# Thread: Partial Derivatives

1. ## Partial Derivatives

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

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

Compute $\displaystyle \frac{\partial}{\partial y}\frac{\partial f}{\partial x}(0,0)$ and $\displaystyle \frac{\partial}{\partial x}\frac{\partial f}{\partial y}(0,0)$ 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 $\displaystyle f_{xy}$ and $\displaystyle f_{yx}$ and then check as to whether or not they are continous at $\displaystyle x,y=0$. If so then thats your answer. Note based on these conditions that $\displaystyle f_{xy}=f_{yx}$.

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

Compute $\displaystyle \frac{\partial}{\partial y}\frac{\partial f}{\partial x}(0,0)$ and $\displaystyle \frac{\partial}{\partial x}\frac{\partial f}{\partial y}(0,0)$ if they exist. (I'll denote these derivatives by $\displaystyle \color{blue}f_{xy}$ and $\displaystyle \color{blue}f_{yx}$ 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 $\displaystyle (x^2-y^2)\arctan\tfrac{x}{y}$ at points where neither x nor y is zero. But if you want to find for example $\displaystyle f_x(0,y)$ (where y≠0) then you have to evaluate the limit given by the definition $\displaystyle f_x(0,y) = \lim_{h\to0}\frac{f(h,y)-f(0,y)}h = \lim_{h\to0}\frac{(h^2-y^2)\arctan\frac hy-0}h$. To find this limit, you'll need to use the fact that $\displaystyle \lim_{x\to0}\frac{\arctan x}x=1$. Leaving out the details, this gives $\displaystyle f_x(0,y) = -y$.

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

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

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