📚 Implicit Differentiation | 隐函数的求导方法
In A-Level Mathematics, most functions we meet are explicit: y is written directly in terms of x, such as y = x² + 3x. However, many important curves are defined by equations where x and y are mixed together, for example x² + y² = 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 has the form F(x, y) = 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?
If we can solve an implicit equation for y, we could differentiate the resulting explicit function. However, this 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。
Using the circle example x² + y² = 25, 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
Once dy/dx is known at a given point, we can write the equation of 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。
Example: A point moves 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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