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?
Best Answer
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.