is the function $f$ differentiable at $(0,0)$?

Question

Let$f(x,y)$ be defined as

$x^2y^2\over\sqrt{x^2+y^2}$ if $(x,y) \ne (0,0)$,

$0$ if $(x,y) = (0,0)$.

is $f$ differentiable at $(0,0)$?

I am trying to prove that it is differentiable by computing the partial derivative and show that the partial derivative is continous at $(0,0)$ and hence the whole domain, but so far I cannot even find a way to show the limit of derivative exist at $(0,0)$.

Could someone show steps of finding that limit? Thanks!

May I ask for a proof without the usage of polar coordinate?

Answer 1

Hmm from what I recall from calculus 2 courses you can show this function in polar coordinates $X = r\cos(\theta)$ $Y=r\sin(\theta)$ Now we can see that $\sqrt(x^2+y+2)$ in polar coordinates will be equal to $r$ Then $f=r^4*\sin^2(\theta)*\cos^2(\theta) / r$ Then you can see the $\lim$ will be $0$ because $r\to 0$ You can use the polar coordinates for the derivatives.

Answer 2

To show that $f$ is differentiable at $(0,0)$ you have to show that $$f(h)=f(0,0)+\nabla f (0,0) \cdot h + o(|h|)$$ for $h \in \Bbb{R}^2$ in a neighbourhood of $(0,0)$ (here $\cdot$ denotes the scalar product). It is natural to put $\nabla f (0,0) = (0,0)$, so that indeed you need to prove $$\lim_{h \to (0,0)} \frac{f(h)-f(0,0)}{|h|} =0$$ Using polar coordinates $h=(R \cos \theta , R \sin \theta)$ you have $$0 \le \frac{f(h)-f(0,0)}{|h|} = \frac{R^4 \cos^2 \theta \sin^2 \theta}{R^2} \le R^2 \to 0$$ and you are done.