Find out whether a function of several variables is differentiable

Question

$$ f= \begin{cases} \frac{xy}{(x^2+y^2)^\frac{1}{4}},\ (x, y) \ne (0, 0) \\ 0,\ (x, y) = (0, 0) \end{cases} $$ I need to find out whether this function is differentiable at $(0, 0)$ and $(1, 0)$ and find $df(x,y)$ where the differential exists. I have already done the following: $$ \frac{\partial f}{\partial x} = \frac{x^2y + 2y^3}{2(x^2+y^2)^\frac{5}{4}}, \\ \frac{\partial f}{\partial y} = \frac{y^2x + 2x^3}{2(x^2+y^2)^\frac{5}{4}} $$ Therefore, both partial derivatives are continuous where $(x,y) \ne (0,0)$, so $f(x,y)$ is differentiable at $(1,0)$, and $dx=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$. I hope, that is right.
The problem is the point $(0,0)$. By definition I found that $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$ where $(x,y)=(0,0)$. But I am not sure if I can say that these partial derrivatives are also continuous because they are just constants and make a conclusion that $f(x,y)$ is also differentiable at $(0, 0)$, and $df(0,0)=0$. I would be grateful if anyone could explain what to do in that second case where $(x,y) = (0,0)$.

Answer 1

Since $\frac{\partial f}{\partial x}(0,0)=\frac{\partial f}{\partial y}(0,0)=0$, then, if $f$ is differentiable at $(0,0)$, then $f'(0,0)$ must be the null function. In other words, either $f$ is not differentiable at $(0,0)$ or $f'(0,0)$ is the null function.

So, let us check whether or not $f'(0,0)$ is the null function. What this means to assert that $f'(0,0)$ is the null function is that$$\lim_{(x,y)\to(0,0)}\frac{\bigl\lvert f(x,y)-f(0,0)\bigr\rvert}{\bigl\lVert(x,y)\bigr\rVert}=0.$$Now, note that\begin{align}\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)}{\lVert(x,y)\rVert}&=\lim_{(x,y)\to(0,0)}\frac{xy}{\sqrt{x^2+y^2}\sqrt[4]{x^2+y^2}}\\&=\lim_{(x,y)\to(0,0)}\frac{xy}{(x^2+y^2)^{3/4}}\\&=0,\end{align}because, if $x=r\cos\theta$ and $y=r\sin\theta$, then$$\frac{xy}{(x^2+y^2)^{3/4}}=\frac{r^2\cos\theta\sin\theta}{r^{3/2}}=\sqrt r\cos\theta\sin\theta.$$

Answer 2

I'm not sure what you mean with "the partial derivatives are just constants"...but observe that for $\;(x,y)\neq(0,0)\;$:

$$f'_x=\frac{y(x^2+y^2)^{1/4}-\frac{x^2y}2(x^2+y^2)^{-3/4}}{\sqrt{x^2+y^2}}=\frac{2x^2y+2y^3-x^2y}{2(x^2+y^2)^{5/4}}=\frac{x^2y+2y^3}{2(x^2+y^2)^{5/4}}\implies$$

$$\implies |f'_x|\le |y|\frac{2x^2+2y^2}{2(x^2+y^2)^{5/4}}\le\frac{|y|}{(y^2)^{1/4}}=\sqrt{|y|}\xrightarrow[(x,y)\to(0,0)]{}0$$

By symmetry, we get that both $\;f'_x,\,f'_y\;$ are continuous at $\;(0,0)\;$ and thus the function's differentiable there.