$\displaystyle -\ln(2x^2+2y^2+1)$
Having a bit of trouble with this , need to find stationary points.
Im not sure i'm differentiating it right.
$\displaystyle \frac{\partial}{\partial x} -\ln(2x^2+2y^2+1) = \frac{1}{(2x^2+2y^2+1)}\cdot \frac{\partial}{\partial x}(2x^2+2y^2+1)$
is that how i go about it?