I was doing an exercise which asks for these partial derivatives, if possible.

\begin{cases} \dfrac{\partial f}{\partial x}(0,0), \dfrac{\partial f}{\partial y}(0,0) \end{cases} Consider this function $f$: \begin{cases} \dfrac{xy}{x^2+y^2} & (x,y) \neq (0,0)\\ 0 & (x,y) = (0,0)\\ \end{cases}

Since I wanna know the derivative at $(0,0)$ is it enough to only calculate the limit for the first equation or do I have to check for the second? Can anyone just tell me which limits to do, please?

  • 1
    $\begingroup$ Does this answer your question? How to check if this function has a partial derivative $\endgroup$ Commented Mar 13, 2020 at 14:29
  • $\begingroup$ @PeterForeman no. Maybe I didn't pick a good function, I just wanna know the method for calculating the partial derivative, if it exists. $\endgroup$ Commented Mar 13, 2020 at 14:31
  • $\begingroup$ By definition, when the RHS limit exists we have$$\frac{\partial f}{\partial x}(x,y)=\lim_{h\to0}\frac{f(x+h,y)-f(x,y)}h$$So for the case $(x,y)=(0,0)$ we need to calculate$$\frac{\partial f}{\partial x}(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}h$$ $\endgroup$ Commented Mar 13, 2020 at 14:35
  • $\begingroup$ To calculate f(h,0) I must use the first equation because h is considered infinitely close to 0 but different than 0 ? @PeterForeman $\endgroup$ Commented Mar 13, 2020 at 14:43
  • $\begingroup$ $h$ is not "infinitely close to $0$" (that's nonsensical) it is an arbitrary non-zero real number. $\endgroup$ Commented Mar 13, 2020 at 14:44

1 Answer 1


We will utilize the limit definitions of partial derivatives, as user Peter Foreman notes in the comments. Both subdomains of the function will be utilized in this process. $$f_x(0,0)=\lim\limits_{h\to0}\frac{f(0+h,0)-f(0,0)}{h}=\lim\limits_{h\to0}\frac{f(h,0)-f(0,0)}{h}$$ We know from the function definitions that $f(0,0)=0$. $$\lim\limits_{h\to0}\frac{f(h,0)-f(0,0)}{h}=\lim\limits_{h\to0}\frac{f(h,0)}{h}$$ Now, as was discussed in the comments, the limit as $h\to0$ does not have $h$ ever equal $0$, only approach it. As such, $h$ will always be a nonzero real number. Consequently, we will have to follow the $(x,y)\ne(0,0)$ subdomain to finish solving the limit. $$\lim\limits_{h\to0}\frac{f(h,0)}{h}=\lim\limits_{h\to0}\frac{\frac{(h)(0)}{h^2+0^2}}{h}=\lim\limits_{h\to0}\frac{\frac{0}{h^2}}{h}=\lim\limits_{h\to0}\frac{0}{h}=0$$ Next, let us find $f_y(0,0)$ utilizing the same method as with $f_x(0,0)$. $$f_y(0,0)=\lim\limits_{h\to0}\frac{f(0,0+h)-f(0,0)}{h}=\lim\limits_{h\to0}\frac{f(0,h)}{h}=\lim\limits_{h\to0}\frac{\frac{(0)(h)}{0^2+h^2}}{h}$$ Clearly, we will come to the same answer as before. $$\lim\limits_{h\to0}\frac{\frac{(0)(h)}{0^2+h^2}}{h}=\lim\limits_{h\to0}\frac{\frac{0}{h^2}}{h}=\lim\limits_{h\to0}\frac{0}{h}=0$$ Thus, $f_x(0,0)=0$ and $f_y(0,0)=0$ as well.


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