# How do you derive y = x^3+(3x)/e^(x^2) using the quotient rule?

Oct 31, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 \left({x}^{2} + \setminus \frac{1 - 2 {x}^{2}}{{e}^{{x}^{2}}}\right)$

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({x}^{3} + \frac{3 x}{{e}^{{x}^{2}}}\right)$
$= \frac{d}{\mathrm{dx}} \left({x}^{3}\right) + \frac{d}{\mathrm{dx}} \left(\frac{3 x}{{e}^{{x}^{2}}}\right)$
$= 3 {x}^{2} + \frac{d}{\mathrm{dx}} \left(\frac{3 x}{{e}^{{x}^{2}}}\right)$
$= 3 {x}^{2} + \frac{{e}^{{x}^{2}} \frac{d}{\mathrm{dx}} \left(3 x\right) - 3 x \frac{d}{\mathrm{dx}} \left({e}^{{x}^{2}}\right)}{{\left({e}^{{x}^{2}}\right)}^{2}}$
$= 3 {x}^{2} + \frac{{e}^{{x}^{2}} \left(3\right) - 3 x \frac{d}{\mathrm{dx}} \left({e}^{{x}^{2}}\right)}{{\left({e}^{{x}^{2}}\right)}^{2}}$
$= 3 {x}^{2} + \frac{{e}^{{x}^{2}} \left(3\right) - 3 x \left(2 x {e}^{{x}^{2}}\right)}{{\left({e}^{{x}^{2}}\right)}^{2}}$
$= 3 {x}^{2} + \frac{{e}^{{x}^{2}} \left(3\right) - 3 x \left(2 x {e}^{{x}^{2}}\right)}{{\left({e}^{{x}^{2}}\right)}^{2}} \frac{{e}^{- {x}^{2}}}{{e}^{- {x}^{2}}}$
$= 3 {x}^{2} + \frac{3 - 3 x \left(2 x\right)}{{e}^{{x}^{2}}}$
$= 3 {x}^{2} + \frac{3 - 6 {x}^{2}}{{e}^{{x}^{2}}}$
$= 3 \left({x}^{2} + \setminus \frac{1 - 2 {x}^{2}}{{e}^{{x}^{2}}}\right)$