Differentiability of $f(x,y)=\frac{xy}{x^2+y^2}$ at $(1,1)$

Question

I tried to show that $f(x,y)=\frac{xy}{x^2+y^2}$ is differentiable at $(1,1)$ by doing the following but I got stuck:

$$\lim_{(h,k)\longrightarrow(0,0)}\frac{f(1+h,1+k)-f(1,1)-f_x(1,1)h-f_y(1,1)k}{\sqrt{h^2+k^2}}$$

$$=\lim_{(h,k)\longrightarrow(0,0)}\frac{\frac{(1+h)(1+k)}{(1+h)^2+(1+k)^2}-\frac{1}{2}-0-0}{\sqrt{h^2+k^2}}$$

$$=\lim_{(h,k)\longrightarrow(0,0)}\frac{\frac{2(1+h)(1+k)-(1+h)^2-(1+k)^2}{2[(1+h)^2+(1+k)^2]}}{\sqrt{h^2+k^2}}$$

$$=\lim_{(h,k)\longrightarrow(0,0)}\frac{-(h-k)^2}{2[(1+h)^2+(1+k)^2]\sqrt{h^2+k^2}}$$

Would appreciate any help from hence forth. Thank you!

Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. — Martin R, Dec 01 '23 at 08:13
$f$ has continuous partial derivatives in a neighborhood of $(1,1)$. — geetha290krm, Dec 01 '23 at 08:23
I checked that it is not differentiable at $(0,0)$. Is that related to differentiability at $(1,1)$? — KHOOS, Dec 01 '23 at 08:24
@geetha290krm That is helpful! Should I abandon my approach using definition of differentiability? — KHOOS, Dec 01 '23 at 08:25
Using your method, it ultimately suffices to show $|h-k|/\sqrt{h^2+k^2} \leq 2$ — Brian Moehring, Dec 01 '23 at 08:25
Quick tip: You might want to use the fact that if $f(t), g(t)$ are both differentiable at some $t=a$, then for $h(t)=\frac{f(t)}{g(t)}$ we have $h'(a)=\frac{f'(a)g(a)-f(a)g'(a)}{g(a)^2}$. This way you only have to compute partial derivatives of polynomials first. — Cartesian Bear, Dec 01 '23 at 08:27
@KHOOS I don't know what you already learned about differentiability. Here, $f$ is the ratio of two polynomials that are differentiable, with the denominator not vanishing around $(1,1)$. Hence the desired conclusion. — mathcounterexamples.net, Dec 01 '23 at 17:48

Sine of the Time · Accepted Answer · 2023-12-05T10:14:17.947

2

$f(x,y)$ is $C^1$ in a neigbourhood of $(1,1)$ and hence it is differentiable. However, if you want to prove it by the definition, note that as $(x,y)\to (1,1)$, the quantity $(1+h)^2+(1+k)^2$ is bigger than $1$. Now use polar coordinates:

$$0\le\left|\frac{(h-k)^2}{[(1+h)^2+(1+k)^2]\sqrt{h^2+k^2}}\right|\le \frac{(h-k)^2}{\sqrt{h^2+k^2}}=\frac{\rho^2(1-\sin 2\vartheta)}{\rho}\le 2\rho \to 0 $$

edited Dec 05 '23 at 10:14

answered Dec 02 '23 at 10:14

Sine of the Time

4,931

1

This is so elegant. Thanks! – KHOOS Dec 03 '23 at 23:55
@KHOOS you're welcome my friend – Sine of the Time Dec 03 '23 at 23:59
It should be noted that pointwise convergence as $\rho \to 0$ isn't sufficient, but in this case, the convergence at the end is uniform with respect to $\theta$, which is sufficient. I personally would insert $$\frac{\rho^2(1-\sin 2\vartheta)}{\rho} \leq 2\rho \to 0$$ to avoid the confusion with the common error. – Brian Moehring Dec 04 '23 at 02:38
@BrianMoehring right, I'll edit it – Sine of the Time Dec 04 '23 at 07:31

Differentiability of $f(x,y)=\frac{xy}{x^2+y^2}$ at $(1,1)$

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