I came across the following problem:

Question: Let \begin{equation*} f : \mathbb{R}^2 \to \mathbb{R}, ~~~ (x,y) \mapsto \begin{cases} 0, & (x,y) = (0,0) \\ \frac{x^3}{x^2 + y^2}, & (x,y) \neq (0,0). \end{cases} \end{equation*} Show that $f$ is partially differentiable everywhere, and determine where $g$ is differentiable.

And the answer (some detail omitted):

Answer: Let $\mathbf{v} = (h,k) \neq (0,0)$. Then the directional derivatives are \begin{equation*} D_{\mathbf{v}}f(\mathbf{0}) = \displaystyle \lim_{t \to 0} \frac{f(\mathbf{0} + t\mathbf{v}) - f(\mathbf{0})}{t} = \displaystyle \lim_{t \to 0} \frac{t^3h^3}{t^3(h^2 + k^2)} = \displaystyle \lim_{t \to 0} \frac{h^3}{h^2 + k^2} = \begin{cases} h, & k=0 \\ \frac{h^3}{h^2 + k^2}, & k \neq 0. \end{cases} \end{equation*} For $\mathbf{v} = \mathbf{e_1} = (1,0)$ and $\mathbf{v} = \mathbf{e_2} = (0,1)$ the partials at the origin are $\frac{\partial f}{\partial x}(0,0) = 1$ and $\frac{\partial f}{\partial y}(0,0) = 0$. For everywhere else, $f$ is rational and the partials exists, and specifically we define them as \begin{equation*} \frac{\partial f}{\partial x} = \begin{cases} \frac{x^4 + 3x^2y^2}{(x^2 + y^2)^2}, & (x,y) \neq (0,0) \\ 0, & (x,y) = (0,0) \end{cases} ~~~~~~~\text{and}~~~~~~~ \frac{\partial f}{\partial y} = \begin{cases} \frac{-2x^3y}{(x^2 + y^2)^2}, & (x,y) \neq (0,0) \\ 0, & (x,y) = (0,0). \end{cases} \end{equation*} The function is differentiable everywhere except at the origin, since if we consider $\mathbf{x}_n = \left(\frac{1}{n}, \frac{1}{n}\right) \to 0$ then \begin{equation*} f(\mathbf{x}_n) = \frac{1}{n^3} \cdot \frac{n^2}{2} = \frac{n^2}{2n^3} = \frac{1}{2n} \to \infty, \end{equation*} as $n \to 0$.

My question is can't we take from the fact the the partials at the origin are different for $\mathbf{e}_1$ and $\mathbf{e}_2$, i.e. $\frac{\partial f}{\partial x}(0,0) \neq \frac{\partial f}{\partial y}(0,0)$, that it is not continuous at the origin. The example at the end is correct, but is it necessary if we can simply see that the partials are different at the origin?


1 Answer 1


The partial derivatives at a point are usually always different. When you are taking partial derivatives you are moving along different axes and so you do not expect to get the same number. Take $f(x,y)=x-y$. It is everywhere differentiable but the partial derivatives are 1 and $-1$

  • $\begingroup$ Makes sense. I suppose we would need to extend this to say that if $\mathbf{v} = (1,1)$ then $D_{\mathbf{v}}f(\mathbf{0}) \neq D_{\mathbf{e}_1}f(\mathbf{0}) + D_{\mathbf{e}_2}f(\mathbf{0})$. Then it would not be continuous at the origin and we wouldn't need a specific example, per se. $\endgroup$
    – jj8989
    Nov 6, 2017 at 14:13
  • 1
    $\begingroup$ You are right, it is not necessary $\endgroup$
    – Gio67
    Nov 6, 2017 at 14:24

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