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

Oct 19, 2016

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

#### Explanation:

The quotient rule is $\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$

So, $f \left(x\right) = {x}^{2} / {e}^{x - 3}$
$\therefore f ' \left(x\right) = \frac{{e}^{x - 3} \frac{d}{\mathrm{dx}} \left({x}^{2}\right) - {x}^{2} \frac{d}{\mathrm{dx}} \left({e}^{x - 3}\right)}{{e}^{x - 3}} ^ 2$
$\therefore f ' \left(x\right) = \frac{{e}^{x - 3} \left(2 x\right) - {x}^{2} \left({e}^{x - 3}\right)}{{e}^{x - 3}} ^ 2$
$\therefore f ' \left(x\right) = \frac{{e}^{x - 3} \left(2 x - {x}^{2}\right)}{{e}^{x - 3}} ^ 2$
$\therefore f ' \left(x\right) = \frac{\left(2 x - {x}^{2}\right)}{{e}^{x - 3}}$
$\therefore f ' \left(x\right) = \frac{x \left(2 - x\right)}{{e}^{x - 3}}$