# Finding partial derivatives

1. Oct 25, 2005

### Odyssey

Greetings, I need help in finding the partials with respect to x and y at (x,y) =/= (0,0) and (x,y) = (0,0)...

Let f(x,y) = { (xy^2-x^2y+3x^3-y^3) / (x^2+y^2) , (x,y) =/= (0,0)
{ 0 (x,y) = (0,0)

There was a hint given: first simplify f(x,y). I guess we have to get rid of the x^2+y^2 in the denominator to get rid of dividing by zero. But then I couldn't do it because of the "3" in the 3x^3.

Thanks for the help!

2. Oct 25, 2005

### mathmike

do you know the quotiant rule

[[g(x) * f'(x)] - [ f(x) * g'(x)]] / g(x)^2

3. Oct 25, 2005

### Odyssey

yes....i differentiated with respect to x and y...but then I'm not sure how to find the partials of x and y at (0,0).

4. Oct 25, 2005

### HallsofIvy

Staff Emeritus
Use the basic definition.
$$f_x(0,0)= lim_{h\rightarrow0}\frac{f(0+h,0)- f(0,0)}{h}$$
$$= lim_{h\rightarrow 0}\frac{\frac{3(h)^3}{(h)^2}}{h}$$.

$$f_x(0,0)= lim_{h\rightarrow0}\frac{f(0,0+h)- f(0,0)}{h}$$
$$= lim_{h\rightarrow 0}\frac{\frac{-(h)^3}}{(h)^2}{h}$$.

That should be easy.