# How do you differentiate g(x) = (1/x^3)*sqrt(x-xe^(x)) using the product rule?

Aug 8, 2016

$\frac{1 - {e}^{x} - x {e}^{x}}{2 {x}^{3} \sqrt{x - x {e}^{x}}} - \frac{\sqrt{x - x {e}^{x}}}{3 {x}^{4}}$

#### Explanation:

First let me rewrite this
$\left({x}^{-} 3\right) \cdot {\left(x - x {e}^{x}\right)}^{\frac{1}{2}}$

$\left(f \left(x\right) \cdot g \left(x\right)\right) p r i m e = f \left(x\right) p r i m e g \left(x\right) + g \left(x\right) p r i m e f \left(x\right)$

$\frac{d}{\mathrm{dx}} \left({x}^{-} 3\right) \cdot {\left(x - x {e}^{x}\right)}^{\frac{1}{2}} =$

$- 3 {x}^{-} 4 \cdot {\left(x - x {e}^{x}\right)}^{\frac{1}{2}} + {x}^{-} 3 \cdot \left(\frac{1}{2}\right) {\left(x - x {e}^{x}\right)}^{- \frac{1}{2}} \frac{d}{\mathrm{dx}} \left(x - x {e}^{x}\right)$

now just focusing on the last part because we need to do the power rule again
$\frac{d}{\mathrm{dx}} \left(x - x {e}^{x}\right) = 1 - {e}^{x} + x {e}^{x}$

now we can place this back in

$- 3 {x}^{-} 4 \cdot {\left(x - x {e}^{x}\right)}^{\frac{1}{2}} + {x}^{-} 3 \cdot \left(\frac{1}{2}\right) {\left(x - x {e}^{x}\right)}^{- \frac{1}{2}} \left(1 - {e}^{x} - x {e}^{x}\right)$

now i rearrange it a bit

$\frac{1 - {e}^{x} - x {e}^{x}}{2 {x}^{3} \sqrt{x - x {e}^{x}}} - \frac{\sqrt{x - x {e}^{x}}}{3 {x}^{4}}$