Total differentiability of $f(x,y)=(x+y)g(x,y)$ where $g(x,y)$ is a continuous function

Question

The question is:

Let $g(x,y)$ be continuous and $g(0,0)=0$, $f(x,y)=(x+y)g(x,y)$, show that $f(x,y)$ is totally differentiable at the origin.

I tried showing it by using the definition of total differentiability:
$f(x_0+v)=f(x_0)+(\text{Jacobian of }f)v +r(v)$
if $\,\lim\limits_{v\to0}\dfrac{r(v)}{\lVert v\rVert}=0\,,\,$ then $f(x,y)$ is totally differentiable at $x_0$.
The problem is, when I calculate $\,r(v)=f(0+v)-f(0,0)-\big(\text{Jacobian of }f\,$ at $0\big)v$.
The Jacobian Matrix includes partial derivatives of $g(x,y)$, and because of this, I cannot take the limit $\,\lim\limits_{v\to0}\dfrac{r(v)}{\lVert v\rVert}$.
I get something like this:
$\lim\limits_{v\to0}\dfrac{r(v)}{\lVert v\rVert}=\dfrac{(v_1+v_2)\left(\frac{\partial}{\partial x}g(v_1,v_2)\right)v_1+(v_1+v_2)\left(\frac{\partial}{\partial y}g(v_1,v_2)\right)v_2}{\lVert v\rVert}$.

How can I get rid of the partial derivatives of $g(x,y)$ ? Is there another way to show the original claim ?

Answer 1

You can calculate the partial derivatives of $f(x,y)$ at $(0,0)$ by using their definitions (you do not need to write the partial derivatives of $g(x,y)$) :

$\dfrac{\partial f}{\partial x}\big(0,0\big)\!=\!\lim\limits_{x\to0}\dfrac{f(x,0)\!-\!f(0,0)}x\!=\!\lim\limits_{x\to0}\dfrac{x\,g(x,0)}x\!=\!g(0,0)\!=\!0$

$\dfrac{\partial f}{\partial y}\big(0,0\big)\!=\!\lim\limits_{y\to0}\dfrac{f(0,y)\!-\!f(0,0)}y\!=\!\lim\limits_{y\to0}\dfrac{y\,g(0,y)}y\!=\!g(0,0)\!=\!0$

Consequently,

$\begin{align}\left|\dfrac{r(x,y)}{\sqrt{x^2+y^2}}\right|&=\dfrac{\big|f(x,y)-f(0,0)\big|}{\sqrt{x^2+y^2}}=\dfrac{\big|x+y\big|\!\cdot\!\big|g(x,y)\big|}{\sqrt{x^2+y^2}}\leqslant\\[3pt]&\leqslant \dfrac{|x|+|y|}{\sqrt{x^2+y^2}} \!\cdot\!\big|g(x,y)\big|=\dfrac{\sqrt{x^2}+\sqrt{y^2}}{\sqrt{x^2+y^2}} \!\cdot\!\big|g(x,y)\big|\leqslant\\[3pt]&\leqslant\dfrac{\sqrt{x^2+y^2}+\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}} \!\cdot\!\big|g(x,y)\big|=\\[3pt]&=2\big|g(x,y)\big|\;.\end{align}$

Since $\,\left|\dfrac{r(x,y)}{\sqrt{x^2+y^2}}\right|\leqslant 2\big|g(x,y)\big|\;$ and $\lim\limits_{(x,y)\to(0,0)}2\big|g(x,y)\big|=0\,,\;$ by applying the Squeeze theorem, it follows that

$\lim\limits_{(x,y)\to(0,0)}\dfrac{r(x,y)}{\sqrt{x^2+y^2}}=0$

that is the function $f(x,y)$ is totally differentiable at $(0,0)$.

Answer 2

Let's make the reasonable guess that we will have $Df(0)=0$. To show that $f$ is differentiable at the origin with this derivative we thus have to show that

$$f(h)-f(0)=o(\lVert h\rVert)$$

as $h\to0$. Now if we write $h=(h_1,h_2)$, then

$$\frac{\lvert f(h)-f(0)\rvert}{\lVert h\rVert}=\frac{\lvert(h_1+h_2)g(h)\rvert}{\lVert h\rVert}\leq\frac{\sqrt{h_1^2+h_2^2}+\sqrt{h_1^2+h_2^2}}{\lVert h\rVert}\lvert g(h)\rvert=2\lvert g(h)\rvert\to0$$

as $h\to0$ since $g$ is continuous and equals zero at the origin. The result follows.