Product of differentiable functions is total differentiable

Question

Given two function $f,g \colon \mathbb{R} \to \mathbb{R}$ that are differentiable in $0$, define the function $$ F(x,y) \colon \mathbb{R}^2 \to \mathbb{R}, \quad (x,y) \mapsto f(x)g(y). $$

I have to proof that $F$ is total differentiable in $(0,0)$. We know that $F(x,0) = f(x)g(0)$ and $F(0,y) = f(0)g(y)$ are partial differentiable because $f$ and $g$ are differentiable in $0$.

So $(\partial_1F)(0,0) = f'(0)g(0)$ and $(\partial_2F)(0,0) = f(0)g'(0)$ but I have some trouble with proving that these partial derivatives are continuous in $(0,0)$. I think I can proof this using the fact that

$$ F(x,y) - F(0,0) - A(x,y) = o(\|(x,y)\|) \qquad \text{when $(x,y) \to (0,0)$} $$

with $A$ the suggested total derivative in $(0,0)$. Can someone help me with this last step?

Answer 1

Since $f$ and $g$ are differentiable, the function$$\begin{array}{rccc}h\colon&\mathbb{R}^2&\longrightarrow&\mathbb{R}^2\\&(x,y)&\mapsto&\bigl(f(x),g(y)\bigr)\end{array}$$is differentiable too: if $(x_0,y_0)\in\mathbb{R}^2$ then $h'(x_0,y_0)$ is the linear map $(x,y)\mapsto\bigl(f'(x_0)x,g'(y_0)y\bigr)$. Now, consider the map$$\begin{array}{rccc}p\colon&\mathbb{R}^2&\longrightarrow&\mathbb{R}^2\\&(x,y)&\mapsto&xy.\end{array}$$It is differentiable too and $F=p\circ h$. Therefore, $F$ is differentiable.

Answer 2

Since $f$ and $g$ are differentiable at 0 then for $x\to0$ and $y\to0$

$$f(x)=f(0)+xf'(0)+o(x)$$ and similarly $$g(y)=g(0)+yg'(0)+o(y)$$ Hence, multiplying the two equalities we get

$$\underbrace{f(x)g(x)}_{F(x,y)}=\underbrace{f(0)g(0)}_{=F(0,0)}+\underbrace{xf'(0)g(0)+yf(0)g'(0)}_{=\nabla F(0,0)\cdot(x,y)}+\underbrace{o(y)(f(0)+xf'(0)+o(x))+xyf'(0)g'(0)+o(x)(g(0)+yg'(0))}_{=o((x,y))}$$ We conclude from the last equality that $F$ is differentiable at $(0,0)$ and $$\nabla F(0,0)=(f'(0)g(0),f(0)g'(0))$$