$\displaystyle f(x,y)=(x+y)/(x^2+2y^2+6)$
find the stationary points.
i'm having trouble finding the partial derivatives. they came out really messy.
Hello,
For example, let's take the one with respect to x.
$\displaystyle \frac{\partial (x+y)}{\partial x}=1$
$\displaystyle \frac{\partial (x^2+2y^2+6)}{\partial x}=2x$
---> $\displaystyle \begin{aligned} \frac{\partial f(x,y)}{\partial x} & =\frac{(x^2+2y^2+6)-2x \cdot (x+y)}{(x^2+2y^2+6)^2} \\
& =\frac{x^2+2y^2+6-2x^2-2xy}{(x^2+2y^2+6)^2} \\
& =\frac{-x^2-2xy+2y^2}{(x^2+2y^2+6)^2} \end{aligned}$
Not much mess that I can see. Just use the quotient rule:
$\displaystyle \frac{\partial f}{\partial x} = \frac{(x^2 + 2y^2 + 6) - 2x(x+y)}{\text{stuff}} = \frac{-x^2 -2xy + 2y^2 + 6}{\text{stuff}}$
$\displaystyle \frac{\partial f}{\partial y} = \frac{(x^2 + 2y^2 + 6) - 4y(x+y)}{\text{stuff}} = \frac{x^2 -4xy - 2y^2 + 6}{\text{stuff}}$
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.