# How do you find (df)/dy and (df)/dx of f(x,y)=(3x^2-2e^y)/(2x+y), using the quotient rule?

Mar 22, 2018

Quotient rule states that
$\frac{d}{\mathrm{dx}} \left(\frac{h}{g}\right) = \frac{g h ' - h g '}{{g}^{2}}$

We can apply that to each of the variables x and y separately:

$f \left(x , y\right) = \frac{h \left(x , y\right)}{g} \left(x , y\right)$ where
$h \left(x , y\right) = 3 {x}^{2} - 2 {e}^{y}$ and $g \left(x , y\right) = 2 x + y$

Therefore,
$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{g \cdot \frac{\mathrm{dh}}{\mathrm{dx}} - h \cdot \frac{\mathrm{dg}}{\mathrm{dx}}}{{g}^{2}}$
$= \frac{\left(2 x + y\right) \left(6 x\right) - \left(3 {x}^{2} - 2 {e}^{y}\right) \left(2\right)}{2 x + y} ^ 2 = \frac{6 {x}^{2} + 6 x y + 4 {e}^{y}}{2 x + y} ^ 2$

$\frac{\mathrm{df}}{\mathrm{dy}} = \frac{g \cdot \frac{\mathrm{dh}}{\mathrm{dy}} - h \cdot \frac{\mathrm{dg}}{\mathrm{dy}}}{{g}^{2}}$
$= \frac{\left(2 x + y\right) \left(- 2 {e}^{y}\right) - \left(3 {x}^{2} - 2 {e}^{y}\right) \left(1\right)}{2 x + y} ^ 2 = \frac{- 4 x {e}^{y} - 2 y {e}^{y} - 3 {x}^{2} + 2 {e}^{y}}{2 x + y} ^ 2$