Cournot oligopoly - first-order condition I am reading an article that has this description of the first-order condition for a Cournot n-firm game:
Take $P(Q) = Q^{-1}$, $\pi_i(q_i, Q) = (Q^{-1} - c_i)q_i$.
Then the first-order condition for an interior profit-maximizing choice of $q_i$ requires that
$$ \frac{\partial \pi_i}{\partial q_i} + \frac{\partial \pi_i}{\partial Q} = Q^{-1} - c_i - q_iQ^{-2} = 0.$$
I am trying to understand why it is OK to simply take $\frac{\partial \pi_i}{\partial Q}$ ignoring the fact that $Q$ is actually a function of $q_i$. If I expand the term so that $Q = q_i + q_{-i}$ and take the partial derivatives $\frac{\partial \pi_i}{\partial q_i} + \frac{\partial \pi_i}{\partial q_{-i}}$, the solution is not the same as the one that is written in the article. Would appreciate any explanation.
Note :$Q = \sum_{i=1}^n q_i$.
Thus the optimisation problem of firm $i$ is:
\begin{align}
max_{x_i\in\mathbb{R}_+}\pi_i(q_i,Q)
\end{align}
where $\pi_i(q_i,Q) = \big(Q^{-1}(q_i;q_{-i}) -c_i\big)q_i$. Assuming interior solution, the first order condition is
\begin{align}
\frac{\partial\pi_i}{\partial q_i} + \frac{\partial \pi_i}{\partial Q}\frac{\partial Q}{\partial q_i} &amp;= 0\\
\implies (Q^{-1} - c_i) + (-1)Q^{-2}q^*_i* 1 &amp;= 0\\
\implies q^*_i &amp;= Q(1-Qc_i) 
\end{align}