# System of equations (multivariable second derivative test)

1. Jul 11, 2014

### jonroberts74

I am doing critical points and using the second derivative test (multivariable version)

1. The problem statement, all variables and given/known data
$$f(x,y) = (x^2+y^2)e^{x^2-y^2}$$
Issue I am having is with the system of equations to get the critical points from partial wrt x, wrt y

3. The attempt at a solution
$$f_{x} = 2xe^{x^2-y^2}+2xe^{x^2-y^2}(x^2+y^2)$$
$$f_{y} = 2ye^{x^2-y^2}-2ye^{x^2-y^2}(x^2+y^2)$$

It is pretty easy to see (0,0) makes both equal to zero

I know +/- 1 = y and x = 0 is a solution as well.

$$(x^2+y^2)$$ is never negative let alone zero [over the reals]
$$e^{x^2-y^2}$$ won't be zero, it will get infinitesimally close to zero for values of square of y > square of x

and 2x+2x=0 only when x = 0

and 2y-2y=0 for all real values of y

how do I arrive at y = +/- 1 [ I can see it in the line just above but then I can use any real value for y]

2. Jul 11, 2014

You should note that you have
\begin{equation*}
\frac{\partial f} {\partial x} = 2x e^{x^2 - y^2}(1 + x^2 + y^2)
\end{equation*}

and similarly for $\frac{\partial f} {\partial y}$.

3. Jul 11, 2014

thank you