1. ## partial dereivatives

$f(x,y)=(x+y)/(x^2+2y^2+6)$

find the stationary points.

i'm having trouble finding the partial derivatives. they came out really messy.

2. Hello,

Originally Posted by dinNA89
$f(x,y)=(x+y)/(x^2+2y^2+6)$

find the stationary points.

i'm having trouble finding the partial derivatives. they came out really messy.
For example, let's take the one with respect to x.

$\frac{\partial (x+y)}{\partial x}=1$

$\frac{\partial (x^2+2y^2+6)}{\partial x}=2x$

---> \begin{aligned} \frac{\partial f(x,y)}{\partial x} & =\frac{(x^2+2y^2+6)-2x \cdot (x+y)}{(x^2+2y^2+6)^2} \\
& =\frac{x^2+2y^2+6-2x^2-2xy}{(x^2+2y^2+6)^2} \\
& =\frac{-x^2-2xy+2y^2}{(x^2+2y^2+6)^2} \end{aligned}

3. Originally Posted by dinNA89
$f(x,y)=(x+y)/(x^2+2y^2+6)$

find the stationary points.

i'm having trouble finding the partial derivatives. they came out really messy.
Not much mess that I can see. Just use the quotient rule:

$\frac{\partial f}{\partial x} = \frac{(x^2 + 2y^2 + 6) - 2x(x+y)}{\text{stuff}} = \frac{-x^2 -2xy + 2y^2 + 6}{\text{stuff}}$

$\frac{\partial f}{\partial y} = \frac{(x^2 + 2y^2 + 6) - 4y(x+y)}{\text{stuff}} = \frac{x^2 -4xy - 2y^2 + 6}{\text{stuff}}$

I haven't worried about the denominators because you're gonna equate each of the partial derivatives to zero so it's only the numerators that matter.

4. Originally Posted by mr fantastic
$\frac{\partial f}{\partial x} = \frac{(x^2 + 2y^2 + 6) - 2x(x+y)}{\text{stuff}} = \frac{-x^2 -2xy + 2y^2 + 6}{\text{stuff}}$

$\frac{\partial f}{\partial y} = \frac{(x^2 + 2y^2 + 6) - 4y(x+y)}{\text{stuff}} = \frac{x^2 -4xy - 2y^2 + 6}{\text{stuff}}$
now i'm having trouble solving the numerators to equal 0.

5. Originally Posted by dinNA89
now i'm having trouble solving the numerators to equal 0.
You have two equations. Add them together and simplify. You get xy = 2. Now substitute y = 2/x, say, into either one of the two equations and solve for x etc.