# Partial derivatives question

• Feb 22nd 2011, 07:38 PM
mathswannabe
Partial derivatives question
consider the function:

(2x(x^2-y^2))/(x^2+y^2) ---> (x,y) /= 0

f(x,y) ={

0 ---> (x,y)= 0

Show that fx (x partial) and fy(y partial) exist at (0,0) and determine their value...

I get an x-partial derivative of (2x^4+8x^2y^2-2y^4)/((x^2 + y^2)^2)

I get a y-partial derivative of (-8yx^3)/((x^2+y^2)^2)

I have done them a million times and just keep getting that - how on earth are they supposed to exist at (0,0)??

I was thinking that the numerator would have a factor of (x^2+y^2)^2 but no such luck?

any ideas?
• Feb 22nd 2011, 07:41 PM
Prove It
This is almost unreadable. Is the function

$\displaystyle \displaystyle \begin{cases}\frac{2x(x^2-y^2)}{x^2+y^2}\,\textrm{if}\,(x,y) \neq (0,0)\\ 0 \,\textrm{if}\,(x,y) = (0,0)\end{cases}$?
• Feb 22nd 2011, 07:43 PM
mathswannabe
that is exactly it ;)
• Feb 23rd 2011, 04:24 AM
HallsofIvy
Do NOT try to find the derivative at (0, 0) by finding the derivative at other points and taking the limit as (x, y) goes to (0, 0). Derivatives are NOT necessarily continuous.

The partial derivative of f(x,y), with respect to x, at (0, 0) is
$\displaystyle \displaytype\lim_{h\to 0}\frac{f(h,0)- f(0,0)}{h}$

The partial derivative of f(x,y), with respect to y, at (0, 0) is
$\displaystyle \displaytype\lim_{h\to 0}\frac{f(0,h)- f(0,0)}{h}$

Here, f(0,0)= 0 and $\displaystyle f(h, 0)= \frac{2h(h^2- 0^2)}{h^2+ 0^2}= \frac{2h(h^2)}{h^2}= 2h$

$\displaystyle f(0, h)= \frac{2(0)(0^2- h^2)}{0^2+ h^2}= 0$