Consider the following function.
$H(x,y) = 4 \ln(x^3 + 4y^2)$
(a) Find $f_{xx}(2,3)$.
(b) Find $f_{yy}(2,3)$.
(c) Find $f_{xy}(2,3)$.
I know that I should use second order partial derivatives to solve this problem.
$f_{xx} = (f_x)_x$
$f_{yy} = (f_y)_y$
$f_{xy} = (f_x)_y$
but I do not know how to use these notations and get the solutions. I would appreciate the help.
$H(x,y)=4 ln(x^3+4y^2)$
So $f_{x}(x,y)=\frac{4}{x^3+4y^2}(3x^2)=\frac{12x^2}{x^3+4y^2}$
And $f_{xx}(x,y)=\frac{12(-x^4+8xy^2)}{(x^3+4y^2)^2}$
Plugging in for $x=2, y=3$ will leave you with $f_{xx}(2,3)=96/121$
The trick is to treat $y$ as a constant. Keep that in mind when you partially differentiate.
$f_{y}(x,y)=\frac{32y}{x^3+4y^2}$ and $f_{yy}(x,y)=\frac{32(x^3-4y^2)}{(x^3+4y^2)^2}$ (treat $x$ as a constant in this case)
Plugging in for $x=2, y=3$ will leave you with $f_{yy}(2,3)=-56/121$
For $f_{xy} = (f_x)_y$, we partially differentiate $f_{x}(=\frac{12x^2}{x^3+4y^2})$ with respect to $y$.
So $f_{xy}=\frac{\partial}{\partial y}(\frac{12x^2}{x^3+4y^2})=-\frac{96x^2y}{\left(x^3+4y^2\right)^2}$
Plugging in for $x=2, y=3$ will leave you with $f_{xy}(2,3)=-72/121$
I think now you can solve $$f_{xx}=\frac{\partial^2 f}{\partial x^2}$$ $$f_{yy}=\frac{\partial^2 f}{\partial y^2}$$ $$f_{xy}=\frac{\partial^2f}{\partial x\partial y}$$
