# Clarification of partial derivative notation $\frac{\partial q(p,q)}{\partial p}$

$$\left.\frac{\partial q(p,q)}{\partial p}\right|_q = 0$$

Where there are three variables, $p$, $q$, and $a$, with $a=pq$.

My problem is with what is meant by "$q(p,q)$", I can see how $q$ can be a function of $a$ and $p$ but I don't understand how $q$ can be a function of $q$.

• " I don't understand how q can be a function of q." Actually, the answer uses that $q$ is a function of $(p,q)$, not only of $q$, namely, $q=f(p,q)$ with $$f(u,v)=v$$ This representation, and other similar ones, are really all there is to understand to the situation (and to the answer you mention). – Did Mar 12 '18 at 6:53
• @Did Ok I see, that clears things up. Thanks. – eugenhu Mar 12 '18 at 7:17

Expanding on @Did's comment (with what I believe is my understanding).

In the equation, the symbol $q$ is being reused as both '$q$ the variable' and '$q(p, q)$ the implicit function (from the relation $a=pq$) returning the value of $q$ given the values of $p$ and $q$'. The value of $q$ given $p$ and $q$ is trivially just $q$, in other words $q(p, q)$ can be thought of as $q(p', q') = f(p', q')$ (dashes added to differentiate the function from its parameters), with:

$$f(p', q') = q'$$

Another way to think of it could be beginning with the more intuitive function $q(p,a)$, which returns the value of $q$ given the values of $p$ and $a$: $$q(p, a) = \frac{a}{p}$$ And substituting $a$ for $a(p,q)$, where $a(p,q) = pq$: $$q(p, a(p, q)) = \frac{a(p, q)}{p} = \frac{pq}{p}=q$$ Noticing that $q(p, a(p, q))$ is now just a function of $p$ and $q$ so redefining $q$ yet again:

$$q(p,q)=q(p,a(p,q))=q$$

So taking the partial derivative of $q(p,q)$ (the function) with respect to $p$, while keeping $q$ (the variable) constant, yields $0$.

($q$ can be kept constant as $a$ is free to vary)

Thought I'd answer my own question providing my understanding along with what's been said in the comments to possibly help provide a clearer explanation for future readers.