finding partial derivative using Rudin limit definition

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!

• 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. Nov 12, 2022 at 7:56
• thank you :) )) Jun 7, 2023 at 17:51

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