How can I show that this function is smooth? I got an assignment which I just can't find the right way to solve and I hope that someone could help me out here. 
It goes like this:
Let $\Omega\in R^n$ be a domain and $b_1,...,b_n:\Omega\to R^n$ smooth mappings (or function, don't know the correct translation into english), so that for every $x\in \Omega$ the vectors $b_1(x),...,b_n(x)$ are linearly independent. 
Let $c_1,...,c_n:\Omega\to R$ be mappings (or functions). 
Show that the function $F(x):=c_1b_1(x)+...+c_n(x)b_n(x)$ is smooth when $c_1,...,c_n$ are smooth. 
 A: Let me start by reducing your problem to a simpler one.
It is not necessary that the functions $b_1$ are linearly independent; the result is true even without that assumption.
If we can show that $\Omega\ni x\mapsto c_1(x)b_1(x)$ is smooth, then $F$ will be smooth as a finite sum of smooth functions.
A function $\Omega\to\mathbb R^n$ is smooth if (and only if) every component function $\Omega\to\mathbb R$ is smooth.
Therefore it is enough to show that if $b,c:\Omega\to\mathbb R$ are smooth, then $f(x)=b(x)c(x)$ is also a smooth function of $x$.
Let us show by induction that $f$ has derivatives of all orders.
First, let $\alpha$ be a multi-index with $|\alpha|=1$.
Then $\partial^\alpha f$ exists and $\partial^\alpha f=(\partial^\alpha b)c+b\partial^\alpha c$ by the product rule.
We only used the product rule in one direction (that of the coordinate $\alpha$).
We learned that $\partial^\alpha f$ is a sum of products of derivatives of $b$ and $c$ (which are smooth functions).
Suppose then that for some integer $m\geq1$ we know that all derivatives $\partial^\alpha f$ of $f$ exist for $|\alpha|\leq m$ and they are sums of products of derivatives of $b$ and $c$.
We want to show that for any $\beta$ with $|\beta|=m+1$, the derivative $\partial^\beta f$ also exists and is of such form.
Since $m\geq1$ and thus $|\beta|\geq2$, we know that $\beta_i\geq1$ for some index $i\in\{1,\dots,n\}$.
Let $\gamma=(0,\dots,0,1,0,\dots,0)$ with a one in the $i$th position.
Now also $\alpha:=\beta-\gamma$ is a multi-index (all components positive).
We have $\partial^\beta f=\partial^{\gamma+\alpha}f=\partial^\gamma(\partial^\alpha f)$.
We can now apply the product rule to each term in the sum ($\partial^\alpha f$ is of sum form), and the product rule also tells that $\partial^\beta f$ is a sum of products of derivatives of $b$ and $c$.
This proves the inductive step.
This is not the only possible approach.
One option is to show that for any $k$ you can approximate $f$ by a polynomial of order $k$ at any point.
You can construct suitable polynomials from the Taylor polynomials of $b$ and $c$.
