What is the derivative of (xe^-x)/(x^3+x)?

Jun 1, 2016

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

Explanation:

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

Applying quotient rule,
${\left(\frac{f}{g}\right)}^{'} = \frac{{f}^{'} \cdot g - {g}^{'} \setminus \cdot f}{{g}^{2}}$

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

We know,
$\frac{d}{\mathrm{dx}} \left(x {e}^{- x}\right) = {e}^{- x} - {e}^{- x} x$
and,
$\frac{d}{\mathrm{dx}} \left({x}^{3} + x\right) = 3 {x}^{2} + 1$

So,
$= \frac{\left({e}^{- x} - {e}^{- x} x\right) \left({x}^{3} + x\right) - \left(3 {x}^{2} + 1\right) x {e}^{- x}}{{\left({x}^{3} + x\right)}^{2}}$

Simplifying it,we get,
$\frac{d}{\mathrm{dx}} \left(\frac{x {e}^{- x}}{{x}^{3} + x}\right) = \frac{- {e}^{- x} {x}^{4} - {e}^{- x} {x}^{2} - 2 {e}^{- x} {x}^{3}}{\setminus {\left({x}^{3} + x\right)}^{2}}$