Let there be $(a,b),(c,d)$ intervalls in $\mathbb{R}$ and $f:(a,b) \rightarrow \mathbb{R}$, $g:(c,d) \rightarrow \mathbb{R}$ differentiable. Show that $\varphi : (a,b) \times (c,d) \rightarrow \mathbb{R}$ with $\varphi (x,y) := f(x)g(y)$ ist also differentiable.
I need a product rule for this i guess but i dient learned a product rule for higher dimensions yet.
A partial answer:
You can compute the partial derivatives of $\phi$ with respect to $x$ and $y$:
$$ \frac{\partial \phi}{\partial x} = f'(x) g(y) $$ for instance. If you knew that these two partials were continuous (i.e., if you'd been given the fact that $f$ and $g$ are not only differentiable, but have continuous derivatives), then it'd follow that $\phi$ is differentiable, for if both partials exist and are continuous in a neighborhood of a point $P \in \Bbb R^2$, then the derivative exists at $P$ as well.
For this particular case, referring to the definition of differentiabilty in $\Bbb R^2$ might actually let you go ahead and prove thing thing (in analogy with the proof of the product rule for 1 variable) without the added assumption of continuity of $f'$ and $g'$, however.