x}^{2}}$. The product rule states that the derivative of the product of the two will equal the following:
$f ' \left(x\right) = \textcolor{red}{\frac{d}{\mathrm{dx}} \left[x\right] \cdot {e}^{- 2 {x}^{2}}} + \textcolor{b l u e}{x \cdot \frac{d}{\mathrm{dx}} \left[{e}^{- 2 {x}^{2}}\right]}$

Let's take a moment to calculate both of the derivatives inside of our product rule statement.
$\textcolor{red}{\frac{d}{\mathrm{dx}} \left[x\right] = 1}$
The next derivative will require the chain rule. Also, remember that the derivative of ${e}^{x}$ is ${e}^{x}$.
color(blue)(d/(dx)[e^(-2x^2)]=d/(dx)[-2x^2]*e^(-2x^2)=-4x(e^(-2x^2))

Now that we know what both our derivatives are equal to, let's plug them back into our equation.

$f ' \left(x\right) = \textcolor{red}{1 \cdot {e}^{- 2 {x}^{2}}} + \textcolor{b l u e}{x \cdot - 4 x \left({e}^{- 2 {x}^{2}}\right)}$
$f ' \left(x\right) = {e}^{- 2 {x}^{2}} - 4 {x}^{2} \left({e}^{- 2 {x}^{2}}\right)$
$f ' \left(x\right) = {e}^{- 2 {x}^{2}} \left(1 - 4 {x}^{2}\right)$
$\textcolor{g r e e n}{f ' \left(x\right) = \frac{1 - 4 {x}^{2}}{{e}^{2 {x}^{2}}}}$