# How do you differentiate f(x)= ( x^2-3x-6 )/ (e^x + 2)  using the quotient rule?

Nov 1, 2016

$f ' \left(x\right) = \frac{{e}^{x} \left(5 x + 3 - {x}^{2}\right) + 4 x - 6}{{e}^{x} + 2} ^ 2$

#### Explanation:

You need to use the quotient rule;
$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$

So with $f \left(x\right) = \frac{{x}^{2} - 3 x - 6}{{e}^{x} + 2}$ we have

$f ' \left(x\right) = \left\{\frac{\left({e}^{x} + 2\right) \left(\frac{d}{\mathrm{dx}} \left({x}^{2} - 3 x - 6\right)\right) - \left({x}^{2} - 3 x - 6\right) \left(\frac{d}{\mathrm{dx}} \left({e}^{x} + 2\right)\right)}{{e}^{x} + 2} ^ 2\right\}$

$\therefore f ' \left(x\right) = \left\{\frac{\left({e}^{x} + 2\right) \left(2 x - 3\right) - \left({x}^{2} - 3 x - 6\right) \left({e}^{x}\right)}{{e}^{x} + 2} ^ 2\right\}$

$\therefore f ' \left(x\right) = \left\{\frac{{e}^{x} \left(2 x - 3\right) + 2 \left(2 x - 3\right) - \left({x}^{2} - 3 x - 6\right) \left({e}^{x}\right)}{{e}^{x} + 2} ^ 2\right\}$

$\therefore f ' \left(x\right) = \left\{\frac{{e}^{x} \left(2 x - 3 - \left({x}^{2} - 3 x - 6\right)\right) + 2 \left(2 x - 3\right)}{{e}^{x} + 2} ^ 2\right\}$

$\therefore f ' \left(x\right) = \frac{{e}^{x} \left(2 x - 3 - {x}^{2} + 3 x + 6\right) + 4 x - 6}{{e}^{x} + 2} ^ 2$

$\therefore f ' \left(x\right) = \frac{{e}^{x} \left(5 x + 3 - {x}^{2}\right) + 4 x - 6}{{e}^{x} + 2} ^ 2$