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And all this is, is saying that when you take a mixed second derivative, it doesn't matter in which order you take the two derivatives.
麦克斯韦关系的本质是,当你考虑混合的二阶导数时，求导的顺序不影响最后的结果,现在，我们利用这些关系。
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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
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What's a function, what's a derivative, what's a second derivative, how to take derivatives of elementary functions, how to do elementary integrals.
什么是函数,什么是导数,什么是二阶导数,如何对初等函数求导,如何进行初等积分
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And I can take this derivative.
我们这样求导。
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We differentiate dE by dr.
用E对r求导。
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Change the independent variable, find the change in the function, take the ratio and that's the derivative.
改变自变量,算出函数的变量,计算比值,这就是求导
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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
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If I give you a function, you know how to take the derivative.
如果我给你一个函数,你就知道如何求导
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Good, take a derivative and Set it equal to zero.
好的,求导,然后呢,令它等于0
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Everyone happy with the way I differentiated?
我这么求导大家都会是吧
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All right, so we differentiate.
好了我们来求导吧
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In other words, the order of taking the derivatives with respect to pressure and temperature doesn't matter And what this will show is that dS/dp dS/dp at constant temperature, here we saw how entropy varies with volume, this is going to show us how it varies with pressure.
换句话说,对温度和压强的求导顺序无关紧要,结果会表明,恒定温度下的，对应我们上面看到的，熵如何随着体积变化，这个式子告诉我们，熵如何随着压强变化。
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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
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The basic idea in solving these equations and integrating is you find one answer, so then when you take enough derivatives, the function does what it's supposed to do.
解决这类方程以及积分的基本思想就是,你求出一个解,然后进行多次求导,求导的结果就满足条件
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If you're taking only one derivative you can add a constant.
如果只进行一次求导,你可以添加一个常数项
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If you took the derivative of this, you will get the velocity at time t, it would be: v=v0+at.
如果你对它求导,你就可以知道 t 时刻的速度,即,v=v0+at
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We're going to differentiate this with respect to q1 and then what?
对q1求导之后还要做什么
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i and j are constant, ignore them when you take derivatives.
和 j 是常矢量,求导的时候可以忽略它们
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You don't want to cancel the d in the derivative.
你不能在求导的时候约掉d
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If you want to know how fast it's moving at a given time, if you want to know the velocity, I just take the derivative of this answer, which is 10-10t.
如果你想知道它在给定时刻的运动有多快,如果你想知道它的速度,我只要对这个式子求导,得到10-10t
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I'm going to have 2 still, and then this S1 is going to become a 1, and this S1 here is going to become a plus B S2, everyone happy with that?
还保留,S1求导后是1,这里的S1会变成B*S2,大家都会吧
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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
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Rule for taking derivatives, if you want to do this mindless application of calculus, you can do it.
按求导法则来做,如果你想解决这个微积分的简单应用,你可以很容易做到
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If you're taking two derivatives you can add a constant and something linear in t.
如果你要二次求导,你可以添加一个一次项和常数项
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But I can add to this some number, like 96, that'll still have the property that if you take two derivatives, you're going to get the same acceleration.
但是我可以在后面加上某个数,比如96,经过两次求导你仍会得到,同样的加速度
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But more and more vectors can be manufactured by taking derivatives.
但是更多的矢量,可以通过求导得到
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v is obtained by taking derivative of this expression.
可以通过这个式子求导得到
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I differentiate again, right?
需要再次求导,是吧
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That was just differentiating.
这就是求导过程
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At any point on the graph you can take the derivative, which will be tangent to the curve at each point, and its numerical value will be what you can call the instantaneous velocity of that point and you can take the derivative over the derivative and call it the acceleration.
在图上的任意一点,你可以进行求导,得到曲线上每一点的切线斜率,所得到的数值,即为该点处的瞬时速度,然后你再求一次导,得出它的加速度
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