And what's fallen out when we do that, because in each case, one of the first derivatives gives us the entropy.

当我们这样做时就得到了结果，因为在这些例子中，一阶导数是熵。

麻省理工公开课 - 热力学与动力学课程节选

So I'm hoping you guys are comfortable with the notion of taking one or two or any number of derivatives.

我希望你们,能习惯求一阶,二阶,或者任意阶导数的概念

耶鲁公开课 - 基础物理课程节选

Okay, good, so what we're going to do is we're going to differentiate this thing to find a first order condition.

好吧,那我们接下来,对它求导后找出一阶条件

耶鲁公开课 - 博弈论课程节选

Above some temperature is going to be positive, below some temperature is going to be negative.

低于这一温度它是负数，我们一般忽略这些高阶项。

麻省理工公开课 - 热力学与动力学课程节选

So the slope of the guess is the first derivative.

因此斜率等于此处的一阶导数。

麻省理工公开课 - 计算机科学及编程导论课程节选

One is that there's four different colors and so you can see red, blue, yellow, green here - four different colors and that's all there are, there aren't more than four.

一是这些梯阶有四种不同颜色,你可以看到红 蓝 黄 绿,恰恰只有四种不同颜色,不多不少

耶鲁公开课 - 生物医学工程探索课程节选

So if we differentiate this object, I'm gonna find a first order condition in a second.

想要求它的导数,先让我想想一阶条件

耶鲁公开课 - 博弈论课程节选

This was a first order condition or a first order necessary condition.

这个只是一阶条件,是必要条件

耶鲁公开课 - 博弈论课程节选

The first one is velocity, the second one is acceleration.

一阶导数叫速度,二阶导数叫加速度

耶鲁公开课 - 基础物理课程节选

Once you can take one derivative, you can take any number of derivatives and the derivative of the velocity is called the acceleration, and we write it as the second derivative of position.

只要你能求一阶导数,你就能求任意阶的导数,速度的导数被称为"加速度",我们把它写成位移的二阶导数

耶鲁公开课 - 基础物理课程节选

So here it is, I've got my first order condition.

到这里我满足了一阶条件

耶鲁公开课 - 博弈论课程节选

So I differentiated this object, this is my first derivative and I set it equal to 0 Now in a second I'm going to work with that, but I want to make sure i'm going to find a maximum and not a minimum, so how do I make sure I'm finding a maximum and not a minimum?

这样我就对它求出导数了,这是一阶导数,令它等于0,一会我们就要计算了,但我先确定一下是最大值还是最小值,我怎么确定是最大值还是最小值呢

耶鲁公开课 - 博弈论课程节选

Well, unfortunately, we know this is not the right answer, because if you take the first derivative, I get 2t.

遗憾的是,我们知道这还不是正确答案,因为如果对它求一阶导数,会得到2t

耶鲁公开课 - 基础物理课程节选

I differentiate a second time and check the sign, so the second order condition, I differentiate this expression again with respect to q1.

我们对它进行二次求导然后看符号,这个式子的二阶导数,就是一阶导数再对q1进行求导

耶鲁公开课 - 博弈论课程节选

Once you've got that, one derivative will give you the velocity, then in a crunch you can eliminate t and put it into this formula.

一旦你得到了这个,求一阶导数就能得到速度,然后你可以消去t,把它代入这个式子

耶鲁公开课 - 基础物理课程节选

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

To make this a first order condition, I'm going to say "at the best response," put a hat over the 1.

为了达到一阶条件,我说在最佳对策下,在S1上写个帽

耶鲁公开课 - 博弈论课程节选

It tells me that the best response to S2 is the ?1 that solves this equation, that solves this first order condition.

我们得出S2的最佳对策是?1,?1是这个方程的解,它满足一阶条件

耶鲁公开课 - 博弈论课程节选