# Show that the two partial derivatives exist at every point but that f is not differentiable at (0, 0) .

Let f : $\mathbb{R^2} → \mathbb{R}$ be given by

$$f(x,y) = \frac{x^3-y^3}{x^2+y^2}, \quad (x,y) \ne (0,0)$$ $$f(x,y) = \quad 0, \quad \quad \, \, (x,y) \ne (0,0)$$

Show that the two partial derivatives exist at every point but that f is not differentiable at (0, 0) .

So i'm thinking, i have to find the partial derivative of x and y and show continuity between the two paths at (0,0)

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

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

are continuous so so long as $(x,y) \ne 0$

Now I have to show...

$$\frac{\partial f(0,0)}{\partial x} = \lim_{k \to 0} \frac{Df(0+k,0)-f(0,0)}{k} = 0 = \frac{\partial f(0,0)}{\partial y}$$

This is where I am lost

• Any thoughts? I think i'm nearly there – Patrick Sep 30 '15 at 8:57

At all points $(x,y)\ne(0,0)$ the function $f$ is differentiable according to the standard rules, since the denominator is nonzero in a full neighborhood of such points.
Now the origin: As $f(0,0)=0$ and $f(r\cos\phi,r\sin\phi)=r(\cos^3\phi-\sin^3\phi)$ when $r>0$ the function $f$ is continuous at the origin.
I guess you have computed $f_x(0,0)=f_y(0,0)=0$. Therefore, if $f$ were differentiable at $(0,0)$ we would have $df(0,0)=0$, and therefore $$\lim_{r\to0+}{f(r\cos\phi,r\sin\phi)\over r}=0\ ,$$ whatever $\phi$. Now let $\phi:={3\pi\over4}$. Then $\lim_{r\to0+}{f(r\cos\phi,r\sin\phi)\over r}$ computes to $-{1\over\sqrt{2}}$. This shows that $f$ is not differentiable at the origin.
• It needs to be continuous at the origin, differentiability at the origin implies that. If we assume that it is differentiable then we must $\lim_{r \to 0^{+}} f(r, \phi)=f(0,\phi)$. – user135520 Sep 30 '15 at 19:40