Cournot's game profit maximisation first order condition question My question is about the last part (I attached the rest for context). I don't understand the final line at all. I thought the first order condition would just be to set the derivative = 0. Very confused! Any help would be greatly appreciated! Thank you :D
To elaborate on @HerrK’s comment:

The product rule of differentiation is
$$\frac{d}{dx}fg=f’g+fg’$$
The text in your question states that you are trying to maximize the quantity
$$[a-Q-c]q_i$$
So, as you predicted, we must differentiate this and find when the derivative equals $0$. If we differentiate this with respect to $q_i$ and use the product rule, setting $f=a-Q-c$ and $g=q_i$, we obtain
$$-\frac{dQ}{dq_i}q_i+[a-Q-c]$$
since the derivative of $a-Q-c$ with respect to $q_i$ is $\frac{dQ}{dq_i}$ (since $a,c$ are constant) and the derivative of $q_i$ w.r.t $q_i$ is $1$. Setting this equal to zero gives
$$-q_i\frac{dQ}{dq_i}+[a-Q-c]=0$$
or, equivalently,
$$a-Q-c=q_i \frac{dQ}{dq_i}$$
which is the last line of the text.