Implicit Differentiation: Key Steps for A-Level Maths | A-Level 数学：隐函数求导的关键步骤

📚 Implicit Differentiation: Key Steps for A-Level Maths | A-Level 数学：隐函数求导的关键步骤

In A-Level mathematics, not all curves can be written in the form y = f(x). Some equations, such as x² + y² = 25, define a relationship between x and y implicitly. Implicit differentiation allows us to find dy/dx without solving for y first. It is a core technique in Pure Mathematics and appears frequently in differentiation and coordinate geometry questions.

在 A-Level 数学中，并非所有曲线都能写成 y = f(x) 的形式。像 x² + y² = 25 这样的方程只是隐式地给出了 x 与 y 的关系。隐函数求导让我们无需先解出 y，就能直接求出 dy/dx。它是纯数学的核心技巧，在微分和坐标几何题目中经常出现。

1. Recognising Implicit Equations | 识别隐式方程

An explicit equation gives y directly in terms of x, for example y = x³ + 2x. An implicit equation mixes x and y, for example x² + y² = 25, xy + sin y = 1, or yeˣ = x + y. In these cases, solving for y may be difficult or impossible.

显式方程直接将 y 表示成 x 的函数，例如 y = x³ + 2x。隐式方程则把 x 和 y 混合在一起，例如 x² + y² = 25、xy + sin y = 1 或 yeˣ = x + y。在这些情况下，解出 y 可能很困难甚至不可能。

• Explicit form: y = f(x) | 显式形式：y = f(x)
• Implicit form: f(x, y) = 0 | 隐式形式：f(x, y) = 0

---

2. Treat y as a Function of x | 把 y 看作 x 的函数

When differentiating an equation involving y, remember that y itself depends on x. The chain rule gives:

对含有 y 的方程求导时，要记住 y 本身依赖于 x。根据链式法则：

d/dx (yⁿ) = n yⁿ⁻¹ · dy/dx

For example, d/dx (y²) = 2y dy/dx, and d/dx (y³) = 3y² dy/dx.

例如，d/dx (y²) = 2y dy/dx，d/dx (y³) = 3y² dy/dx。

---

3. Differentiate Both Sides Term by Term | 逐项对两边求导

Take the equation x² + y² = 25. Differentiating each term with respect to x:

以方程 x² + y² = 25 为例，对每一项关于 x 求导：

2x + 2y · dy/dx = 0

Then solve for dy/dx:

然后解出 dy/dx：

dy/dx = -x/y

Notice how the constant 25 differentiates to 0, and the y² term produces a factor dy/dx.

注意常数 25 的导数为 0，而 y² 项会产生因子 dy/dx。

---

4. Use the Chain Rule for Composite Functions | 对复合函数应用链式法则

For functions such as sin y, eʸ or ln y, multiply the ordinary derivative by dy/dx:

对于 sin y、eʸ 或 ln y 这类函数，求导数后要乘以 dy/dx：

Function	Derivative with respect to x
sin y	cos y · dy/dx
eʸ	eʸ · dy/dx
ln y	(1/y) · dy/dx

---

5. Apply the Product Rule for xy Terms | 对 xy 项应用乘积法则

If a term contains x times y, such as xy or x²y, use the product rule because both factors depend on x. For xy:

如果某一项含有 x 乘以 y，如 xy 或 x²y，需要同时使用乘积法则，因为两个因子都依赖于 x。对于 xy：

d/dx (xy) = y + x · dy/dx

For x²y:

对于 x²y：

d/dx (x²y) = 2xy + x² · dy/dx

Keep the dy/dx terms visible and do not treat y as a constant.

不要丢失 dy/dx 项，也不要将 y 当作常数处理。

---

6. Apply the Quotient Rule for Rational Terms | 对分式项应用商法则

If a fraction such as x/y appears, use the quotient rule. Since y is a function of x:

如果出现像 x/y 这样的分式，需要使用商法则。由于 y 是 x 的函数：

d/dx (x/y) = (y − x · dy/dx) / y²

In some questions it is simpler to multiply both sides of the original equation by a common denominator before differentiating. This avoids an extra quotient rule step.

有些题目可以先在等式两边乘以公分母来化简，再求导，这样可省去商法则的步骤。

---

7. Collect dy/dx Terms and Factorise | 整理 dy/dx 项并提取公因式

After differentiating, group all terms containing dy/dx on one side of the equation and all remaining terms on the other side. Then factor out dy/dx and divide by its coefficient.

求导后，将所有含 dy/dx 的项移到方程一侧，

Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）