# Partial Derivatives of Discontinuous Fcn?

1. Sep 27, 2010

### absci2010

f(x,y)= xy2/(x2+y2) if (x,y)$$\neq$$(0,0)
=0 if (x,y)=(0,0)
Show that the partial derivatives of x and y exist at (0,0).

This may be a really stupid question, but would the partial derivatives of x and y at (0,0) just be 0? I tried taking that partial derivatives of xy2/(x2+y2) and got:
df/dx=[(x2+y2)(y2)-xy2(2x)]/(x2+y2)2
and
df/dy=[(x2+y2)(2xy)-xy2(2y)]/(x2+y2)2
which i don't believe cancel out.

2. Sep 28, 2010

### HallsofIvy

You are asked for the partial derivatives at (0, 0) so use the definition directly:

$$\displaytype\frac{\partial f}{\partial x}(0, 0)= \lim_{h\to 0}\frac{f(h, 0)- f(0, 0)}{h}$$
$$\displaytype = \lim_{h\to 0}\frac{\frac{h(0)}{h^2+ 0}}{h}= \lim_{h\to 0} 0= 0$$

Why do you say this function is discontinuous?

3. Sep 28, 2010

### absci2010

Thanks so much!
I realized after I posted the question that it's not discontinuous. Oops. >.<