Calculus: Partial Derivatives Question

Question

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.

Answer 1

$H(x,y)=4 ln(x^3+4y^2)$

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 $f_{xy}(2,3)=-72/121$

Answer 2

I think now you can solve $$f_{xx}=\frac{\partial^2 f}{\partial x^2}$$ $$f_{yy}=\frac{\partial^2 f}{\partial y^2}$$ $$f_{xy}=\frac{\partial^2f}{\partial x\partial y}$$