# How do you find the derivative of  (x^(1/2))(e^(-x)) ?

Oct 2, 2016

${e}^{- x} \left[\frac{1 - 2 x}{2 \sqrt{x}}\right]$

#### Explanation:

The rule for deriving a product is

$\left(f \cdot g\right) ' = f ' g + f g '$

$f \left(x\right) = {x}^{\frac{1}{2}} = \sqrt{x}$

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

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

$g ' \left(x\right) = {e}^{- x} \cdot \frac{d}{\mathrm{dx}} \left(- x\right) = {e}^{- x} \left(- 1\right) = - {e}^{- x}$

I used the power rule $\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$ for $f ' \left(x\right)$ and the composite rule $\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$ for g'(x)#

Now that we have all the elements, let's plug everything into the formula:

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

Which we can rearrange factoring ${e}^{- x}$:

${e}^{- x} \left[\frac{1}{2 \sqrt{x}} - \sqrt{x}\right]$

and again, if you prefer,

${e}^{- x} \left[\frac{1 - 2 x}{2 \sqrt{x}}\right]$