Everyone knows from calculus that if you're trying to find a function about which you know only the derivative, you can always add a constant to one person's answer without changing anything.

学过微积分的人都知道,如果你想根据已知的导数,求出其原函数,你总是可以给某人的答案,随便加一个常数,且不影响结果

耶鲁公开课 - 基础物理课程节选

I would say, okay, this guy wants me to find a function which reduces to the number a when I take two derivatives, and I know somewhere here, this result, which says that when I take a derivative, I lose a power of t.

出题者想要我找出一个函数,它在经过两次求导后得到数字a,我知道这里的某个地方,这个结论告诉我,我每求一次导,t就降一次幂

耶鲁公开课 - 基础物理课程节选

By putting it in a function bug if I find a bug and I change my program I can just run the function again.

把我输入的值放到一个函数里,如果以后我在程序中发现了一个,并对程序进行更改的话，我可以直接。

麻省理工公开课 - 计算机科学及编程导论课程节选

Identify the best responses of each player as a function of the others and find out where they intersect.

把每个人的最佳对策看成别人策略的函数,然后找出函数的交点

耶鲁公开课 - 博弈论课程节选

Now remember, we went through before how it's a state function but to calculate it, you'd need to find a reversible path, along which you can figure this out.

请记住，我们不需要知道它是怎样的一个态函数,只计算就可以了，你需要找到一条可逆路径，沿着这条路径就能算得这个结果。

麻省理工公开课 - 热力学与动力学课程节选

If I gave you the location of a particle as a function of time, you can find the velocity by taking derivatives.

如果我给出物体的位移是时间的函数,你可以通过求导来得到速度

耶鲁公开课 - 基础物理课程节选

Now conversely, how do we find Firm 2's best response as a function of q1?

反过来如何由q1求公司2最佳对策的方程

耶鲁公开课 - 博弈论课程节选

Find player I's payoff is a function of what player II's effort is.

参与人I的收益是参与人II付出的函数

耶鲁公开课 - 博弈论课程节选

We have Player I's payoff as a function of the two efforts and now I want to find out what is Player I's best efforts given a particular choice of S2? Yeah.

参与人I的收益是两人付出的函数,那么对于某个给定的S2,怎么计算参与人I的最佳对策呢

耶鲁公开课 - 博弈论课程节选

Then, to find the meaning of b, we take one derivative of this, dx/dt, that's velocity as a function of time, and if you took the derivative of this guy, you will find as at+b. That's the velocity of the object.

接下来,为了弄清b的含义,我们取它的一阶导数,dx/dt,得到速度作为时间的函数,如果你对它求导的话,你会得到at+b,这就是物体的速度

耶鲁公开课 - 基础物理课程节选

What I did was I differentiated this fairly simple function with respect to q1 and since I want to find a maximum, what I'm going to do is I'm going to set this thing equal to 0.

只不过是对q1求导了,既然我们要求出最大值,只需要令导数等于0就可以了

耶鲁公开课 - 博弈论课程节选