I'm getting $\displaystyle f_{xx}$ and $\displaystyle f_{yy}$ having opposite signs (-ve & +ve) for $\displaystyle f(x,y) = e^{5x^2-20x-9y^2-36y}$, I get point = (2, -2), and $\displaystyle f_x \left (i.e. \frac{\partial f}{ \partial x} \right ) = (10x - 20) \cdot e^{5x^2 -20x -9y^2 -36y} $

$\displaystyle f_{xx} = \frac{\partial (f_x)}{ \partial x}$

$\displaystyle = 10 \cdot e^{5x^2-20x-9y^2-36y} \{10 (x - 2)^2 + 1 \}$

at point 2, -2 , $\displaystyle f_{xx} = 10 \cdot e^{5 \cdot 4-40-36+72} \{10(2-2)^2 + 1 \}$

$\displaystyle = 10e^{16}$

$\displaystyle f_y (i.e. \frac{\partial f}{ \partial y})$

$\displaystyle = (-18y -36) \cdot e^{5x^2-20x-9y^2-36y}$

$\displaystyle f_{yy} = \frac{\partial (f_y)}{ \partial y}$

$\displaystyle f_{yy} = -18 \{[(y + 2) \cdot e^{5x^2-20x-9y^2-36y} \times (-18y -36)] + e^{5x^2-20x-9y^2-36y} \times 1 \}$

$\displaystyle f_{yy} = -18 \cdot e^{5x^2-20x-9y^2-36y} \{-18(y + 2)^2 + 1 \}$

$\displaystyle f_{yy} = -18 \cdot e^{5x^2-20x-9y^2-36y} (-18y^2 - 72y - 72 + 1 )$

$\displaystyle f_{yy} = -18 \cdot e^{5x^2-20x-9y^2-36y} (-18y^2 - 72y - 71 )$

at point 2,-2 $\displaystyle f_{yy} = -18 \cdot e^{5x^2-20x-9y^2-36y} (-18 \cdot 4 + 144 - 71 )$

$\displaystyle = -18e^{16}$

so $\displaystyle f_{xx} = 10 \cdot e^{16}$ and $\displaystyle f_{yy} = -18 \cdot e^{16}$ This gives a -ve sign for delta.

I was taught that $\displaystyle f_{xx}$ and $\displaystyle [f_{yy}$ should have the same sign but i'm not getting that here.