I tried to explain this in Section 4.2, but it appears my message was lost. Here, I am going to clarify the notation a bit. When specifying the inputs to a function, I am going to use square brackets. Evaluation of a function will employ parentheses. That is, $f\!\left[x\right]$ should be read “$f$— a function of $x$” while $f\!\left(y\right)$ should be read “$f$— a function of one variable evaluated at $y$.”

For example, though Wilfred Kaplan states in

*Advanced Calculus (3*that

^{rd}ed.)If $z=f\!\left(x,y\right)$ and $x=g\!\left(u,v\right)$, $y=g\!\left(u,v\right)$, then $$ \frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial u} $$Kaplan also clarifies on the following page that $z\!\left[u,v\right]=f\!\left(g\!\left(u,v\right),h\!\left(u,v\right)\right)$ “is the function whose derivative with respect to $u$ is denoted by ${\partial z}/{\partial u}$.”

That is, when $z$ is evaluated at $\left(g\!\left(u,v\right),h\!\left(u,v\right)\right)$ the result is an entirely new function. Kaplan’s chain rule could be written more clearly (if pedantically)

If $z\!\left[x,y\right]=f\!\left(x,y\right)$ and $x\!\left[u,v\right]=g\!\left(u,v\right)$, $y\!\left[u,v\right]=h\!\left(u,v\right)$, then $$ \frac{\partial z^*\!\left[u,v\right]}{\partial u}=\frac{\partial z\!\left[x,y\right]}{\partial x}\frac{\partial x\!\left[u,v\right]}{\partial u}+\frac{\partial z\!\left[x,y\right]}{\partial y}\frac{\partial y\!\left[u,v\right]}{\partial u} $$ where $z^*\!\left[u,v\right]=z\!\left(g\!\left(u,v\right),h\!\left(u,v\right)\right)=f\!\left(g\!\left(u,v\right),h\!\left(u,v\right)\right)$Likewise, though Standish and Keen write ${\partial P}/{\partial q_i}$ this is not entirely proper. What they mean is not $P\!\left[Q^d\right]$ but $p\!\left[\bf q\right]=P\!\left(\sum_j{q_j}\right)$. Even if market clearance requires $Q_d=\sum_j{q_j}$, Standish and Keen cannot so effortlessly slip between $P\!\left[Q^d\right]$ and $p\!\left[\bf q\right]$. They are completely different functions.

Their response suggests that this critique is overly deferential to Standish and Keen. Even if we clean up their response to read,

Standard Neoclassical pedagogy teaches that, regardless of market structure, an individual firm in an industry will maximize its profits by equating its marginal revenue (the derivative of its revenue to its output) to its marginal cost (the derivative of its total cost of production with respect to its output): \begin{equation} \frac{\partial\pi_i\!\left[\bf q\right]}{\partial q_i}=\frac{\partial\mathrm{TR}_i\!\left[\bf q\right]}{\partial q_i}-\frac{\partial\mathrm{TC}_i\!\left[q_i\right]}{\partial q_i}\label{eq:eq1} \end{equation} where $\mathrm{TR}_i\!\left[\bf q\right]=P\!\left(\sum_j{q_j}\right)\cdot q_i$ and $P\!\left[Q^d\right]$ is inverse demand.it makes no general sense “irrespective of market structure” as opposed to Cournot oligopoly in particular. It is not clear that Standish and Keen refer to any actual “Neoclassical pedagogy.” For example, textbook economics teaches that total revenues for a perfectly price-discriminating monopoly are given by $$ \mathrm{TR}\!\left[q\right]=\int_0^q{P\!\left(Q^d\right)dQ^d} $$ There is no obvious way to reconcile this result with the claims of Standish and Keen. It may be that Standish and Keen are correct and “Neoclassical pedagogy” contradicts textbook theory. Alternatively, Standish and Keen themselves may be confused. Their premise

*itself*is extraordinary and so it is incumbent upon Standish and Keen to support it clearly. At some point they must back up their perplexing idea that “Standard Neoclassical pedagogy teaches that, regardless of market structure” there exists any single formula for total revenues.

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