📚 IB Math: Implicit Differentiation Step-by-Step Guide with Examples | IB数学：隐函数微分法的操作步骤与实例
Implicit differentiation is a powerful technique in IB Mathematics (Analysis and Approaches HL and Applications and Interpretation HL) used to find derivatives when variables are not explicitly separated. This guide walks through the steps, key rules, and worked examples that align with the IB syllabus.
隐函数微分法是IB数学（分析与方法HL、应用与解释HL）中的一项重要技能，适用于变量未明确分离的函数。本指南将分步骤讲解操作方法、关键法则，并提供符合IB考纲的实例。
1. What Is Implicit Differentiation? | 什么是隐函数微分法？
ases, we cannot easily solve for y in terms of x, so we differentiate both sides with respect to x and apply the chain rule to terms involving y.
显函数具有形如 y = f(x) 的形式，例如 y = x² + 3x。隐函数则将 x 和 y 混合在一个方程中，例如 x² + y² = 25 或 x³ + y³ = 6xy。此时我们无法轻松解出 y，因此需要对等式两边同时对 x 求导，并对含有 y 的项应用链式法则。
2. The Core Rule: Chain Rule on y | 核心规则：对 y 运用链式法则
When differentiating a function of y with respect to x, we must multiply by dy/dx. For example, the derivative of y² with respect to x is 2y ⋅ dy/dx. Similarly, the derivative of sin(y) is cos(y) ⋅ dy/dx. This step appears in almost every implicit differentiation problem.
当对 y 的函数关于 x 求导时，我们必须乘以 dy/dx。例如，y² 对 x 的导数是 2y ⋅ dy/dx。类似地，sin(y) 的导数是 cos(y) ⋅ dy/dx。这一步骤几乎出现在所有隐函数求导问题中。
d/dx [ f(y) ] = f'(y) ⋅ dy/dx
3. Step-by-Step Procedure | 分步操作流程
Follow these steps for any implicit differentiation question.
遇到任何隐函数求导问题，均可按照以下步骤操作。
- Step 1: Differentiate every term on both sides of the equation with respect to x.
- 步骤1： 将等式两边每一项分别对 x 求导。
- Step 2: Apply the product rule and chain rule where needed. Whenever you differentiate a y-term, multiply by dy/dx.
- 步骤2： 在需要时应用乘积法则和链式法则。每当你对含有 y 的项求导时，乘以 dy/dx。
- Step 3: Collect all terms with dy/dx on one side of the equation and all other terms on the other side.
- 步骤3： 将所有含有 dy/dx 的项移到等式一边，其余项移到另一边。
- Step 4: Factor out dy/dx and solve for it.
- 步骤4： 提取公因子 dy/dx，并解出 dy/dx。
- Step 5: Simplify the expression if possible. Substitute given coordinates if the question asks for the slope at a point.
- 步骤5： 尽可能化简表达式。若题目要求某点处的斜率，则将坐标代入。
4. Worked Example 1: Circle | 实例1：圆方程
Find dy/dx for the circle x² + y² = 25.
求圆 x² + y² = 25 的 dy/dx。
Differentiate each term with respect to x:
将每一项分别对 x 求导：
2x + 2y ⋅ dy/dx = 0
Then solve for dy/dx:
然后解出 dy/dx：
dy/dx = −x / y
Note that the derivative depends on both x and y, not just x. This is typical for implicit functions.
注意导数同时依赖于 x 和 y，而不只取决于 x。这正是隐函数的典型特点。
5. Worked Example 2: Product Rule | 实例2：使用乘积法则
Given x³ + y³ = 6xy, find dy/dx.
已知 x³ + y³ = 6xy，求 dy/dx。
Differentiate both sides. The right side 6xy requires the product rule:
对等式两边求导。右侧 6xy 需要使用乘积法则：
3x² + 3y² ⋅ dy/dx = 6y + 6x ⋅ dy/dx
Collect dy/dx terms:
整理含有 dy/dx 的项：
3y² ⋅ dy/dx − 6x ⋅ dy/dx = 6y − 3x²
Factor out dy/dx:
提取 dy/dx：
dy/dx (3y² − 6x) = 6y − 3x²
Therefore:
因此：
dy/dx = (6y − 3x²) / (3y² − 6x)
We can simplify by factoring out 3:
可提取公因数 3 进行化简：
dy/dx = (2y − x²) / (y² − 2x)
6. Worked Example 3: Finding Slope at a Point | 实例3：求某点处的斜率
Find the slope of the tangent line to x² + xy + y² = 7 at the point (1, 2).
求曲线 x² + xy + y² = 7 在点 (1, 2) 处的切线斜率。
Differentiate implicitly. The term xy needs the product rule:
进行隐函数求导。xy 项需使用乘积法则：
2x + (y + x ⋅ dy/dx) + 2y ⋅ dy/dx = 0
Now substitute x = 1 and y = 2:
代入 x = 1, y = 2：
2(1) + 2 + 1 ⋅ dy/dx + 2(2) ⋅ dy/dx = 0
Simplify:
化简：
2 + 2 + dy/dx + 4 ⋅ dy/dx = 0
4 + 5 ⋅ dy/dx = 0
dy/dx = −4/5
So the slope at (1, 2) is −4/5.
因此，点 (1, 2) 处的斜率为 −4/5。
7. Second Derivatives | 隐函数的二阶导数
To find the second derivative d²y/dx², differentiate dy/dx with respect to x, treating y as an implicit function and substituting the known expression for dy/dx where necessary.
要求二阶导数 d²y/dx²，需将 dy/dx 对 x 再次求导，把 y 视为隐函数，并在必要时代入已求得的 dy/dx 表达式。
For example, from dy/dx = −x/y for x² + y² = 25, differentiate using the quotient rule:
例如，对于 x² + y² = 25 有 dy/dx = −x/y，使用商法则求导：
d²y/dx² = −( y − x ⋅ dy/dx ) / y²
Substitute y = √25 − x² or keep the implicit form. Simplify as required by the question.
可代入 y = √25 − x²，也可保留隐函数形式，按题目要求化简。
8. Common IB Questions and Tips | 常见IB题型与技巧
In IB exams, implicit differentiation often appears as part of a longer question involving tangent lines, normals, stationary points, or rates of change. Here are some typical patterns.
在IB考试中，隐函数微分法通常作为综合题的一部分，涉及切线、法线、驻点或变化率。以下是一些常见题型。
- Finding the gradient of a tangent at a given point — substitute the point after differentiating.
- 求给定点处切线的斜率 —— 求导后代入该点坐标。
- Finding where a curve has a horizontal or vertical tangent — set dy/dx = 0 or the denominator = 0.
- 求曲线有水平或垂直切线的位置 —— 令 dy/dx = 0 或分母为 0。
- Using implicit differentiation in related rates problems (e.g., expanding circles or moving ladders).
- 在关联变化率问题中运用隐函数微分法（如膨胀的圆、滑动的梯子）。
- Proving identities involving derivatives and using them to find stationary points.
- 证明含导数的恒等式，并利用它们求驻点。
9. Horizontal and Vertical Tangents | 水平与垂直切线
rovided the denominator is not zero. Similarly, the tangent is vertical when the denominator of dy/dx is zero and the numerator is non-zero. Always check these conditions carefully in IB questions.
对于隐式定义的曲线，当 dy/dx = 0 且分母不为零时，切线为水平；当 dy/dx 的分母为零且分子不为零时，切线为垂直。在IB题目中务必仔细检查这些条件。
ontal tangent occurs when x = 0, so y = ±5. Vertical tangent occurs when y = 0, so x = ±5.
考虑 x² + y² = 25，有 dy/dx = −x/y。水平切线出现于 x = 0，此时 y = ±5；垂直切线出现于 y = 0，此时 x = ±5。
10. Logarithmic Implicit Differentiation | 隐式对数微分法
For functions of the form y = f(x)g(x), take natural logs on both sides first, then differentiate implicitly. This is a standard IB HL technique.
对于形如 y = f(x)g(x) 的函数，可先对等式两边取自然对数，再隐式求导。这是IB HL中的标准技巧。
Example: y = xˣ. Take ln on both sides:
例：y = xˣ。两边取自然对数：
ln y = x ln x
Differentiate implicitly with respect to x:
对 x 隐式求导：
(1/y) ⋅ dy/dx = ln x + 1
Therefore:
因此：
dy/dx = y (ln x + 1) = xˣ (ln x + 1)
11. Worked Example 4: Trigonometric Implicit Function | 实例4：含三角函数的隐函数
Find dy/dx if sin(x + y) = y² cos x.
求 sin(x + y) = y² cos x 的 dy/dx。
Differentiate both sides. Use the chain rule and product rule:
对等式两边求导，运用链式法则和乘积法则：
cos(x + y) ⋅ (1 + dy/dx) = 2y ⋅ dy/dx ⋅ cos x + y² ⋅ (−sin x)
Expand and collect dy/dx terms:
展开并合并 dy/dx 项：
cos(x + y) + cos(x + y) ⋅ dy/dx = 2y cos x ⋅ dy/dx − y² sin x
Bring dy/dx terms together:
将含 dy/dx 的项集中到一边：
cos(x + y) − 2y cos x ⋅ dy/dx + cos(x + y) ⋅ dy/dx = − y² sin x
Factor dy/dx:
提取 dy/dx：
dy/dx [ cos(x + y) − 2y cos x ] = − y² sin x − cos(x + y)
Therefore:
因此：
dy/dx = ( − y² sin x − cos(x + y) ) / ( cos(x + y) − 2y cos x )
12. Final Tips for IB Exams | IB考试最后提示
Always show the chain rule step when differentiating y-terms. Do not skip the dy/dx factor, or you will lose method marks. Check whether the question asks for dy/dx in terms of x and y, or at a specific point. Practice enough past paper questions to become fluent.
对含 y 的项求导时，务必展示链式法则步骤，不要漏写 dy/dx，否则会丢失方法分。注意题目要求的是用 x 和 y 表示 dy/dx，还是在特定点处的值。多练历年真题，做到熟练自如。
Implicit differentiation is a key skill that opens the door to many advanced topics, including related rates, optimization with constraint equations, and parametric curves. Master it early, and you will find IB HL calculus significantly more manageable.
隐函数微分法是通向许多高级主题的关键技能，包括关联变化率、带约束条件的优化问题以及参数曲线。尽早掌握它，你会发现IB HL微积分变得更加轻松。
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