📚 Implicit Differentiation | 隐函数的求导方法
25. To find the derivative dy/dx from such an equation, we use a technique called implicit differentiation. This article explains the method step by step, with worked examples, common pitfalls, and exam-style tips.
在A-Level 数学中，我们遇到的大多数函数都是显函数：y 可以写成关于 x 的表达式，例如 y = x² + 3x。然而，许多重要曲线是由 x 和 y 混合在一起的方程定义的，例如 x² + y² = 25。要从这样的方程求出导数 dy/dx，我们需要用一种称为“隐函数求导”的方法。本文将逐步讲解这一方法，并结合典型例题、常见错误与备考提示。
1. What is an Implicit Function? | 什么是隐函数？
An explicit function has the form y = f(x). For instance, y = x³ − 4x is explicit because y is isolated on one side. An implicit equation = 0, meaning x and y are connected without solving for one variable in terms of the other.
显函数的形式为 y = f(x)。例如，y = x³ − 4x 就是显函数，因为 y 被单独放在等式一边。隐式方程的形式为 F(x, y) = 0，也就是 x 和 y 相互关联，而没有把其中一个变量解成另一个变量的表达式。
Common examples of implicit equations include:
常见的隐式方程包括：
- x² + y² = 25
- x³ + y³ = 6xy
- xy + sin y = x²
Sometimes we can solve for y explicitly, but often it is difficult or even impossible. Even when it is possible, the explicit form may involve square roots and require splitting the curve into multiple branches.
有时我们可以解出 y，但通常很困难甚至不可能；即使有可能，显式形式也可能包含根号，并且需要把曲线分成多个分支。
2. Why Do We Need Implicit Differentiation? | 为什么需要隐微分？
his is not always convenient. For example, the circle x² + y² = 25 can be written as y = ±√25 − x², which gives two separate functions. Differentiating both pieces separately is messy.
如果我们能从隐式方程中解出 y，就可以对显式函数求导。然而，这并不总是方便的。例如，圆 x² + y² = 25 可以写成 y = ±√25 − x²，这给出两个分开的函数。分别对这两部分求导很麻烦。
Implicit differentiation lets us work directly with the original equation. It also provides a unified way to find slopes, tangents, second derivatives, and related rates without needing to change the form of the equation.
隐微分使我们能直接在原方程上求导。它也为求斜率、切线、二阶导数及相关变化率提供了一种统一的方法，而不需要改变方程的形式。
3. Key Rule: The Chain Rule | 关键规则：链式法则
Implicit differentiation relies on one essential idea: if y is a function of x, then when we differentiate an expression involving y, we must multiply by dy/dx. This is exactly the chain rule.
隐微分依赖于一个核心想法：如果 y 是 x 的函数，那么当我们对包含 y 的表达式求导时，必须乘以 dy/dx。这正是链式法则。
d/dx [f(y)] = f '(y) · dy/dx
For example:
例如：
- d/dx (y²) = 2y · dy/dx
- d/dx (sin y) = cos y · dy/dx
- d/dx (eʸ) = eʸ · dy/dx
Why do we multiply by dy/dx? Because y is not the independent variable; y depends on x, so differentiating y² creates an “inner derivative” of y with respect to x.
为什么要乘以 dy/dx？因为 y 不是自变量；y 依赖于 x，所以对 y² 求导时会产生一个关于 x 的“内层导数” dy/dx。
4. General Steps for Implicit Differentiation | 隐微分的一般步骤
The following steps can be applied to most implicit differentiation problems.
大多数隐函数求导问题都可以用以下步骤解决。
- Differentiate every term with respect to x. Treat x as the usual variable, but treat y as an unknown function of x.
- Use the chain rule for y-terms. Whenever you differentiate a term containing y, multiply by dy/dx.
- Use the product or quotient rule when x and y appear multiplied or divided together.
- Collect all terms with dy/dx on one side of the equation.
- Factor out dy/dx and solve for it. Simplify if possible, using the original equation when needed.
- 对每一项关于 x 求导。把 x 当作普通变量，把 y 当作 x 的未知函数。
- 对含 y 的项使用链式法则。每当对含有 y 的项求导时，都要乘以 dy/dx。
- 当 x 和 y 相乘或相除时，使用乘积法则或商法则。
- 把所有含 dy/dx 的项移到方程的一边。
- 提取公因式 dy/dx 并解出它。如有可能，利用原方程进行化简。
5. Worked Example 1: x² + y² = 25 | 例题1：x² + y² = 25
Let us differentiate the equation of a circle of radius 5.
我们来对一个半径为 5 的圆的方程求导。
Step 1: Differentiate both sides with respect to x.
第一步：对两边关于 x 求导。
2x + 2y · dy/dx = 0
Step 2: Solve for dy/dx.
第二步：解出 dy/dx。
2y · dy/dx = −2x
dy/dx = −x/y
At the point (3, 4), the slope of the tangent is −3/4. Notice that dy/dx still contains y; this is normal for implicit differentiation.
在点 (3, 4) 处，切线斜率为 −3/4。注意 dy/dx 中仍然含有 y；这在隐微分中是正常现象。
6. Worked Example 2: x³ + y³ = 6xy | 例题2：x³ + y³ = 6xy
This is the famous “folium of Descartes”. It cannot easily be solved for y in terms of x, so implicit differentiation is essential.
这是著名的“笛卡尔叶形线”。它不容易解出 y 关于 x 的表达式，所以隐微分是必不可少的。
Differentiate term by term with respect to x.
逐项关于 x 求导。
3x² + 3y² · dy/dx = 6y + 6x · dy/dx
The right-hand side requires the product rule: d/dx (xy) = y + x · dy/dx.
右边需要使用乘积法则：d/dx (xy) = y + x · dy/dx。
Now collect terms involving dy/dx.
现在收集含 dy/dx 的项。
3y² · dy/dx − 6x · dy/dx = 6y − 3x²
dy/dx (3y² − 6x) = 6y − 3x²
Therefore:
因此：
dy/dx = (2y − x²) / (y² − 2x)
We divided the numerator and denominator by 3 to simplify.
我们同时除以 3 来化简分子和分母。
7. Second Derivatives | 二阶导数
To find the second derivative d²y/dx², we differentiate dy/dx again with respect to x. Remember that any y in the answer must again be treated as a function of x.
为了求二阶导数 d²y/dx²，我们需要再对 dy/dx 关于 x 求导。注意，答案中出现的任何 y 都必须再次被当作 x 的函数。
, we had dy/dx = −x/y. Differentiate both sides with respect to x using the quotient rule.
以圆 x² + y² = 25 为例，我们已经求得 dy/dx = −x/y。用商法则对两边关于 x 求导。
d²y/dx² = − [ (1)(y) − (x)(dy/dx) ] / y²
Substitute dy/dx = −x/y.
代入 dy/dx = −x/y。
d²y/dx² = − [ y + x²/y ] / y² = − (y² + x²) / y³
Since x² + y² = 25, this simplifies to:
因为 x² + y² = 25，所以可以化简为：
d²y/dx² = −25 / y³
This illustrates how the original equation can be used to simplify second derivatives.
这个例子说明，可以利用原方程来化简二阶导数。
8. Tangents and Normals to Implicit Curves | 隐式曲线的切线与法线
f the tangent using y − y₁ = m (x − x₁). The normal slope is the negative reciprocal of m.
一旦求出某点处的 dy/dx，我们就可以用 y − y₁ = m (x − x₁) 写出切线方程。法线斜率是 m 的负倒数。
Example: For x² + y² = 25 at (3, 4), we found m = −3/4. The tangent equation is:
例如：对于 x² + y² = 25，在点 (3, 4) 处，我们求得 m = −3/4。切线方程为：
y − 4 = −3/4 (x − 3)
Simplifying gives y = −3/4 x + 25/4. The normal slope is 4/3, so the normal equation is:
化简得 y = −3/4 x + 25/4。法线斜率为 4/3，所以法线方程为：
y − 4 = 4/3 (x − 3)
This technique works for any implicit curve, even when the equation cannot be solved explicitly.
这种方法适用于任何隐式曲线，即使方程无法显式解出。
9. Introduction to Related Rates | 相关变化率简介
Implicit differentiation is also useful when x and y are both functions of time t. Differentiating with respect to t creates derivatives like dx/dt and dy/dt.
当 x 和 y 都是时间 t 的函数时，隐微分也非常有用。关于 t 求导会得到像 dx/dt 和 dy/dt 这样的导数。
along the circle x² + y² = 25. At a certain moment x = 3, y = 4 and dx/dt = 2. Find dy/dt.
例如：一个点沿圆 x² + y² = 25 运动。某一时刻 x = 3，y = 4，且 dx/dt = 2。求 dy/dt。
Differentiate both sides with respect to t.
对两边关于 t 求导。
2x · dx/dt + 2y · dy/dt = 0
Substitute the known values:
代入已知数值：
2(3)(2) + 2(4) · dy/dt = 0
12 + 8 · dy/dt = 0
dy/dt = −3/2
The negative sign shows that y is decreasing while x is increasing.
负号表示当 x 增大时 y 在减小。
10. Common Mistakes and Exam Tips | 常见错误与考试提示
Students frequently lose marks on implicit differentiation for avoidable reasons. Here are the most common errors:
学生在隐函数求导中经常因为一些可以避免的原因失分。以下是最常见的错误：
- Forgetting the chain rule on y-terms: d/dx (y²) is 2y · dy/dx, not 2y.
- Misusing the product rule: d/dx (xy) = y + x · dy/dx, not simply dy/dx.
- Losing signs when moving terms across the equation.
- Not simplifying at the end, especially when a common factor can cancel.
- Confusing dy/dx with dx/dy: Always ask which variable is being treated as the independent one.
- 忘记对含 y 的项使用链式法则：d/dx (y²) 应为 2y · dy/dx，而不是 2y。
- 错误使用乘积法则：d/dx (xy) = y + x · dy/dx，而不只是 dy/dx。
- 移项时弄错正负号。
- 最后不化简，尤其是在可以约去公因式的时候。
- 混淆 dy/dx 与 dx/dy：要始终清楚哪个变量是自变量。
Exam tip: After finding dy/dx, check whether your answer can be rewritten using the original equation. In simple cases, the derivative should reduce to a familiar explicit form.
考试提示：求出 dy/dx 后，检查答案是否能利用原方程改写。在简单情况下，导数应该能化简为熟悉的显式形式。
11. Summary and Practice | 总结与练习
Implicit differentiation is a powerful tool for finding derivatives of curves defined by equations that mix x and y. The key is to remember the chain rule: differentiate every y-term with respect to x and multiply by dy/dx. Then collect and solve for dy/dx.
隐微分是求由 x 和 y 混合定义的曲线导数的有力工具。关键在于记住链式法则：对每个含 y 的项在关于 x 求导时乘以 dy/dx，然后整理并解出 dy/dx。
Try these short exercises to check your understanding:
试试下面这些简短练习，检查你的理解：
- Given xy = 1, find dy/dx. Answer: dy/dx = −y/x = −1/x².
- Given x + sin y = 1, find dy/dx. Answer: dy/dx = −1/cos y.
- Find the slope of x² + xy + y² = 7 at the point (2, 1). Answer: dy/dx = −(2x + y) / (x + 2y), so the slope is −5/4.
- 已知 xy = 1，求 dy/dx。答案：dy/dx = −y/x = −1/x²。
- 已知 x + sin y = 1，求 dy/dx。答案：dy/dx = −1/cos y。
- 求 x² + xy + y² = 7 在点 (2, 1) 处的斜率。答案：dy/dx = −(2x + y) / (x + 2y)，所以斜率为 −5/4。
Mastering implicit differentiation will help you solve a wide range of calculus problems, from tangents and normals to related rates and higher-order derivatives. Practise with both simple and mixed-product equations, and always check your final derivative.
掌握隐函数求导将帮助你解决广泛的微积分问题，从切线、法线到相关变化率和高阶导数。练习时既要包含简单的方程，也要包含乘积混合的方程，并且一定要检查最终得到的导数。
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