📚 Implicit Differentiation | 隐微分法
a technique that allows us to differentiate term by term without first solving for y. Implicit differentiation treats y as an unknown function of x and applies the chain rule whenever we differentiate a y‑term. It is a core A‑ of tangents and normals, locating stationary points, and handling related rates problems.
当曲线由 x 与 y 的方程给出且无法方便地整理成 y = f(x) 的形式时，我们需要一种无需预先解出 y 就能逐项求导的技巧。隐微分法将 y 视为 x 的未知函数，并在对 y 项求导时始终使用链式法则。这是 A‑level 阶段的核心技能，用于求出切线与法线的斜率、寻找驻点以及处理相关变化率问题。
1. What Is Implicit Differentiation? | 什么是隐微分法？
³ = 6xy. Instead of isolating y, we differentiate both sides of the equation with respect to x, remembering that y is a function of x. Every time we differentiate a term in y, we multiply by dy/dx (the chain rule).
隐微分法是在 x 与 y 由方程隐式定义（例如 x² + y² = 25 或 x³ + y³ = 6xy）时求 dy/dx 的方法。我们不将 y 分离出来，而是对等式两边关于 x 求导，并牢记 y 是 x 的函数。每次对含有 y 的项求导时，都要乘以 dy/dx（链式法则）。
2. Explicit vs Implicit Functions | 显函数与隐函数
An explicit function gives y directly, like y = √25 − x² ritten as a single explicit function without splitting into multiple branches. Implicit form is more natural for circles, ellipses, and many curves encountered in coordinate geometry.
显函数直接给出 y，例如 y = √25 − x²。隐式方程则将 x 和 y 混合在一起，如 x² + y² = 25，而且常常无法在不拆分成多个分支的情况下写成一个单一的显函数。对于圆、椭圆以及坐标几何中常见的许多曲线，隐式形式更为自然。
3. The Chain Rule Connection | 链式法则的联系
get 3y², then multiply by dy/dx. Hence d/dx (y³) = 3y² dy/dx. The same idea applies to any function of y: d/dx [f(y)] = f'(y) dy/dx.
隐微分法的关键在于链式法则。如果我们需要对 x 求导 y³，先对 y 求导得到 3y²，再乘以 dy/dx。因此 d/dx (y³) = 3y² dy/dx。这一思路适用于任何关于 y 的函数：d/dx [f(y)] = f'(y) dy/dx。
4. Basic Steps for Implicit Differentiation | 隐微分法的基本步骤
Step‑by‑step: (1) Differentiate every term with respect to x. For x‑only terms, use standard rules. For y‑only terms, attach dy/dx. For mixed terms like xy, apply the product rule and remember to attach dy/dx to the y factor. (2) After differentiation, collect all terms containing dy/dx on one side of the equation. (3) Factor out dy/dx. (4) Solve for dy/dx by dividing through by the remaining factor.
逐步操作：(1) 对每一项关于 x 求导。仅含 x 的项使用常规求导法则。仅含 y 的项要附上 dy/dx。对于 xy 这样的混合项，使用乘法法则并记得对 y 因子附上 dy/dx。(2) 求导后，将所有包含 dy/dx 的项移到等式的一边。(3) 提取公因子 dy/dx。(4) 用剩余因子除过去，解出 dy/dx。
5. Differentiating y with Respect to x | 对 x 求导 y
d/dx (y) = dy/dx. For powers: d/dx (yⁿ) = n yⁿ⁻¹ dy/dx. For exponentials and logs: d/dx (eʸ) = eʸ provided y > 0. Trigonometric functions follow the same idea: d/dx (sin y) = cos y dy/dx.
最简单的情况是对 y 自身求导：d/dx (y) = dy/dx。对于幂函数：d/dx (yⁿ) = n yⁿ⁻¹ dy/dx。对于指数函数与对数函数：d/dx (eʸ) = eʸ dy/dx；在 y > 0 时 d/dx (ln y) = (1/y) dy/dx。三角函数遵循相同思路：d/dx (sin y) = cos y dy/dx。
6. Worked Example 1: Circle Equation | 例题1：圆的方程
g gives 2y dy/dx = −2x, so dy/dx = −x/y. This result makes sense: the gradient of a circle's tangent depends on both coordinates, and it is undefined when y = 0 (vertical tangents).
已知 x² + y² = 25，对两边求导：2x + 2y dy/dx = 0。整理得 2y dy/dx = −2x，因此 dy/dx = −x/y。这个结果是合理的：圆的切线斜率依赖于两个坐标，并且当 y = 0 时无定义（此时切线垂直）。
7. Worked Example 2: Product Rule in Implicit Form | 例题2：隐函数中的乘法法则
uating both sides: 3x² + 3y² dy/dx = 6x dy/dx + 6y. Collect dy/dx terms: 3y² dy/dx − 6x dy/dx = 6y − 3x². Factor: dy/dx (3y² − 6x) = 6y − 3x², hence dy/dx = (6y − 3x²) / (3y² − 6x) = (2y − x²) / (y² − 2x).
对 x³ + y³ = 6xy 求导。左边得到 3x² + 3y² dy/dx。右边需要使用乘法法则：d/dx (6xy) = 6[x dy/dx + y × 1] = 6x dy/dx + 6y。令两边相等：3x² + 3y² dy/dx = 6x dy/dx + 6y。整理含有 dy/dx 的项：3y² dy/dx − 6x dy/dx = 6y − 3x²。提取公因子：dy/dx (3y² − 6x) = 6y − 3x²，因此 dy/dx = (6y − 3x²) / (3y² − 6x) = (2y − x²) / (y² − 2x)。
8. Finding the Gradient of a Curve | 求曲线的梯度
, at the point (3, 3) we get dy/dx = (2·3 − 3²) / (3² − 2·3) = (6 − 9) / (9 − 6) = −3/3 = −1. The tangent line is then y − 3 = −1(x − 3), or y = −x + 6.
一旦用 x 和 y 表示出 dy/dx，代入所求点的坐标即可求出梯度。例如，在曲线 x³ + y³ = 6xy 上，点 (3, 3) 处的 dy/dx = (2·3 − 3²) / (3² − 2·3) = (6 − 9)/(9 − 6) = −3/3 = −1。于是切线方程为 y − 3 = −1(x − 3)，即 y = −x + 6。
9. Second Derivative via Implicit Differentiation | 通过隐微分求二阶导数
oth sides: d²y/dx² = − [ (1·y − x·dy/dx) / y² ] using the quotient rule. Substitute the known dy/dx = −x/y and simplify. This technique is often tested to confirm the nature of stationary points on implicit curves.
为求 d²y/dx²，从一阶导数的表达式出发，再次隐式求导。举例，若 dy/dx = −x/y，对两边再次求导：使用商法则得 d²y/dx² = − [ (1·y − x·dy/dx) / y² ]。代入已知的 dy/dx = −x/y 并化简。这一技巧常用于确定隐式曲线上驻点的性质，考试中时有出现。
10. Common Mistakes to Avoid | 常见错误
Many students forget to apply the chain rule when differentiating y‑ of 2y dy/dx. Another frequent error is failing to use the product rule for mixed terms like xy or x²y. Also watch out for sign errors when moving terms; factorisation of dy/dx must be done carefully, and the final expression should be simplified if possible.
许多学生在对 y 项求导时忘记使用链式法则，写成了 d/dx (y²) = 2y 而不是 2y dy/dx。另一个常见错误是对 xy 或 x²y 这类混合项没有使用乘法法则。同时要留意移项时的符号错误；提取 dy/dx 时必须谨慎操作，最终表达式如果可能应进行化简。
11. Exam Tips for Edexcel | Edexcel 考试技巧
In Edexcel Pure Mathematics exams, implicit differentiation questions typically award method marks for correct application of the chain and product rules. Always show the step where you multiply by dy/dx. Even if the final answer has a minor slip, clear working with dy/dx attached correctly can secure most of the marks. Familiarity with quotient rule and rearranging into a neat form is also helpful for subsequent parts.
在 Edexcel 纯数考试中，隐微分题目通常对链式法则和乘法法则的正确使用给予方法分。务必展示乘以 dy/dx 的那个步骤。即便最终答案有小错误，只要在推导中清晰地正确附上 dy/dx，就可以拿到大部分分数。熟悉商法则并将其重排为整洁形式对后续小问也很有帮助。
12. Practice Question | 练习题
pression for dy/dx in terms of x and y. Hence determine the gradient of the curve at the point (1, 2). (Hint: differentiate implicitly, then substitute.)
给定曲线 y² = x³ + 3x²，求用 x 和 y 表示的 dy/dx 表达式。进而确定曲线在点 (1, 2) 处的梯度。（提示：隐式求导，然后代入。）
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