The Chain Rule
AP Calculus BC Topic 3 7:58 English narration · English + 中文 subtitles burned in
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Think of two gears joined together.
想象两个啮合在一起的齿轮。
Turn the first one, and it drives the second.
转动第一个，它就带动第二个。
If the second turns three times as fast as the first, and the first turns twice as fast as your hand, then the second turns six times as fast as your hand.
如果第二个转得比第一个快三倍， 而第一个又比你的手快两倍，那么第二个就比你的手快六倍。
The two rates multiply.
两个变化率相乘。
That is the whole idea of this unit.
这就是本单元的全部核心。
When one function sits inside another, you differentiate the outside, and then you multiply by the rate of the inside.
当一个函数嵌在另一个函数里面时，你先对外层求导， 然后再乘以内层的变化率。
Unit two differentiated single functions.
第二单元对单个函数求导。
Unit three differentiates functions built inside other functions.
第三单元要对嵌在其他函数里面的函数求导。
Let's begin.
让我们开始吧。
Before any rule, read the expression.
在用任何法则之前，先读懂这个式子。
Real exam questions mix several rules, so read it from the outside in, and ask four questions.
真正的考题会混合好几条法则，所以要从外向内读， 并且问四个问题。
Is it a sum?
它是加减吗？
Then differentiate term by term.
那就逐项求导。
Is it a product or a quotient?
它是乘积或商吗？
Then use that rule, and expect the chain rule inside it.
那就用那条法则， 并且预期里面还会用到链式法则。
Is it something inside something?
它是"某个东西套在另一个东西里"吗？
Then it is a composite, and the chain rule leads.
那就是复合函数，由链式法则主导。
Is it given as an equation that is not solved for y?
它是一个没有解出 y 的方程吗？
Then differentiate implicitly.
那就用隐函数求导。
Name the outermost operation, apply its rule, and work inward.
先说出最外层的运算，用它的法则，然后一层一层往里推进。
Here is the rule.
法则是这样的。
The derivative of the outer function, evaluated at the inside, times the derivative of the inside.
外层函数的导数，在内层处取值，再乘以内层的导数。
Say it in plain words.
用大白话说一遍。
Differentiate the outside and leave the inside alone.
对外层求导，内层原样不动。
Then multiply by the derivative of the inside.
然后再乘以内层的导数。
Take the sine of x squared.
以 x 平方的正弦为例。
The outside is sine, so its derivative is cosine of x squared.
外层是正弦，所以它的导数是 x 平方的余弦。
The inside is x squared, so its derivative is two x.
内层是 x 平方，所以它的导数是 2x。
Multiply them.
把它们乘起来。
That second factor is the piece students forget, so always ask two questions.
第二个因子正是学生最容易漏掉的部分，所以永远要问两个问题。
What is the inside?
内层是什么？
And what is its derivative?
它的导数是什么？
The same rule in Leibniz notation is easier to remember.
同一条法则用莱布尼茨记号写出来更好记。
Let y depend on u, and let u depend on x.
设 y 依赖于 u，而 u 依赖于 x。
Then the rate of y with respect to x equals the rate of y with respect to u, times the rate of u with respect to x.
那么 y 对 x 的变化率，等于 y 对 u 的变化率，乘以 u 对 x 的变化率。
The middle variable appears to cancel, like a fraction.
中间那个变量看起来像分数一样被约掉了。
It is not really cancelling, but the memory trick is reliable, and it extends to as many layers as you need.
它并不是真的在约分，但这个记忆技巧很可靠， 而且你需要多少层，它就能延伸多少层。
A very common exam question gives you a table of values instead of formulas.
有一类非常常见的考题给你的是一张数值表，而不是公式。
It lists f, g, and both of their derivatives at a few points, then asks for the derivative of the composite at one point.
表里列出 f、g 以及它们在几个点处的导数， 然后问复合函数在某一点处的导数。
Do not panic.
不要慌。
The rule tells you exactly which numbers to look up.
法则会告诉你到底该查哪些数。
Read the inside first.
先读内层。
Find g at that point.
查出 g 在那一点的值。
Then look up the derivative of f at that answer.
然后查 f 的导数在这个结果处的值。
Then look up the derivative of g at the original point.
再查 g 的导数在原来那一点的值。
Finally, multiply the two numbers.
最后把这两个数乘起来。
Nothing else is needed.
除此之外什么都不需要。
Work one through.
我们完整做一道。
Differentiate the quantity two x squared plus one, all raised to the fifth power.
对"2x 平方加 1"的五次方求导。 先读结构。
Read the structure. The outer function is something to the fifth power, and the inside is two x squared plus one.
外层是"某个东西的五次方"， 内层是 2x 平方加 1。
Differentiate the outside first. Five times the inside to the fourth power.
先对外层求导，得到 5 乘以内层的四次方。
Then multiply by the derivative of the inside, which is four x.
再乘以内层的导数，也就是 4x。
Tidy it up, and the answer is twenty x times the inside to the fourth power.
整理一下，答案就是 20x 乘以内层的四次方。
Some curves are not given as y equals something.
有些曲线并不是以"y 等于某个式子"的形式给出的。
Take the equation of a circle, x squared plus y squared equals twenty five.
以圆的方程为例，x 平方加 y 平方等于 25。
You could solve for y, but you would get two branches and a square root.
你可以解出 y，但会得到两个分支和一个根号。
Implicit differentiation avoids all of that.
隐函数求导完全避开了这些。
Treat y as a function of x, and differentiate both sides with respect to x.
把 y 看作 x 的函数，然后对方程两边同时关于 x 求导。
Every time you differentiate a y term, attach the derivative of y with respect to x.
每当你对一个含 y 的项求导时， 就要附上 y 对 x 的导数。
That factor is just the chain rule at work.
这个因子正是链式法则在起作用。
Then solve for it.
然后解出它。
Here we get minus x over y, which works on the whole circle at once.
这里我们得到负的 x 比 y，它一次就适用于整个圆。
Use it.
用一下。
Find the tangent to that circle at the point three, four.
求这个圆在点 (3， 4) 处的切线。
Substitute both coordinates into the formula.
把两个坐标都代入公式。
The slope there is minus three quarters.
那里的斜率是负四分之三。
Write the tangent in point slope form, and you are done.
用点斜式写出切线，就完成了。
There is a free check here.
这里有一个免费的检验。
The radius to that point has slope four thirds, and our tangent has slope minus three quarters.
到那一点的半径斜率是三分之四， 而我们的切线斜率是负四分之三。
Those two are negative reciprocals, so the tangent is perpendicular to the radius, exactly as geometry promises.
这两者互为负倒数，所以切线垂直于半径， 正如几何学所保证的那样。
Now an exam skill that appears almost every year.
现在讲一个几乎每年都会出现的考试技巧。
The question says, Show that the derivative equals some given expression.
题目说：证明导数等于某个给定的表达式。
Because the target is printed for you, the marks are not for the answer.
因为目标结果已经印在题目上了，所以分数不是给答案的，而是给过程的。
They are for the working. So show every step.
所以每一步都要写出来。
Differentiate both sides.
两边同时求导。
Use the product rule on any mixed term that contains both x and y.
对任何同时含 x 和 y 的混合项用乘积法则。
Then collect every term that carries the derivative of y onto one side.
然后把所有带有 y 的导数的项都收集到一边。 把它提出来。
Factor it out.
再作除法。
Divide. Here, the algebra is the answer, so a correct final line with no working earns almost nothing.
在这里，代数过程就是答案，所以只写一个正确的最后一行、没有过程，几乎得不到分。
Then comes the follow up part.
接着就是后续小问。
Where is the tangent horizontal, and where is it vertical?
切线在哪里水平？ 在哪里竖直？
Look at your fraction.
看你的那个分式。
The tangent is horizontal where the slope is zero, so set the top of the fraction to zero.
切线水平的地方就是斜率为零的地方，所以令分子等于零。
The tangent is vertical where the slope is undefined, so set the bottom to zero. One warning.
切线竖直的地方就是斜率无定义的地方，所以令分母等于零。
Whatever point you find must also satisfy the original equation, so always substitute it back and check that the point really lies on the curve.
有一个提醒。 你求出的任何点都必须同时满足原方程，所以一定要代回去检验，确认这个点真的在曲线上。
A function and its inverse are mirror images across the line y equals x.
一个函数和它的反函数关于直线 y 等于 x 互为镜像。
That reflection swaps rise and run, so the slope of the inverse is the reciprocal of the slope of the original, at the matching point.
这个反射把纵向变化和横向变化互换了， 所以在对应点上，反函数的斜率是原函数斜率的倒数。
Watch which point.
要注意是哪一个点。
The derivative of the inverse at a value equals one over the derivative of the original, evaluated where the original produced that value.
反函数在某个值处的导数，等于 1 除以原函数的导数，而原函数的导数要在 "原函数取到这个值的那一点"处求值。
Suppose f of two is five, and the derivative of f at two is three.
假设 f 在 2 处的值是 5，f 的导数在 2 处是 3。
Then five, two lies on the inverse, and the derivative of the inverse at five is one third.
那么点 (5， 2) 在反函数上，反函数在 5 处的导数就是三分之一。
The same reciprocal idea gives the inverse trigonometric derivatives.
同样的倒数思想给出了反三角函数的导数。
There are three you should know by heart.
有三个你应该背下来。
Arc sine gives one over the square root of one minus x squared.
反正弦的导数是 1 除以根号下 1 减 x 平方。
Arc tangent gives one over one plus x squared.
反正切的导数是 1 除以 1 加 x 平方。
Arc secant gives one over the absolute value of x, times the square root of x squared minus one.
反正割的导数是 1 除以 x 的绝对值再乘以根号下 x 平方减 1。
None of them look trigonometric, which is the surprise.
它们看上去都不像三角函数，这正是让人意外的地方。
Then combine them with the chain rule whenever the input is itself a function.
然后，只要输入本身是一个函数，就把它们和链式法则结合起来用。
Last idea.
最后一个概念。
Differentiate the derivative, and you get the second derivative.
对导数再求一次导，就得到二阶导数。
Repeat, and you get higher orders.
继续重复，就得到更高阶的导数。
Learn all three notations, because exams use them interchangeably.
三种记号都要学会，因为考试会混用它们。
The meaning matters more than the symbol.
含义比符号更重要。
The first derivative measures how the function is changing.
一阶导数衡量函数如何变化。
The second derivative measures how the slope itself is changing.
二阶导数衡量斜率本身如何变化。
It drives concavity in unit five, and acceleration in motion problems.
它在第五单元决定凹凸性，在运动问题中给出加速度。
For an implicit second derivative, differentiate your expression for the first derivative again, then substitute the first derivative back in wherever it appears.
要求隐函数的二阶导数，就把一阶导数的表达式再求一次导， 然后在出现一阶导数的地方把它代回去。
Three marks students lose.
学生最常丢分的三个地方。
First, the inner derivative.
第一，内层导数。
The chain rule has two factors, and the second one is the one that goes missing.
链式法则有两个因子，第二个正是最常漏掉的那个。
Second, in implicit work, attach it every time you differentiate a y term, not just the first time.
第二，在隐函数求导中，每次对含 y 的项求导时都要附上它，不只是第一次。
Third, when you find a horizontal or vertical tangent, check the point actually lies on the curve before you write it down.
第三，当你找到水平或竖直切线时，写下答案之前先检验这个点确实在曲线上。