I'm having trouble with a certain multi-variable calculus question. $$ f(x,y) = \begin{cases} \large\frac{2xy^2}{x^2 + y^4}, & \text{$(x,y)\neq 0$} \\[2ex] 0, & \text{$(x,y) = 0$} \end{cases}$$

I need to show that both $\large\frac{∂f}{∂x}$ and $\large\frac{∂f}{∂y}$ exist everywhere.

I can easily manage to find both partial derivatives, but I'm not really sure what the question means when it asks to show that they "exist everywhere".

Any help would be appreciated, thanks.

  • 4
    $\begingroup$ In future, please don't accept answers too quickly. Waiting a couple of hours before choosing an accepted answer is good practice, because it lets different people give their own unique perspective, and then you can pick and choose which perspective was most useful to you. Good luck with your assignment. $\endgroup$ Mar 16, 2017 at 3:35

1 Answer 1


The potential problem is at the origin. But note that

$$f_x(0,0)=\lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to 0}\frac{\frac{2h(0^2)}{h^2+0^4}-0}{h}=0$$


$$f_y(0,0)=\lim_{h\to 0}\frac{f(0,h)-f(0,0)}{h}=\lim_{h\to 0}\frac{\frac{2(0)(h^2)}{0^2+h^4}-0}{h}=0$$

Therefore, $f_x(0,0)=f_y(0,0)$. For $x^2+y^2>0$, we can simply note that $f(x,y)$ is composition of differentiable functions with

$$\begin{align} \frac{\partial f(x,y)}{\partial x}&=\frac{2y^2(y^4-x^2)}{(x^2+y^4)^2}\\\\ \frac{\partial f(x,y)}{\partial y}&=\frac{4xy(x^2-y^4)}{(x^2+y^4)^2} \end{align}$$

Hence, we see that

$$\begin{align} \frac{\partial f(x,y)}{\partial x}=\begin{cases}\frac{2y^2(y^4-x^2)}{(x^2+y^4)^2}&,x^2+y^2>0\\\\ 0&,x=y=0 \end{cases} \end{align}$$

$$\begin{align} \frac{\partial f(x,y)}{\partial y}=\begin{cases}\frac{4xy(x^2-y^4)}{(x^2+y^4)^2}&,x^2+y^2>0\\\\ 0&,x=y=0 \end{cases} \end{align}$$

NOTE: While the first partial derivatives, $f_x$ and $f_y$, exist everywhere, neither is continuous at the origin.

  • $\begingroup$ So will this mean that $f(x,y)$ will be differentiable when $(x,y) \neq 0$ and not for $(x,y) = 0$? $\endgroup$ Mar 16, 2017 at 3:51
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    $\begingroup$ @NathanLowe, THEOREM: if all partial derivatives exist at a point and are continuous, then the function is differentiable there. But mere existence of the the derivatives there isn't enough to guarantee differentiability. On the other hand, just because some of the partial derivatives there are discontinuous, does not necessarily guarantee that the function fails to be differentiable there. So, you have to do a bit more work to find out whether or not it's differentiable at $(x:=0,y:=0)$. $\endgroup$ Mar 16, 2017 at 3:56
  • $\begingroup$ That is, the answer to your question is that it's not immediately clear whether or not $f$ is differentiable is at $(x:=0,y:=0)$ or not just from the information you've got. $\endgroup$ Mar 16, 2017 at 3:57
  • $\begingroup$ @NathanLowe That is a good question and Goblin addressed it aptly. We know that $f$ is differentiable for $x^2+y^2>0$, but would need to do additional analysis to see if $f$ is differentiable at $(0,0)$. Are you familiar with the definition of differentiability of a multivariable function? If not, here is a reference. $\endgroup$
    – Mark Viola
    Mar 16, 2017 at 4:08
  • $\begingroup$ @Dr.MV and goblin, So using what I know from the question, as both partial derivatives exist everywhere, can we use limits to determine if the derivatives are continuous at (0,0) to find if the function is differentiable at (0,0)? $\endgroup$ Mar 16, 2017 at 4:23

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