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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

  • $\begingroup$ Hint: Use product rule and rearrange. $\endgroup$
    – Herr K.
    Commented Apr 16, 2019 at 20:39

1 Answer 1


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.

  • $\begingroup$ I do not deserve you. I couldn't ask for a better answer. Thank you soooo much. You are a legend :) $\endgroup$ Commented Apr 17, 2019 at 7:12
  • $\begingroup$ 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) $\endgroup$ Commented Apr 17, 2019 at 8:33

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