Find the value of D(a,b) using the second partial test for
g(x,y) = e^(3x^2 − 6y^2 + 36y)
I found the critical point which is at (0,3) but i cant seem to be able to find the value for D(a,b).
The answer I got after multiple attempts is -72e^54 which is incorrect.
The first partial derivatives are
$f_x=6x\cdot e^{3x^2-6y^2+36y}, f_y=(-12y+36)\cdot e^{3x^2-6y^2+36y}$
Now you have to apply the product rule to get $f_{xx}$ and $f_{yy}$.
$f_{xx}=6\cdot e^{3x^2-6y^2+36y}+6x\cdot 6x\cdot e^{3x^2-6y^2+36y}$
$f_{yy}=-12\cdot e^{3x^2-6y^2+36y}+(-12y+36)^2\cdot e^{3x^2-6y^2+36y}$
$f_{xy}=6x\cdot (-12y+36)\cdot e^{3x^2-6y^2+36y}$
$d=f_{xx}\cdot f_{xy}-f_{xy}^2$
Finally you have to insert the values of the critical point.
$f_{xy}(0,3)=0$.
The part $6x\cdot 6x\cdot e^{3x^2-6y^2+36y}$ of $f_{xx}$ is $0$ as well.
The rest I´ve calculated here. I get $D(0,3)=-72\cdot e^{108}$