我们再求一次导数，也就是对导数求导。
Let's sneak in one more derivative here, which is to take the derivative of the derivative.
角加速度等于，角速度的导数。
And angular acceleration is the derivative of angular velocity.
基本上是的，那就是方向导数。
And that's basically, yes, that's the directional derivative.
它是收入Pq相对于数量的导数。
It is the derivative of revenue pq with respect to quantity.
但都是使用导数的逼近公式。
But, it's the usual approximation formula using the derivative.
垂直于梯度的方向上,方向导数为零。
The directional derivative in a direction that's perpendicular to the gradient is basically zero.
速度向量，是位置向量关于时间的导数。
So, the velocity vector is the derivative of a position vector with respect to time.
因此，一个多变量的函数没有通常的导数。
So, a function of several variables doesn't have the usual derivative.
那就是，每个小盒子里烟雾总量的导数和。
Well, that will be the sum of the derivatives of the amounts of smoke inside each little box.
我们还学过加速度，也就是速度向量的导数。
And, we've also learned about acceleration, which is the derivative of velocity.
如果把这个和方程比较,我能得到g的导数。
If I match this with equation two then that will tell me what the derivative of g should be.
以摆线为例，1，-，cos的导数是什么？
If we take the example of the cycloid, well, what's the derivative of one minus cos?
对于任意两分量，混合偏导数相等。
For every pair of components the mixed partials must be the same.
临界点是，偏导数都为零的点。
So, critical points, remember, are the points where all the partial derivatives are zero.
但是，当我不想替换的时候，我可以先求导数。
And when I don't really want to actually substitute arctangents everywhere maybe I would rather deal with the derivatives.
我们叫他f的导数。
我们首先要做的事是,求偏导数，fx是多少？
What we would start doing immediately is taking the partial derivatives. What is f sub x?
事实上，这个方程给出的是，u对t的偏导数。
And, in fact, what the equation will give us is the derivative of u with respect to t.
那么dx和dy前面的系数就是偏导数。
今天的话题是导数。
首先考虑第一个分量，求出它关于x的偏导数。
Ry-Qz Well, so we have to take the first component, Ry minus Qz, and take its partial with respect to x.
这一因素被称为是“二阶导数”，或者是变化率的变化率。
This factor has been dubbed "the second derivative", or the rate of change of the rate of change.
未知量是一个函数，这个方程，将把函数的偏导数联系起来。
So the unknown is a function, and the equation will relate the partial derivatives of that function to each other.
基本上，答案就是，我们不可以用二次导数来验证这个情况。
Basically, the answer for us is that we don't have a second derivative test in this situation.
扭矩除以转动惯量,就会引起角加速度，也就是角速度的导数。
Torque divided by moment of inertia is what will cause the angular acceleration, namely the derivative of angular velocity.
我们还学过微分的链式法则，也就是用其他量来代替这些偏导数。
So, we've learned about differentials and chain rules, which are a way of repackaging these partial derivatives.
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。
And what's fallen out when we do that, because in each case, one of the first derivatives gives us the entropy.
这个类似于，力除以质量等于加速度，就是速度对t的导数。
That is the analog of force divided by mass equals acceleration, which is d over dt of velocity.
这个斜率就是此方向的方向导数,好了，我想我说的已经尽量图形化了。
And,the slope is going to be the directional derivative in that direction OK, I think that's as graphicas I can get.
在我们开始之前，因为我们需要的是导数，取方程两边的导数。
Before we do that, since what we're after is the derivative, let's take the derivative of both sides.
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