<p>To elaborate on @HerrK’s comment:</p>

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

CC BY-SA 4.0
2019-04-17T07:12:29.873Z
I do not deserve you. I couldn't ask for a better answer. Thank you soooo much. You are a legend :)
seekingknowledge111
CC BY-SA 4.0
2019-04-17T08:33:59.270Z
If I wanted to find the second derivative would it be...
LHS: dQ/dx and
RHS: dQ/dx + qi*(d2Q/dqi2)? 
(sorry I'm new to this website and don't know how to format stuff, the d2Q/dqi2 just means the double derivative)
seekingknowledge111
CC BY-SA 4.0
2019-04-16T23:44:50.977Z
2019-04-16T23:44:50.977Z
Franklin Pezzuti Dyer
<p><a href="https://i.sstatic.net/pJ3QJ.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/pJ3QJ.png" alt="enter image description here"></a></p>

<p>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</p>

CC BY-SA 4.0
2019-04-16T20:39:05.810Z
Hint: Use [product rule](https://en.wikipedia.org/wiki/Product_rule) and rearrange.
Herr K.
<microeconomics><game-theory><regulation>
Cournot's game profit maximisation first order condition question
seekingknowledge111
# Cournot's game profit maximisation first order condition question

(Asked by: username_0 on 2019-04-16)

[![enter image description here](https://i.sstatic.net/pJ3QJ.png)](https://i.sstatic.net/pJ3QJ.png)

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

**Tags:** microeconomics, game-theory, regulation

> **Comments:**
> **username_1**: Hint: Use [product rule](https://en.wikipedia.org/wiki/Product_rule) and rearrange.  

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## 1 answers

### Answer by username_2 on 2019-04-16. Score: 4

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.

> **Comments:**
> **username_0**: I do not deserve you. I couldn't ask for a better answer. Thank you soooo much. You are a legend :)  
> **username_0**: If I wanted to find the second derivative would it be... LHS: dQ/dx and RHS: dQ/dx + qi*(d2Q/dqi2)?  (sorry I'm new to this website and don't know how to format stuff, the d2Q/dqi2 just means the double derivative)  

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