📚 The Chain Rule for Two Parameters | 两个参数的链式法则
In multivariable calculus, the chain rule extends to functions of several variables that depend on other variables. Here we focus on the case where a function z depends on two intermediate variables x and y, which in turn depend on one or two parameters. This topic is crucial for IB Mathematics: Analysis and Approaches, and it reappears in related rates, implicit differentiation and partial differential equations.
在多变量微积分中，链式法则扩展到依赖于其他变量的多元函数。这里我们关注一种情形：函数 z 依赖于两个中间变量 x 和 y，而 x 和 y 又依赖于一个或两个参数。本主题对于 IB 数学分析与方法至关重要，并会再次出现在相关变化率、隐函数求导和偏微分方程中。
1. From Single-Variable to Multivariable Chain Rule | 从一元链式法则到多元链式法则
f the composite function is the product of the two derivatives. The rate of change of y with respect to x is the rate of change of y with respect to u multiplied by the rate of change of u with respect to x. Every dependency is a single chain, so a single product is enough.
在一元微积分中，如果 y = f(u) 且 u = g(x)，那么复合函数的导数就是两个导数的乘积。y 关于 x 的变化率，等于 y 关于 u 的变化率乘以 u 关于 x 的变化率。每条依赖关系都只是一条单链，所以一个乘积就够了。
Chain rule is needed because a change in x first causes a change in u, which then causes a change in y. Writing that idea algebraically gives the familiar result below, where the two fractions are ordinary derivatives and not partial derivatives.
我们需要链式法则，是因为 x 的变化先引起 u 的变化，再引起 y 的变化。把这个想法写成代数形式就得到下面熟悉的结果，其中两个分式都是常导数，而不是偏导数。
dy/dx = (dy/du) · (du/dx)
ntermediate variables instead of one. A small change in the parameter t changes x and changes y; each of those changes feeds into z. Because there are two routes from t to z, two products must be added together rather than one product alone.
现在假设 z = f(x, y) 依赖于两个中间变量，而不是一个。参数 t 的微小变化会改变 x，也会改变 y；这两处变化都会传递到 z。由于从 t 到 z 有两条路径，必须把两个乘积相加，而不是只留一个乘积。
2. The Core Idea: Summing Along Dependency Paths | 核心思想：沿依赖路径求和
the product of the rate at which x changes with t and the partial rate at which z changes with x. The other path runs t → y → z, and its contribution is defined in the same way.
z 关于 t 的总变化率，是沿每条依赖路径的贡献之和。一条路径是 t → x → z，它的贡献是 x 随 t 变化的速率与 z 随 x 变化的偏变化率的乘积。另一条路径是 t → y → z，它的贡献以同样的方式定义。
This "sum of products" structure is the signature of the multivariable chain rule. When a variable can be reached by several routes, the total effect is obtained by adding the effects of all routes; nothing is double counted because each route uses a different partial derivative.
这种"乘积之和"的结构正是多元链式法则的标志。当一个变量可以通过多条路径到达时，总效应就是把所有路径的效应相加；由于每条路径使用的是不同的偏导数，因此不会重复计算。
le gives dz/dt = y · dx/dt + x · dy/dt, which is exactly the product rule for derivatives.
该公式也是乘积法则的严格推广。如果 z = xy，且 x = g(t)、y = h(t)，链式法则给出 dz/dt = y · dx/dt + x · dy/dt，这恰好就是导数的乘积法则。
3. Statement of the Chain Rule for One Parameter | 单个参数的链式法则陈述
ve is given as follows. All partial derivatives are evaluated at the point (x, y) = (g(t), h(t)).
如果 z = f(x, y) 可微，且 x = g(t)、y = h(t) 是 t 的可微函数，那么 z 是 t 的可微函数，其全导数如下所示。所有偏导数都在点 (x, y) = (g(t), h(t)) 处取值。
dz/dt = (∂z/∂x) · (dx/dt) + (∂z/∂y) · (dy/dt)
are partial derivatives, while dx/dt and dy/dt are ordinary derivatives, because x and y each depend only on the single variable t.
这通常称为 z 关于 t 的全导数。请特别注意公式中两类不同的导数：∂z/∂x 和 ∂z/∂y 是偏导数，而 dx/dt 和 dy/dt 是常导数，因为 x 和 y 各自只依赖于单变量 t。
d y are substituted. Writing ∂z/∂t would be wrong here, because t is not an argument of f.
一个有用的检验方法是问记号本身是否讲得通：写 dz/dt 是合理的，因为把 x 和 y 代入之后，z 确实只是 t 的函数。这里写 ∂z/∂t 则是错误的，因为 t 并不是 f 的自变量。
4. Tree Diagrams and the Cancelling Mnemonic | 树形图与"约分"记忆法
A dependency (tree) diagram makes the correct formula almost automatic. Put z at the top, x and y on the middle level, and t at the bottom, then draw an edge for every genuine dependency. Every path from t up to z contributes one product of derivatives along its edges.
依赖关系（树形）图几乎能让正确的公式自动出现。把 z 放在顶层，x 和 y 放在中间层，t 放在底层，然后为每一对真实存在的依赖关系画一条边。从 t 向上到 z 的每条路径，都贡献一个沿该路径各边导数之积。
- Path t → x → z gives the term (∂z/∂x) · (dx/dt). | 路径 t → x → z 给出项 (∂z/∂x) · (dx/dt)。
- Path t → y → z gives the term (∂z/∂y) · (dy/dt). | 路径 t → y → z 给出项 (∂z/∂y) · (dy/dt)。
- If there is no edge, there is no term. | 如果没有边，就没有对应的项。
he pattern, not a proof, and it can be misleading when the wrong derivative type is used. Never cancel a ∂ against a d as though they were identical objects.
莱布尼茨记号提示了一种机械化的速记：∂z/∂x 中的 dx 似乎与 dx/dt 中的 dx 相消。这仅是记住公式模式的记忆法，而不是证明；当导数类型用错时，它会产生误导。切勿把 ∂ 与 d 当作完全相同的对象来约分。
5. Worked Example: One Parameter | 例题：单个参数
ind dz/dt using the chain rule, then verify the answer by direct substitution. This is the standard structure of an IB or A-Level extended-response question.
设 z = x²y + y³，其中 x = t²、y = 2t + 1。用链式法则求 dz/dt，然后通过直接代入验证答案。这正是 IB 或 A-Level 长答题的标准结构。
Step 1: compute the partial derivatives, treating the other variable as a constant. Step 2: compute the ordinary derivatives dx/dt and dy/dt. Step 3: substitute everything into the chain rule.
第一步：求偏导数，求 ∂z/∂x 时把 y 视为常数，求 ∂z/∂y 时把 x 视为常数。第二步：求常导数 dx/dt 和 dy/dt。第三步：把全部结果代入链式法则。
∂z/∂x = 2xy, ∂z/∂y = x² + 3y², dx/dt = 2t, dy/dt = 2
dz/dt = 2xy · 2t + (x² + 3y²) · 2 = 4txy + 2x² + 6y²
2t + 1) + (2t + 1)³, and differentiating term by term produces exactly the same polynomial. Agreement between the two methods is strong evidence that no term was dropped.
代入 x = t²、y = 2t + 1 后得到 10t⁴ + 4t³ + 24t² + 24t + 6。也可以先直接代入：z = t⁴(2t + 1) + (2t + 1)³，逐项求导会得到完全相同的多项式。两种方法结果一致，是有力证据表明没有漏掉任何一项。
6. Extending to Two Parameters | 扩展到两个参数
(s, t)) is a function of two variables, so it has two partial derivatives, one with respect to each parameter.
现在让中间变量依赖于两个参数 s 和 t：x = g(s, t)、y = h(s, t)。于是 z = f(g(s, t), h(s, t)) 是两个变量的函数，因此它有两个偏导数，分别关于每个参数。
∂z/∂s = (∂z/∂x) · (∂x/∂s) + (∂z/∂y) · (∂y/∂s)
∂z/∂t = (∂z/∂x) · (∂x/∂t) + (∂z/∂y) · (∂y/∂t)
Nothing conceptually new has happened. The one-parameter formula is applied twice, once while holding t constant and once while holding s constant. The tree diagram simply gains one more level at the bottom, and each path from s or t up to z again contributes one product.
概念上并没有新东西。单参数公式被使用了两次：一次保持 t 不变，一次保持 s 不变。树形图只是在底层多了一个层次，而从 s 或 t 向上到 z 的每条路径同样贡献一个乘积。
For the one-parameter version the final derivative is ordinary because z becomes a function of t alone. For the two-parameter version the final derivatives are partial, because after substitution z still depends on both s and t.
在单参数版本中，最终导数是常导数，因为代入后 z 只成为 t 的函数。在双参数版本中，最终导数是偏导数，因为代入后 z 仍然同时依赖于 s 和 t。
7. Worked Example: Two Parameters | 例题：两个参数
², and the four needed partial derivatives of the intermediate variables are listed below.
设 z = x²y，其中 x = s² + t、y = s t²。求 ∂z/∂s 和 ∂z/∂t。z 的偏导数为 ∂z/∂x = 2xy、∂z/∂y = x²，所需的四个中间变量偏导数如下所列。
∂x/∂s = 2s, ∂x/∂t = 1, ∂y/∂s = t², ∂y/∂t = 2st
chain rule and then substituting x = s² + t and y = s t² gives compact factorised answers.
应用双参数链式法则，再代入 x = s² + t、y = s t²，就得到简洁的因式分解结果。
∂z/∂s = 4s x y + t² x² = t²(s² + t)(5s² + t)
∂z/∂t = 2xy + 2st x² = 2st(s² + t)(s² + 2t)
hich differentiating directly with respect to s and to t reproduces the same expressions. Examiners expect to see the substitution performed at the end, and factorising is usually rewarded.
两个结果都可以通过先代入来检验：z = s t²(s² + t)²，然后直接对 s 和 t 求导会得到相同的表达式。考官期望看到最后完成代入，而且因式分解通常会得到加分。
8. Three or More Intermediate Variables | 三个或更多中间变量
ifferentiable functions of t, then the total derivative has one term for each intermediate variable, and the number of terms always equals the number of variables on the middle level of the tree diagram.
这个模式可以原样推广。如果 z = f(x, y, w)，且 x、y、w 都是 t 的可微函数，那么全导数对每个中间变量各有一项，项数总等于树形图中间层变量的个数。
dz/dt = (∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt) + (∂z/∂w)(dw/dt)
eaches z both directly and through y, so the correct derivative ∂z/∂x must include both routes and is not simply f with respect to its first slot.
若同一个变量在树形图的两个不同层出现，就需要额外的记录工作。例如在 z = f(x, y) 且 y = g(x, t) 时，参数 x 既直接到达 z，又通过 y 到达 z，因此正确的 ∂z/∂x 必须包含这两条路径，而不只是对 f 的第一个位置求导。
depend on t. Each path from t to z multiplies four derivatives together, and all such paths are summed.
学生还应能处理三层结构：z = f(x, y)，而 x = g(u, v)、y = h(u, v)，且 u、v 本身依赖于 t。从 t 到 z 的每条路径都要把四个导数相乘，然后把所有这样的路径相加。
9. Related Rates and Implicit Differentiation | 相关变化率与隐函数求导
dr/dt, because the only dependency path is t → r → V. Every rate in a related rates question is one edge of a tree diagram.
相关变化率问题实际上是伪装起来的链式法则问题。如果球的体积为 V = (4/3)πr³，而半径随时间增长，那么 dV/dt = 4πr² · dr/dt，因为唯一的依赖路径是 t → r → V。相关变化率问题中的每一个速率都是树形图的一条边。
s y as a differentiable function of x, differentiating both sides with respect to x gives the identity below, where the second term is produced by the chain rule acting on y.
隐函数求导来自同一条法则。如果 F(x, y) = 0 把 y 定义为 x 的可微函数，那么两边关于 x 求导就得到下面的恒等式，其中第二项正是链式法则作用于 y 产生的。
Fₓ + Fy · dy/dx = 0 ⇒ dy/dx = − Fₓ / Fy
For a surface F(x, y, z) = 0 the same argument gives ∂z/∂x = − Fₓ / Fz and ∂z/∂y = − Fy / Fz, provided Fz he standard IB and first-year university tool for finding tangent planes and normal lines.
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