Question

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

Answer 1

Quotient rule states that
`d/dx(h/g) = (gh' - hg')/(g^2)`

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

`f(x, y) = (h(x, y)) / g(x, y)` where
`h(x, y) = 3x^2 - 2e^y` and `g(x, y) = 2x+y`

Therefore,
`(df)/dx = (g cdot (dh)/dx - h cdot (dg)/dx )/(g^2)`
`= ((2x+y)(6x) - (3x^2 - 2e^y) (2))/(2x+y)^2 = (6x^2+6xy + 4e^y)/(2x+y)^2`

`(df)/dy = (g cdot (dh)/dy - h cdot (dg)/dy )/(g^2)`
`= ((2x+y)(-2e^y) - (3x^2 - 2e^y) (1))/(2x+y)^2 = (-4xe^y - 2ye^y - 3x^2 + 2e^y)/(2x+y)^2`