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

And, of course, see that either way we do that we'll have an equality.

利用两种不同的顺序求二阶导数,就可以得到一个等式。

麻省理工公开课 - 热力学与动力学课程节选

Normally, I will give you a function and tell you to take any number of derivatives.

通常情况下,我会给你一个函数,然后让你求任意阶的导数

耶鲁公开课 - 基础物理课程节选

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

Let's sneak in one more derivative here, which is to take the derivative of the derivative.

我们再求一次导数,也就是对导数求导

耶鲁公开课 - 基础物理课程节选

Of course, you can take a function and take derivatives any number of times.

当然,你可以随意拿一个函数,对它求任意阶的导数

耶鲁公开课 - 基础物理课程节选

And all we did, further, is take that second derivative.

总而言之，我们所做的就是求二阶导数。

麻省理工公开课 - 热力学与动力学课程节选

Now let's take it in the other order.

我们用另一种顺序求二阶偏导数。

麻省理工公开课 - 热力学与动力学课程节选

And then we can take the derivative with respect to temperature, it's just R over molar volume minus b.

这样我们求,压强对温度的偏导数，结果等于R除以摩尔体积V杠减去b的差。

麻省理工公开课 - 热力学与动力学课程节选

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

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

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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

So d/dT of dA/dV, just like this.

即对dA/dV求对温度T的偏导数。

麻省理工公开课 - 热力学与动力学课程节选

If I take the second derivative I get 2, but I want to get a and not 2.

如果我再求二阶导数就能得到2,但我想要a不想得到2

耶鲁公开课 - 基础物理课程节选

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

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This way when I take two derivatives, there will be no t left.

这样的话,当我求二阶导数时,t就消失了

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You can take a derivative of the derivative and you can get the acceleration vector, will be d^2r over dt^2, and you can also write it as dv over dt.

你可以对导数再求一次导,你就可以得到加速度矢量,也就是 d^2 r / dt^2,你也可以写成 dv / dt

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Mathematically complete problem is that you can find the function x of t by saying that the second derivative of the function is equal to -k over m times the function.

这个数学问题就是,你可以求出函数 x,只要令函数的二阶导数,等于 -k 除以 m 再乘以函数

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That means, when I take two derivatives, I want to get a, then you should know enough calculus to know it has to be something like at^, and half comes from taking two derivatives.

也就是说,当我想求二阶导时,得到了a,你应该有足够的微积分知识,才能知道必须有类似at^的项,而这个1/2则是因为求了两次导数

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Take a derivative of S1.

求出S1的导数

耶鲁公开课 - 博弈论课程节选

You can actually take derivatives of a vector with time.

你可以直接求位矢对时间的导数

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You can add something else to the answer without invalidating it, and that is anything with one power of t in it, because if you take one derivative it'll survive, but if you take two derivatives, it'll get wiped out.

你可以往答案后面加上其它东西,而且不影响结果,也就是任何t的一次项,因为你求一次导数之后它还在,但如果你求两次导数,这一项就不存在了

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