1. ## Partial Derivatives

Let $f(x,y)=x^2\arctan\frac{x}{y}-y^2\arctan\frac{x}{y}$ if $xy\neq0$
Let $f(x,y)=0$ if $xy=0$

Compute $\frac{\partial}{\partial y}\frac{\partial f}{\partial x}(0,0)$ and $\frac{\partial}{\partial x}\frac{\partial f}{\partial y}(0,0)$ if they exist.

Ive proven that f(x,y) is continuous at 0, but I'm not sure which function I should be using to calculate the derivative, any help?

Thanks

2. Originally Posted by polymerase
Let $f(x,y)=x^2\arctan\frac{x}{y}-y^2\arctan\frac{x}{y}$ if $xy\neq0$
Let $f(x,y)=0$ if $xy=0$

Compute $\frac{\partial}{\partial y}\frac{\partial f}{\partial x}(0,0)$ and $\frac{\partial}{\partial x}\frac{\partial f}{\partial y}(0,0)$ if they exist.

Ive proven that f(x,y) is continuous at 0, but I'm not sure which function I should be using to calculate the derivative, any help?

Thanks
Start by computing that $f_{xy}$ and $f_{yx}$ and then check as to whether or not they are continous at $x,y=0$. If so then thats your answer. Note based on these conditions that $f_{xy}=f_{yx}$.

3. Originally Posted by polymerase
Let $f(x,y)=x^2\arctan\frac{x}{y}-y^2\arctan\frac{x}{y}$ if $xy\neq0$
Let $f(x,y)=0$ if $xy=0$

Compute $\frac{\partial}{\partial y}\frac{\partial f}{\partial x}(0,0)$ and $\frac{\partial}{\partial x}\frac{\partial f}{\partial y}(0,0)$ if they exist. (I'll denote these derivatives by $\color{blue}f_{xy}$ and $\color{blue}f_{yx}$ respectively.)

Ive proven that f(x,y) is continuous at 0, but I'm not sure which function I should be using to calculate the derivative, any help?
For a function where the defining formula breaks down at some points, you have to go back to the definition of the derivative at those points.

For the function given above, you can use the usual rules of differentiation to find the partial derivatives of $(x^2-y^2)\arctan\tfrac{x}{y}$ at points where neither x nor y is zero. But if you want to find for example $f_x(0,y)$ (where y≠0) then you have to evaluate the limit given by the definition $f_x(0,y) = \lim_{h\to0}\frac{f(h,y)-f(0,y)}h = \lim_{h\to0}\frac{(h^2-y^2)\arctan\frac hy-0}h$. To find this limit, you'll need to use the fact that $\lim_{x\to0}\frac{\arctan x}x=1$. Leaving out the details, this gives $f_x(0,y) = -y$.

To find $f_x(0,0)$, you again have to go back to the definition of the derivative, but this time it's easier: $f_x(0,0) = \lim_{h\to0}\frac{f(h,0)-f(0,0)}h = \lim_{h\to0}\frac{0-0}h = 0$. So in fact the formula $f_x(0,y) = -y$ still works when y=0.

To calculate the second partial derivative $f_{xy}(0,0)$ you again have to go back to the definition: $f_{xy}(0,0) = \lim_{k\to0}\frac{f_x(0,k)-f_x(0,0)}k = \lim_{k\to0}\frac{-k-0}k = -1$.

For the other mixed partial derivative $f_{yx}(0,0)$, *if it exists*, you need to use the same procedure, but this time finding $f_y(x,0)$ and $f_y(0,0)$ first, and then using $f_{yx}(0,0) = \lim_{h\to0}\frac{f_y(h,0) - f_y(0,0)}h$.