Let $\displaystyle f(x,y)=\frac{-2xy}{(x^2+y^2)^2}$ and $f(0)=0$. Show that all partial derivatives of $f$ wrt $x$ exist at all points of $\mathbb{R^2}$

My Try:

Regarding $f_x$, it is clear that $f_x$ exists for all $(x,y)\neq 0$. I proved that $f_x$ exists at $(0,0)$ using the limit definition. But this problem asks to show the existence of all other $f_{xx},f_{xxx},...$. How do I show it? It is not possible to calculate all. Is there a general method to show it? Any help is appreciated.

  • $\begingroup$ What is the meaning of "wrt"? $\endgroup$ – Gustavo Sep 12 '15 at 2:38
  • 1
    $\begingroup$ With respect to $\endgroup$ – Extremal Sep 12 '15 at 2:59
  • $\begingroup$ Do you want all the derivatives, or just the ones involving $x$ (as you've written)? $\endgroup$ – Alex M. Sep 14 '15 at 17:47

Consider a point $(x_0,y_0)\in{\mathbb R}^2$ with $y_0\ne0$. Then there is a $2$D neighborhood $U$ of this point within which the formula $$f(x,y)={-2xy\over (x^2+y^2)^2}$$ applies. A simple induction proof using the quotient rule then shows that for all $n\geq0$ one has $${\partial^n f\over\partial x^n}(x,y)={p_n(x,y)\over (x^2+y^2)^{2+n}}\qquad\bigl((x,y)\in U\bigr)\ ,$$ whereby $p_n(x,y)$ is a certain polynomial. It follows that $f$ has partial derivatives with respect to $x$ of all orders at $(x_0,y_0)$.

It remains to consider points $(x_0,0)\in{\mathbb R}^2$. The partial function $\psi(x):=f(x,0)$ is then $\equiv 0$ in a full neighborhood $\>]x_0-h,x_0+h[\>$ of $x_0$, so that we obtain $${\partial^n f\over\partial x^n}(x_0,0)=\psi^{(n)}(x_0)=0\ .$$


First of all, if $y=0$ the function is identically 0 and hence differentiable in $x$.

Fix a value of $y\not=0$ and consider the function $g_y(x)=\frac{1}{x^2+y^2}$. This is a rational function that has no singularities, thus it is in $C^{\infty}(\mathbb R)$. Since $$f(x,y)=y\frac{d}{dx}g_y(x),$$ it follows that $f(x,y)$ is infinitely differentiable w.r.t. $x$.


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