In order to calculate the marginal revenue, some derivative rules are applied, but I don't see which ones.
I don't think you need this, but this is given: $Q = a-bP$ and $P = (a-Q)/b$ $\mbox{total revenue} = P\times Q = aP-bP^2$
and below the chain rule is applied, it seems, right? But I can't get to see what parts of the formula of the chain rule correspond with the elements of the equation below.
$P$ and $Q$ can both be written as functions of each other but...
the solution I don't get: MO = marginal revenue = $ \frac{ \partial (P\times Q)} { \partial Q} = P \times \frac{\partial Q }{\partial Q} + Q \times \frac {\partial P} {\partial Q}$
My MathJax equation doesn't work for some reason, but /partial means the partial derivative, so the partial derivative of $P\times Q$ is calculated for $Q$ ...
The revenue can be written as $R=P\cdot Q(\color{blue}P)=P\cdot (a-bP) $. Here you have to apply the $\textbf{product rule}$. $Q$ depends on $P$ here.
$\text{First summand}$: The first factor $P$ is differentiated w.r.t $P$ and ismultiplied by the second factor $Q(P)$:
$$\frac{\partial P }{\partial P}\cdot Q(P)=1\cdot (a-bP)$$
$\text{Second summand}$: The second factor ($Q(P)$) is differentiated w.r.t $P$ and is multiplied by the first factor $P$:
$$\frac{\partial Q(P) }{\partial P}=-b\Rightarrow \frac{\partial Q(P) }{\partial P}\cdot P=-bP $$
The sum gives the marginal revenue:
$$MR=\frac{\partial P }{\partial P}\cdot Q(P)+\frac{\partial Q(P) }{\partial P}\cdot P=1\cdot (a-bP)-bP=a-2bP$$