# How do you use the quotient rule to find the derivative of y=(1-x*e^x)/(x+e^x) ?

Sep 24, 2014

$y = f \frac{x}{g} \left(x\right) = \frac{1 - x {e}^{x}}{x + {e}^{x}}$

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

$g ' \left(x\right) = 1 + {e}^{x}$

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

Substitute in the values for $f \left(x\right) , f ' \left(x\right) , g \left(x\right) , \mathmr{and} g ' \left(x\right)$

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

FOIL

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

Distribute the negative

$= \frac{- {x}^{2} {e}^{x} - x {e}^{x} - x {e}^{2 x} - {e}^{2 x} - 1 + x {e}^{x} - {e}^{x} + x {e}^{2 x}}{x + {e}^{x}} ^ 2$

Combine like terms

$= \frac{- {x}^{2} {e}^{x} - {e}^{x} - {e}^{2 x} - 1}{x + {e}^{x}} ^ 2$