来自《耶鲁大学开放课程：博弈论 第14集 第14集》

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  So what we're going to do
  我们现在要做的
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  is we're going to differentiate this thing with respect to Q1.
  是继续做关于Q1的求导
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  So differentiating with respect to Q1--let's do it up here --,
  求以Q1为自变量的导数 我们就在这写吧
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  we get [A -C]/2 from this term
  由此式我们能得到[A-C]/2
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  and from the –B Q12/2 we're going to get
  通过–BQ12/2我们能得到
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  --the two's are going to cancel rather pleasantly
  这两项恰好能相消
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  --so we're going to get B Q1 from that term.
  从此式中我们能得到BQ1
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  Everyone with me so far?
  能跟得上我吧
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  I'm going through this in sort of slow steps,
  我再放缓点节奏解释一遍
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  I agree it's not exciting
  这个不怎么好玩
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  but I want to make sure I don't make a mistake.
  但我要确保我没有犯错误
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  So to turn this into a first order condition
  想把它作为一阶导数的条件