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I am solving this problem -

$$f(x)={xy\over {\sqrt{x^2+y^2}} },(x,y)\ne(0,0) \text{ and } 0, (x,y)=(0,0)$$

I am able to show that this function is continuous at $(0,0)$ and the partial derivatives exist at $(0,0)$. But when i find the partial derivatives $f_x,f_y$ i get $y^2\over (x^2+y^2)^{3\over 2}$. But this doesn't exist at $(0,0)$. So how does this add up ?

Aman Mittal
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    The partial derivatives are not continuous in $(0,0)$. – Daniel Fischer Oct 05 '13 at 13:37
  • How can it be continuous at $(0,0)$ when f(it) doesn't exist at that point?

    HINT: substitution rule.

     – Don Larynx Oct 05 '13 at 14:13
  • @DonLarynx : But the function itself is continuous. And the Partial derivatives do exist. – Aman Mittal Oct 05 '13 at 14:26
  • @DanielFischer : but the partial derivates exist at $(0,0)$ as $$f_x=\lim_{h\rightarrow 0} {f(h,0)-f(0,0)\over h}=0$$ – Aman Mittal Oct 05 '13 at 14:27
  • @AmanMittal Yes, the partial derivatives exist, $$\frac{\partial f}{\partial x}(0,0) = \frac{\partial f}{\partial y}(0,0) = 0.$$ And they also exist for all points $(x,y) \neq (0,0)$. Outside the origin, the partial derivatives are smooth, but in the origin, they are not continuous. Fairly standard situation. – Daniel Fischer Oct 05 '13 at 14:30
  • @DanielFischer : But if something is not continuous at the origin it means that it doesn't exist there. no ? – Aman Mittal Oct 05 '13 at 17:25
  • No. I can define $$h(x) = \begin{cases} -1 &, x < 0\ 0 &, x = 0\ 1 &, x > 0\end{cases}$$ That is defined everywhere, but it's not continuous at $0$. The situation with the partial derivatives is a bit more complicated, but also for that phenomenon we have one-dimensional examples, $$g(x) = \begin{cases} x^2\sin \frac1x &, x \neq 0\ 0 &, x = 0\end{cases}$$ is differentiable everywhere, but $g'$ is not continuous at $0$. – Daniel Fischer Oct 05 '13 at 17:28
  • @DanielFischer : That was an excellent eplaination. Thanks !! so, when i put $(0,0) \ in f_x$ it has to do with discontinuity of $f_x$ although partial derivative exists. I hope i got it right this time :) Thanks a lot again. Please write this as an answer, i'll accept it . – Aman Mittal Oct 05 '13 at 17:36

1 Answers1

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The original definition of the partial derivatives is on a point-by-point basis. In case the limit exists, the partial derivative with respect to$x$ of a function $f \colon \mathbb{R}^2 \to \mathbb{R}$ in a point $(x,y)$ is defined as

$$\lim_{h\to 0} \frac{f(x+h,y) - f(x,y)}{h}.$$

In the origin, we have to find the partial derivatives of $f$ by the definition. Using the difference quotient

$$\frac{f(x,0) - f(0,0)}{x} = \frac{1}{x}\left(\frac{x\cdot 0}{\sqrt{x^2+0^2}} - 0\right) = \frac0x = 0,$$

we determine $$\frac{\partial f}{\partial x}(0,0) = 0; \quad \frac{\partial f}{\partial y}(0,0) = 0$$

similarly. In points $(x,y) \neq (0,0)$, we have $f$ given by an analytic expression in a neighbourhood of the point, hence there we can obtain the partial derivatives using the chain and quotient rules in a uniform way, obtaining

$$\frac{\partial f}{\partial x}(x,y) = \frac{y^3}{(x^2+y^2)^{3/2}};\quad \frac{\partial f}{\partial y}(x,y) = \frac{x^3}{(x^2+y^2)^{3/2}}.$$

Thus we see that the partial derivatives exist everywhere, but they are not continuous in the origin, we have $\frac{\partial f}{\partial x}(0,y) = \pm1$ for $y \neq 0$, and $\frac{\partial f}{\partial y}(x,0) = \pm1$ for $x\neq 0$.

We can have the same phenomenon also in the one-dimensional setting, the function

$$g(x) = \begin{cases}x^2\sin \tfrac{1}{x} &, x \neq 0\\ \quad 0 &, x = 0 \end{cases}$$

is differentiable on all of $\mathbb{R}$, but its derivative

$$g'(x) = \begin{cases}2x\sin \tfrac1x - \cos \tfrac1x &, x \neq 0\\ \qquad 0 &, x = 0 \end{cases}$$

is not continuous at $0$.

Daniel Fischer
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