Check if $\frac{x^3+y^3}{|x|+|y|}$ is differentiable at $(0,0)$

Question

We have the function $$f(x,y) = \begin{cases} \frac{x^3+y^3}{|x|+|y|},& (x, y) \ne (0,0) \\ 0, & (x,y) = (0,0) \end{cases}$$

My work so far:

I tried checking if $f$ is continuous at $(0,0)$:

When $x >0, y>0:$

$$\lim\limits_{(x,y)\rightarrow(0,0)} f(x) = \frac{(x+y)(x^2-xy+y^2)}{x+y} = 0$$

When $x >0, y<0:$ we can rewrite $y$ as $-y$, with $y>0$

$\lim\limits_{(x,y)\rightarrow(0,0)} f(x) = \frac{x^3-y^3}{x+y}$

Which I can not solve.

I calculated the partial derivatives:

\begin{align*}\frac{\partial{f}}{\partial{x}}(0,0)&=\lim\limits_{x\to 0} \frac{f(x,0)-f(0,0)}{x} =\lim\limits_{x\to 0}\frac{x^3}{|x|}=\lim\limits_{x\to 0} x^2 \mathrm{sgn}(x)= 0\\ \frac{\partial{f}}{\partial{x}}(x,y)&=\frac{3x^2(|x|+|y|)-(x^3+y^3)\mathrm{sgn}(x)}{(|x|+|y|)^2} \end{align*}

and analogously for $y$.

I tried checking if the partial derivatives are continuous and failed.

I tried with the definition:

$$\lim\limits_{(x,y)\rightarrow(0,0)} \frac{f(x,y)-f(0,0)-T(0,0)}{||(x,y)||} = 0,$$ where $T(x,y) = 0$.

we get $$\lim\limits_{(x,y)\rightarrow(0,0)} \frac{x^3+y^3}{(|x|+|y|)\sqrt{x^2+y^2}}$$

And I got stuck here.

Answer 1

Yes, the partial derivatives exist at $(0,0)$ and they are both equal to $0$. Therefore either $f$ is not differentiable at $(0,0)$, or it is differentibale there and its derivative there is the null function.

And what actually occurs is that $f'(0,0)$ is the null function. In fact, if $\varphi\colon\mathbb R^2\longrightarrow\mathbb R$ is the null function, then\begin{align}f'(0,0)=\varphi&\iff\lim_{(x,y)\to(0,0)}\frac{\bigl\lvert f(x,y)-f(0,0)-\varphi(x-0,y-0)\bigr\rvert}{\bigl\lVert(x,y)\bigr\rVert}=0\\&\iff\lim_{(x,y)\to(0,0)}\frac{\lvert x^3+y^3\rvert}{\bigl(\lvert x\rvert+\lvert y\rvert\bigr)\sqrt{x^2+y^2}}=0.\end{align}But, if $x=r\cos\theta$ and $y=r\sin\theta$, then$$\lvert x^3+y^3\rvert\leqslant2r^3$$and$$\bigl(\lvert x\rvert+\lvert y\rvert\bigr)\sqrt{x^2+y^2}\geqslant r^2,$$since $\lvert\cos\theta\rvert+\lvert\sin\theta\rvert\geqslant\cos^2\theta+\sin^2\theta=1$. Therefore,$$\frac{\lvert x^3+y^3\rvert}{\bigl(\lvert x\rvert+\lvert y\rvert\bigr)\sqrt{x^2+y^2}}\leqslant r\to_{r\to0}0.$$

Answer 2

Using your calculation for the partial derivative, if we can show that this is continuous then the function is $C^1$ and thus the derivative exists at $(0,0$). Thus we need to show that:

$$\frac{\partial{f}}{\partial{x}}(x,y)=\frac{3x^2(|x|+|y|)-(x^3+y^3)\mathrm{sgn}(x)}{(|x|+|y|)^2}$$

Note that we just want to ensure this is continuous around $(0,0)$. When $y=0$, this is equivalent to

$$\frac{\partial{f}}{\partial{x}}\biggr\rvert_{0,0}=\frac{3x^2|x|-(x^3)\mathrm{sgn}(x)}{x^2}=\frac{3x\mathrm{sgn}(x)-x\mathrm{sgn}(x)}{x^2}$$

the absolute value of which is at most $\frac{2}{x}$ which converges to $0$. Thus since both partial derivatives are continuous at $(0,0)$ (because, as you noted, the $y$ case is equivalent), the function is differentiable there.