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Given the function $$f(x,y) = \begin{cases} \frac{x^3y^2}{x^4 + y^4} & (x,y) \neq (0,0) \\ 0 & (x,y) = (0,0) \end{cases}$$ How would you prove (or disprove) the following statements:

  • $f$ has all partial derivatives at $(0,0)$.
  • $f$ has all directional derivatives at $(0,0)$.
  • $f$ has a differential at $(0,0)$.
  • $f$ has continuous partial derivatives in a neighborhood of $(0,0)$.

For the first one I just used the formula for directional derivative and substituted in the standard basis $(1,0)$ and $(0,1)$. The limit formula is $$\begin{array}\ D_xf(0) & = \lim_{t \to 0}\frac{f(0 + tv) - f(0)}{t} \\ & = \frac{v_1^3v_2^2}{v_1^4 + v_2^4} \end{array} $$ for $v = (v_1, v_2)$. And so clearly substituting in $(1,0)$ and $(0,1)$ gives that both the partial derivatives are both zero. This also proves the second point that all directional derivatives exist at $(0,0)$.

For the third point, as both partial derivatives are $0$, the differential at $(0,0)$ would be be $(0 \;\; 0)$ and this would mean that the directional derivative in any direction at $(0,0)$ would be $0$ as well, which it's clearly not.

For the third point I'm inclined to say that because the function is not differentiable at $0$ and all partial derivatives exist then if the partial derivatives would be continuous. Therefore, as we've proved that it's not differnetiable, then all the partial derivatives can't be continuous.

Is my reasoning right on these questions?

A follow up question from the last part: If a function is differentiable at a particular point then this means it's partials are continuous at that point, but does this mean all it's partials are continuous in some neighborhood of that point? In other words how would you prove the last part more generally? (If say it was true).

Thanks!

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Well done. If $f$ were differentiable at $(0,0)$, then $Df(0,0)=0$ (the null vector), since $\partial_1 f(0,0)=\partial_2 f(0,0)=0$. But you should verify easily that the limit $$ \lim_{(x,y) \to (0,0)} \frac{f(x,y)}{\sqrt{x^2+y^2}} $$ is not zero, and therefore $f$ is not differentiable at $(0,0)$. Finally, it is a theorem that the (existence and) continuity of partial derivatives around a point implies the differentiability at that point.

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