Derivative of total revenue I have a maths related question and I was hoping that you could help me out. I'm trying to understand why, when differentiating the total revenue function, one is required to differentiate implicitly.
The total revenue formula is $R=PQ$, where $R$ is total revenue, $P$ is the unit price and $Q$ is the quantity. I know how to differentiate implicitly... $\frac{dR}{dP}=Q+P\frac{dQ}{dP}$ ... but I don't quite understand why it's needed. That is, why don't we treat $Q$ like a constant, giving $\frac{dR}{dP}=Q$.
I also know that the revenue function can be expressed as $R(Q)=P(Q)\times Q$, so does this tell me that, because $P$ is a function of $Q$, I cannot treat it as a constant? What is it about this function that requires me to use implicit differentiation? Thanks in advance!
 A: really this is just differentiation, so no need to get too confused here. many relations encountered in science can be expressed in the form:
$$
y = f(x) \tag{1}
$$
however a more symmetrical expression is sometimes more appropriate;
$$
g(x,y) = 0 \tag{2}
$$
it is trivial (though un-necessary) to rewrite a relation in form 1 as a relation of the second type, viz:
$$
h(x,y) = f(x)-y = 0
$$ 
however it may not be so simple to translate form 2 into form 1, and often this can only be done under certain restrictions.
using form 2, suppose $x$ and $y$ are both functions (in form 1) of a parameter $t$.
then differentiation by $t$ gives:
$$
\frac{dg}{dt} = \frac{\partial g}{\partial x} \frac{dx}{dt} + \frac{\partial g}{\partial y} \frac{dy}{dt} \tag{3}
$$
a special case of this is when the role of the parameter $t$ is taken by the variable $x$. in this case $\frac{dx}{dt}=1$ and (3) becomes
$$
\frac{dg}{dx} = \frac{\partial g}{\partial x} + \frac{\partial g}{\partial y} \frac{dy}{dx} \tag{34}
$$
it is usually the distinction between $\frac{\partial g}{\partial x}$ and $ \frac{dg}{dx}$ which causes difficulty for beginners. 
