let $f(x,y)=\frac{xy}{x^2+y^2}$ and $f(0,0)=0$

Now $f_1$ denotes the partial derivative of $f$ with respect to the 1st coordinate (x).

$$f_1=\frac{x^2y+y^3-2x^2y}{(x^2+y^2)^2}, f_1(0,0)=0$$

We can see that $f_1$ exists everywhere.

Let's get back to the question. Question asks me to show that directional derivatives don't exist at the origin. I couldn't get why wouldn't they exist?

I have $f_1(0,0)=0$ and $f_2(0,0)=0$ Define $Df_B(0,0)=Df_B(f_1(0,0),f_2(0,0))$ Where $B$ is the direction and $Df$ is the gradient vector.

By theorem $Df_B(x,y)=B*Df(x,y)$ where $*$ is the dot product.

Then the directional derivatives of all directions is $0$. Why wouldn't they exist? Can anyone give a counter example / proof / intuition / hint ?

  • $\begingroup$ Okay Edit : The proof that $(*)b*DF=D_bF$ depends on assuming that $DbF$ exists. So I take my word back. But I still need a rigorous proof $\endgroup$
    – math31
    Commented Dec 3, 2016 at 21:29
  • $\begingroup$ No, the theorem is that IF $f$ is differentiable at $(0,0)$ THEN the directional derivatives equal that. But here $f$ is not differentiable at $(0,0)$ $\endgroup$
    – zhw.
    Commented Dec 4, 2016 at 0:28

1 Answer 1


Take a direction $u_\theta=(\cos \theta, \sin \theta)$.

You have $$\frac{f((0,0)+ h u_\theta)-f(0,0)}{h}=\frac{\cos \theta \sin \theta}{h}$$ Hence the directional derivative (limit of above for $h \to 0$) cannot exists except for $\theta \in \{k\frac{\pi}{2} \ ; \ k \in \mathbb Z\}$, I.e. in the $x$ or $y$ direction.


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