How to prove product and quotient of smooth functions is smooth I'm trying to prove the following problem:

Let $A\subset\mathbb{R}^n$ be open.
If $f, g: A \rightarrow \mathbb{R}$ are smooth, show that $fg$ and $f/g$ is smooth. (For the quotient case, $g$ is nonzero on $A$.)

I'm not sure how to show the smoothness of $fg$ and $f/g$. (In this context, "smoothness" refers to $C^\infty$. That is, it means the function has continuous partial derivative of all orders.) For cases like $f+g$, I used induction to prove that partial derivative of order $n\geq 1$ of $(f+g)$ is sum of partial derivative of order $n \geq 1$ of $f$ and $g$ respectively. But for case of product and quotient, I am not sure how to show they are smooth. I think induction is again a possible solution, but it is quite mind-boggling to write down formula for the partial derivative of some order $k$ for $fg$ and $f/g$ as product rule and quotient rule for derivative have a very complicated formula as order goes up.
Could someone help me?
 A: Here is a proof for the first part. It is just tedious induction.
Multi index notation is useful in this context.
Suppose $f,g$ are smooth. You need to show that $D^\alpha(fg)$
is differentiable for all multi indices $\alpha$.
The proposition is that for any $\alpha$ there are constants $\gamma_\beta$ such that
$D^\alpha(fg) = \sum_{\beta \le \alpha} \gamma_\beta (D^{\beta}f) (D^{\alpha-\beta}g)$. If this is true then $D^\alpha(fg)$ is differentiable for any $\alpha$ and hence $fg$ is smooth.
Note that for any $\alpha$ with $|\alpha| = 1$, we have $D^\alpha(fg) = (D^\alpha f)g+f (D^\alpha g)$, hence $fg$ is differentiable and the formula
holds true for $|\alpha|=1$.
Suppose the formula is true for any $\alpha$ with $|\alpha| = n$, then it is clear that $D^{\alpha+e_k}(fg) = \sum_{\beta \le \alpha} \gamma_\beta ( (D^{\beta+e_k}f) (D^{\alpha-\beta}g) + (D^{\beta}f) (D^{\alpha+e_k-\beta}g) )$ is differentiable, and we can write $D^{\alpha+e_k}(fg) = \sum_{\beta \le \alpha+e_k} \delta_\beta (D^{\beta}f) (D^{\alpha+e_k-\beta}g)$ for some constants $\delta_\beta$. Hence the formula holds for any multi index $\alpha$ with $|\alpha|=n+1$.
For the quotient, it seems easier to establish that ${1 \over g}$ is smooth and then combine with the above result. Here are a few details.
For any $\alpha$ with $|\alpha|=1$ we have $D^\alpha {1 \over g^k} = - {k \over g^{k+1}} D^\alpha g$. We may surmise that for any $\alpha$ with $|\alpha | = n$ that $D^\alpha {1 \over g} = \sum_{\beta \le \alpha} \gamma_\beta {1 \over g^{2+|\alpha|-|\beta|}} D^\beta g$.
Imitating the process above shows that this formula holds for
$|\alpha|=n+1$ as well.
Since ${f \over g} = f {1 \over g}$ we have the desired result.
A: $f/g$ cannot be smooth if $g$ is simply assumed to be nonzero. Put $f(x)=x^{100}$ and $g(x)= 3x^2-2x$. We see that $f/g$ is not smooth in a neighborhood of $\frac{1}{3}$. If all derivatives of $g$ are nonzero, then $fg$ and $f/g$ are smooth by product rule.
