How do you differentiate f(x) = (x/4)(e^(-x/4)) using the product rule?

Nov 25, 2015

The product rule is:

For $f \left(x\right) = u \left(x\right) \cdot v \left(x\right)$, the derivative can be built as follows:

$f ' \left(x\right) = u ' \left(x\right) \cdot v \left(x\right) + u \left(x\right) \cdot v ' \left(x\right)$

Let's say that in your example, $u \left(x\right) = \frac{x}{4}$ and $v \left(x\right) = {e}^{- \frac{x}{4}}$.

Build the derivatives for $u \left(x\right)$ and $v \left(x\right)$ first:

$u \left(x\right) = \frac{x}{4} \textcolor{w h i t e}{\times \times} \implies u ' \left(x\right) = \frac{1}{4}$
$v \left(x\right) = {e}^{- \frac{x}{4}} \textcolor{w h i t e}{\times x} \implies v ' \left(x\right) = - \frac{1}{4} {e}^{- \frac{x}{4}}$

So, let's compute the derivative of $f \left(x\right)$:

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

$\textcolor{w h i t e}{\times \times} = \left(\frac{1}{4} - \frac{x}{16}\right) {e}^{- \frac{x}{4}}$