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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.

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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} &= 0\\ \implies (Q^{-1} - c_i) + (-1)Q^{-2}q^*_i* 1 &= 0\\ \implies q^*_i &= Q(1-Qc_i) \end{align}

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The gist/shortened and generalized version of the above answer:

In the context where $Q = \sum_i q_i$ the equation $$ \frac{\partial \pi_i}{\partial q_i} + \frac{\partial \pi_i}{\partial Q} = \frac{\partial \pi_i}{\partial q_i} + \frac{\partial Q}{\partial q_i}\frac{\partial \pi_i}{\partial Q} $$ holds as $$ \frac{\partial Q}{\partial q_i} = 1. $$

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