Thread: Help with differentiating a ln.

$\displaystyle -\ln(2x^2+2y^2+1)$

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?

Originally Posted by el123

$\displaystyle -\ln(2x^2+2y^2+1)$

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?

Yes, except you lost a negative .....

ok gotcha.Thanks.