The Chain Rule
AP Calculus AB Topic 3 6:49 English narration · English + 中文 subtitles burned in
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Look closely at this function: the sine of x squared.
仔细看这个函数：x 平方的正弦。
It is really two functions, one inside the other.
它其实是两个函数，一个套在另一个里面。
First, square the input. Then take the sine of the result.
先把输入平方，再对结果取正弦。
A function built inside another function is called a composite.
一个套在另一个函数里面的函数，叫做复合函数。
To differentiate it, we need a new tool — the chain rule — which peels the layers one at a time, from the outside in.
要对它求导，我们需要一个新工具——链式法则——它把这些层从外到里，一层一层地剥开。
Today we master the chain rule — the tool for differentiating functions nested inside functions — and see how it unlocks implicit and inverse derivatives too.
今天我们来掌握链式法则——它是对嵌套函数求导的工具—— 并看看它如何同样解开隐函数和反函数的导数。
Let's begin.
让我们开始吧。
A composite function feeds one function into another.
复合函数把一个函数送进另一个函数。
Start with the input x.
从输入 x 开始。
The inner function acts first — here, squaring gives x squared.
内函数先起作用—— 这里，平方得到 x 平方。
Then the outer function acts on that result — the sine.
然后外函数作用在这个结果上——也就是正弦。
We write it as f of g of x: the outer f, wrapped around the inner g.
我们把它写成 f g x：外面的 f，套住里面的 g。
Reading this structure — which is inside, which is outside — is the whole battle.
读懂这个结构——谁在里面，谁在外面—— 就是整场仗的关键。
Here is the chain rule.
这就是链式法则。
The derivative of f of g of x equals f prime of g of x, times g prime of x.
f g x 的导数，等于 f 撇 g x，乘以 g 撇 x。
In words: take the derivative of the outer function, leaving the inside untouched — then multiply by the derivative of the inside.
用一句话说： 对外函数求导，里面保持不动——然后再乘以里面的导数。
That last factor, the inner derivative, is the piece students always forget.
最后那个因子， 也就是内导数，是学生总会忘的那一部分。
For the sine of x squared, the outer derivative is cosine of x squared, and the inner derivative is two x.
对 x 平方的正弦来说，外导数是 x 平方的余弦， 内导数是二 x。
Multiply them.
把它们相乘。
There is a second way to see it — Leibniz notation.
还有第二种看法——莱布尼茨记号。
Let y depend on u, and u depend on x.
设 y 依赖于 u，u 依赖于 x。
Then d y d x equals d y d u, times d u d x.
那么 d y d x 等于 d y d u，乘以 d u d x。
The intermediate step, d u, appears to cancel, just like fractions.
中间那一步 d u，看起来像分数一样约掉了。
This form is perfect for table questions: when the exam gives you values of the functions and their derivatives, just read them off and multiply.
这个形式最适合表格题：当考试给你各函数及其导数的值时，直接读出来相乘就行。
Let's differentiate the quantity two x squared plus one, all raised to the fifth power.
我们来对 二 x 平方 加一，整体的五次方求导。
The outer function is something to the fifth; the inner is two x squared plus one.
外函数是"某个东西的五次方"； 内函数是 二 x 平方 加一。
Outer derivative first: five times the inside, to the fourth power.
先求外导数：五乘以里面那一块的四次方。
Then times the inner derivative, which is four x.
再乘以内导数，也就是 四 x。
Multiply and tidy up: twenty x, times the quantity, to the fourth power.
相乘并整理：二十 x，乘以那一块的四次方。
Some curves are given implicitly — an equation in x and y, not solved for y.
有些曲线是隐式给出的——一个关于 x 和 y 的方程，并没有解出 y。
Take the circle: x squared plus y squared equals twenty-five.
看这个圆：x 平方 加 y 平方 等于二十五。
To find the slope, differentiate both sides with respect to x.
要求斜率，就对两边同时关于 x 求导。
But y is secretly a function of x, so each y-term needs the chain rule — differentiate, then multiply by d y d x.
但 y 其实是 x 的函数，所以每一个含 y 的项都要用链式法则——先求导，再乘以 d y d x。
That gives two x, plus two y times d y d x, equals zero.
这就得到 二 x，加上 二 y 乘以 d y d x，等于零。
Now solve: d y d x equals minus x over y.
现在解出：d y d x 等于负的 x 比 y。
Now use it.
现在来用它。
Find the tangent to that circle at the point three, four.
求那个圆在点 三，四 处的切线。
Our slope formula gives minus x over y, which is minus three over four.
我们的斜率公式给出负的 x 比 y， 也就是负四分之三。
So the tangent line is y minus four, equals minus three-quarters, times the quantity x minus three.
所以切线是 y 减四，等于负四分之三，乘以 x 减三。
And notice: this tangent is perpendicular to the radius — exactly as geometry promises.
并且注意：这条切线与半径垂直——正如几何所保证的那样。
Always look for a sanity check like that.
永远要找这样的检验。
That circle was the easy case: no x y term, so no product rule and nothing to factor.
那个圆是最简单的情况：没有 x y 这一项，所以既不需要乘积法则，也没有东西可以提取。
A real exam question says, show that d y d x equals a given expression.
真正的考题会说：证明 d y d x 等于某个给定的表达式。
Take x squared, plus x y, plus y squared, equals seven.
取 x 平方，加 x y，加 y 平方， 等于七。
Differentiate every term, using the product rule on x y.
对每一项求导，在 x y 上使用乘积法则。
Then collect the d y d x terms on one side, factor d y d x out, and divide.
然后把所有含 d y d x 的项收到一边， 把 d y d x 提出来，再相除。
Every one of those steps must be on the page: the algebra is what earns the marks.
这每一步都必须写在卷面上：得分的正是这些代数步骤。
The follow-up is nearly always the same.
接下来的追问几乎总是同一个。
Where is the tangent horizontal, and where is it vertical?
切线在哪里是水平的，在哪里是竖直的？
Horizontal means the slope is zero, and a fraction is zero only when its numerator is zero: so two x plus y equals zero.
水平意味着斜率为零， 而一个分数只有在分子为零时才为零：所以 二 x 加 y 等于零。
Vertical means the slope is undefined, which happens when the denominator is zero: x plus two y equals zero.
竖直意味着斜率不存在， 这发生在分母为零的时候：x 加 二 y 等于零。
Two lines of work, two easy marks.
两行计算，两分轻松到手。
The chain rule also links a function to its inverse.
链式法则也把一个函数和它的反函数联系起来。
If g undoes f, then their derivatives are reciprocals: g prime of x equals one over f prime, at the matching point.
如果 g 抵消 f，那么它们的导数互为倒数： g 撇 x 等于 f 撇 分之一，在对应的点上取值。
The graphs are mirror images across the line y equals x, so their slopes at matching points are reciprocals.
两个图象关于直线 y 等于 x 互为镜像， 所以它们在对应点的斜率互为倒数。
For example, if f of two is five and f prime of two is three, then g prime of five is one third.
举个例子，如果 f 二 等于五，f 撇 二 等于三， 那么 g 撇 五 就等于三分之一。
The same idea gives the inverse trigonometric derivatives — three to know.
同样的思路给出反三角函数的导数——有三个要记住。
The derivative of arcsine is one over the square root of one minus x squared.
反正弦的导数是 一减 x 平方 的平方根 分之一。
The derivative of arctangent is one over one plus x squared.
反正切的导数是 一加 x 平方 分之一。
And arcsecant is one over the absolute value of x, times the square root of x squared minus one.
反正割是 x 的绝对值 乘以 x 平方减一 的平方根，分之一。
When the input is itself a function, combine them with the chain rule.
当输入本身是一个函数时，就把它们和链式法则结合起来。
Real problems mix rules, so always read the structure first, from the outside in.
真实的问题会混用多种法则，所以永远先读清结构，从外到里。
Is it a sum?
是和吗？
Differentiate term by term.
逐项求导。
A product or a quotient?
是乘积或商吗？
Use that rule — and expect a chain rule inside.
用那条法则——并且预期里面还要用链式法则。
Something nested inside something?
是某个东西套在某个东西里面吗？
Chain rule.
用链式法则。
Given implicitly?
是隐式给出的吗？
Implicit differentiation.
用隐函数求导。
Name the outermost operation, apply its rule, and work your way in.
先说出最外层的运算，套用它的法则，再一层层往里做。
Finally, we can keep going.
最后，我们可以继续下去。
Differentiate the derivative, and you get the second derivative — written f double prime.
对导数再求一次导，就得到二阶导数——写作 f 两撇。
It measures how the slope itself is changing.
它衡量的是斜率本身如何变化。
That idea drives concavity, and acceleration, in the units ahead.
这个思想会在后面的单元里推动凹凸性和加速度。
Just differentiate again, carefully, and substitute where needed.
只要再小心地求一次导，并在需要时代入即可。
Before you go, three marks to protect.
结束之前，三个要守住的分。
First, the chain rule — never forget the inner derivative, the most-missed factor on the whole exam.
第一，链式法则——永远别忘内导数， 这是整场考试最常漏掉的因子。
Second, for implicit differentiation, show every algebra step: attach d y d x every time you differentiate a y, then solve.
第二，隐函数求导时，写出每一步代数：每次对 y 求导都要乘上 d y d x，再求解。
Third, an inverse function's slope is the reciprocal of the original's, at the matching point.
第三，反函数的斜率，是原函数在对应点斜率的倒数。
Get these right, and this topic is yours.
把这些做对，这个专题就是你的了。