$\displaystyle f(x,y)=\frac{x^{2}y+xy^{2}}{x^{2}+y^{2}} , if \ \ (x,y)\neq(0,0)

and \ \ 0 \ \ otherwise

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- May 5th 2009, 06:05 PMsilversandis f differentiable
$\displaystyle f(x,y)=\frac{x^{2}y+xy^{2}}{x^{2}+y^{2}} , if \ \ (x,y)\neq(0,0)

and \ \ 0 \ \ otherwise

$ - May 5th 2009, 06:35 PMHallsofIvy
It is clearly differentiable, by the usual differentiation formulas, for any point other than (0,0):

$\displaystyle \frac{\partial f}{\partial x}= \frac{(2xy+ y^2)(x^2+ y^2)- (x^2y+ xy^2)(2x)}{x^2+ y^2}= \frac{2xy^3+ y^4- x^2y^2}{(x^2+ y^2)^2}$

$\displaystyle \frac{\partial f}{\partial y}= \frac{(x^2+ 2xy)(x^2+ y^2)- (x^2y+ xy^2)(2y)}{x^2+ y^2}= \frac{x^4+ 2xy^3- x^2y^2}{(x^2+ y^2)^2}$

Since the partial derivatives are continuous, the function is differentiable. Further, the limits, as (x, y) go to 0, are 0.

So the only question is at (0,0).

$\displaystyle \frac{f(h,0)- f(0,0)}{h}= 0/h= 0$

$\displaystyle \frac{f(0,h)- f(0,0)}{h}= 0/y= 0$

Since those match the limits of the partial derivatives as (x, y) goes to (0,0), the function is also differentiable at (0,0).

Again, that shows that the partial derivatives exist and are continuous at (0,0). That implies that the function is differentiable at (0,0). - May 5th 2009, 06:52 PMsilversand
I am not sure with the continuous bit.

as (x,y) goes to 0, the denominator should be zero? - May 10th 2009, 07:08 AMHallsofIvy
As far as (x,y) not (0,0) is concerned, the partial derivatives give rational function with NON-zero denominator. A rational function is always continuous at such a point.

As far as (0,0) is concerned, I showed that the partial derivatives at (0,0) are both 0.

To see what the limit of the derivatives is at (0,0), it is best to change to polar coordinates.

$\displaystyle \frac{\partial f}{\partial x}=\frac{2xy^3+ y^4- x^2y^2}{(x^2+ y^2)^2}$

In polar coordinates, $\displaystyle x= r cos(\theta)$, $\displaystyle y= r sin(\theta)$ and $\displaystyle r^2= x^2+ y^2$

The derivative becomes

$\displaystyle \frac{2r^4 cos(\theta)sin^3(\theta)+ r^4sin^4(\theta)- r^4cos^2(\theta)sin^2(\theta)}{r^2}$$\displaystyle = r^2(cos(\theta)sin^2(\theta)+ sin^4(\theta)- cos^2(\theta)sin^2(\theta))$

Because of the $\displaystyle r^2$ factor, the limit, as r goes to 0, is 0,**no matter what $\displaystyle \theta$ is**. Because, in polar coordinates, r alone measures the distance from (0,0), that is enough to show that the limit is 0.

I'll let you do the same for the other partial derivative.