Prove $D(fg)=D(f)D(g)res(f,g)^2$ where D is the discriminant and res the resultant 
Let $R$ be a commutative ring and $R[x]$ the polynomial ring over $R$. Let $f,g \in R[x]$. We let $D(fg)$ denote the discriminant of $fg$ and $\operatorname{res}(f,g)$ the respective resultant.
I need to prove that the equation
$$D(fg)=D(f)D(g)\operatorname{res}(f,g)^2$$
holds.

So far I fiddled around to no avail. My ideas so far are to use the properties of the definition of the Sylvester-Matrix that is used in $D(f)$ and $\operatorname{res}(f)$ and the determinant.
The definitions of $D(f)$ and $\operatorname{res}(f,g)$ I am working with are as follows:
$D(f)=(-1)^{n(n-1)/2}\operatorname{res}(f,f')$
and
$\operatorname{res}(f,g)=\det(S(f,g))$. $S(f,g)$ is the Sylvester-Matrix of $f$ and $g$, $f'$ denotes the derivative of $f$.
Help is much appreciated.

Finally, the answer:
We're gonna use two theorems which I am not gonna prove. They are the following:
(1) For f and g in $K[x]$ with $f=\prod_{i=1}^{n} (x-a_i)$ and $g=\prod_{i=1}^{m} (x-b_i)$ it follows that $res(f,g)=\prod_{i=1}^{n}\prod_{j=1}^{m} (a_i-b_i)$
(2) For $f=\prod_{i=1}^{n} (x-a_i) \in K[x]$ it follows that $D(f)=\prod_{1\leq i <j \leq n}(a_i-a_j)^2$
Now in the field extension $R'$ we have $f=\prod_{i=1}^{n} (x-a_i)$ and $g=\prod_{i=1}^{m} (x-b_i)$. 
For $1\leq k \leq  n+m$ define $c_k:=a_k$ for $k\leq n$ and $c_k=b_{k-n}$ for $k\geq n+1$. 
It follows: $D(fg)=\prod_{1\leq k <l \leq n+m}(c_k-c_l)^2=\prod_{1\leq k <l \leq n}(a_k-a_l)^2*\prod_{1\leq k <l \leq m}(b_k-b_l)^2*\prod_{\substack{1\leq k \leq n \\ 1\leq l \leq m}}(a_k-b_l)^2=D(f)*D(g)*res(f,g)^2$
