I was just reading the definition of Rudin on partial derivative given $f:\mathbb{R}^n \to \mathbb{R}^m$, $e_1,\dots,e_n$ and $u_1,\dots ,u_m$ be basis for the spaces, we have $(D_jf_i)(x)=\lim_{t\to 0}\frac{f_i(x+te_j)-f_i(x)}{t}$. Where $f_i(x)=f(x)\cdot u_i$

Now let $f(x,y)=\frac{xy}{x^2+y^2}$ and $f(0,0)=0$, if I want to use the definition to evaluate partial of $x$ at $(x,y)\neq 0$, I have $D_1f_i(x,y)=\lim_{t\to 0}\frac{f_i(x+t,y)-f_i(x)}{t}$, what exactly is the component function $f_i$ here? Since $m=1$ I'm assuming $f_i=f(x,y)$ and we have

$\lim_{t\to 0}\frac{f_i(x+t,y)-f_i(x)}{t}=\lim_{t\to 0}\frac{\frac{(x+t)(y)}{(x+t)^2+y^2}-\frac{xy}{x^2+y^2}}{t}$, I've played a lot with this fraction over here but I cant get anywhere, and everything online is evaluating this function at origin. I'm not sure what I am missing here. If anyone can help me understand this? Thanks a lot!

  • 3
    $\begingroup$ You're essentially going to end up proving the quotient rule, so look at how that proof goes. Also, $y$ is just a fixed number here. And yes, here $m=1$ so there's only 1 component function, namely $f$ itself. $\endgroup$
    – peek-a-boo
    Nov 12, 2022 at 7:56
  • $\begingroup$ thank you :) )) $\endgroup$ Jun 7, 2023 at 17:51

1 Answer 1


Let $(x,y)\neq(0,0)$, we need to find $\dfrac{\partial f}{\partial x}(x,y)$ using the definition of partial derivative.
\begin{equation} \dfrac{\partial f}{\partial x}(x,y)=\lim_{t \to 0}\dfrac{f(x+t,y)-f(x,y)}{t}=\lim_{t \to 0}\dfrac{1}{t}\left[\dfrac{(x+t)y}{(x+t)^2+y^2}-\dfrac{xy}{x^2+y^2}\right]= \end{equation} \begin{equation} =\lim_{t \to 0}\dfrac{1}{t}\dfrac{y(x+t)(x^2+y^2)-xy[(x+t)^2+y^2]}{(x^2+y^2)[(x+t)^2+y^2]} \end{equation} Now, to simplify the calculations, we can use asymptotic behaviour: when $t \to 0$, the small powers of t prevail; so we obtain: \begin{equation} \lim_{t \to 0}\dfrac{1}{t}\dfrac{y(x+t)(x^2+y^2)-xy[(x+t)^2+y^2]}{(x^2+y^2)[(x+t)^2+y^2]}=\lim_{t \to 0}\dfrac{1}{t}\dfrac{yt(x^2+y^2)-2x^2yt}{(x^2+y^2)[(x+t)^2+y^2]}= \\=\dfrac{y(x^2+y^2)-2xy}{(x^2+y^2)^2} \end{equation} Note that using the quotient rule you obtain the same result.


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