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Let $f$ be a function $$ f(x,y)=\begin{cases}\quad0&(x,y)=(0,0)\\\dfrac{x^3y^2}{\left(x^2+y^2\right)^2} & (x,y)\neq(0,0). \end{cases}$$ I know what $f$ is continous at the point because the limit of $f$ when $(x,y)\to(0,0)$ exist: did go to the origin by the $x$- and $y$-axis and all lines through origin. Since the limit exists, then $f$ is continuous and the partial derivatives exist in this point. But why isn't $f$ differentiable in $(0,0)$?

Cameron Williams
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Kim
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  • What does it mean for $f$ to be differentiable at $(0, 0)$? Are you familiar with the definition? – Marcin Łoś Feb 23 '14 at 17:12
  • that the function must be continous on the domain and that the limit of the difference quotient of the function must exist? – Kim Feb 23 '14 at 17:17
  • It's not clear whether you already proved that $f$ is not differentiable but expected $f$ to be differentiable from the existance of partial derivates and continuity, or you are simply asking to prove that $f$ is not differentiable. – dani_s Feb 23 '14 at 17:17
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    For the function to be differentiable the partial derivatives have to be continuous at that point. Are the partial derivatives continuous at the origin? @boywholived You should reconsider your comment. Future people might search this question and find this MSE thread. How delighted would they be to find your comment then? – Mark Fantini Feb 23 '14 at 17:18
  • @Fantini Not really, that's sufficient, but not necessary. – Marcin Łoś Feb 23 '14 at 17:21
  • @Kim Indeed, the limit of the difference quotient must exist. What you need to do is to show it does not exist. Can you do that? – Marcin Łoś Feb 23 '14 at 17:22
  • @MarcinŁoś You're right, but that is much easier to check than cranking up the definition. – Mark Fantini Feb 23 '14 at 17:24
  • http://calculus.subwiki.org/wiki/Existence_of_partial_derivatives_not_implies_differentiable – hrkrshnn Feb 23 '14 at 17:24
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    @Fantini Well yes, but it can only prove differentiability, not lack thereof. Failure of a sufficient condition does not imply the fact does not hold. It's sufficient for function to be constant to be continuous, but there surely are nonconstant continuous functions. – Marcin Łoś Feb 23 '14 at 17:27
  • @boywholived Existance does not imply differentiability, existance and continuity, on the other hand, does. In fact, it's enough for one partial derivative to be continuous IIRC. – Marcin Łoś Feb 23 '14 at 17:32
  • I think you also need the partial derivatives to exist in a neighborhood of the point – dani_s Feb 23 '14 at 17:35
  • i am getting that the limit of the difference quotient exist :/ @MarcinŁoś – Kim Feb 23 '14 at 17:35
  • Please show us your work in that case. – Marcin Łoś Feb 23 '14 at 17:39
  • @MarcinŁoś. By reading the question I think the OP believes that existence of continuity of f and existence of partial derivatives at a point implies differentiability. I just wanted to point out to the OP that it is not true, nothing more. – hrkrshnn Feb 23 '14 at 18:03
  • @boywholived I may have missed the intent of your comment. Sorry. – Marcin Łoś Feb 23 '14 at 18:14
  • @MarcinŁoś Could you help me with this? I did get stuck and confused. – Kim Feb 23 '14 at 18:23

1 Answers1

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As you have noticed, $f$ is continuous at $(0, 0)$, and partial derivatives exist:

$$ \left(\frac{\partial f}{\partial x}\right)_{0, 0}=\lim_{h\to 0}\frac{f\left(h, 0\right)-f\left(0, 0\right)}{h}=\frac{0-0}{h}=0 $$

Similarily, $D_y f(0, 0) = 0$. The definition of differentiability implies existence of the following limit: $$ \lim_{(x, y)\to(0,0)} \frac{f(x, y) - f(0, 0) - x \frac{\partial f}{\partial x}\left(0, 0\right) - y\frac{\partial f}{\partial y}\left(0, 0\right)}{\left|(x, y)\right|}= \lim_{(x, y)\to(0,0)}\frac{x^3y^2}{\left(x^2+y^2\right)^{5/2}} $$

I claim it does not exist. Let $x=r\cos \theta$, $y=r\sin \theta$ and let $r\to 0$. The above limit becomes $$ \lim_{r\to 0} \frac{r^5\cos\theta\sin\theta}{\left(r^2\cos^2\theta+r^2\sin^2\theta\right)^{5/2}}= \lim_{r\to 0}\frac{r^5\cos\theta\sin\theta}{r^5}=\lim_{r\to 0}\cos\theta\sin\theta $$

which clearly depends on $\theta$.

Or, if you prefer, you can just explicitly choose two different directions and show they give different answers, e.g. $(x, y) = (0, t)$, $(x, y) = (t, t)$.

Marcin Łoś
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  • how did you get |(x,y)| to (x^2+y^2)^(5/2)`? Should it not be (x^2+y^2)^(1/2) since the 3 term from right is equal to zero? and why do you have | .. | on the numerator ? – Kim Feb 23 '14 at 19:09
  • $|(x,y)$ is indeed $(x^2+y^2)^{1/2}$, but the $f$ itself has $(x^2+y^2)^2$ in the denumerator. And the absolute value in the numerator, while correct, is not necessary, and possibly misleading, I'll remove it. – Marcin Łoś Feb 23 '14 at 19:11
  • oh, i did forget the denumerator in f :) now everything is clear! thanks again @MarcinŁoś – Kim Feb 23 '14 at 19:13