Example 6.10.2: Implicit Differentiation | 例题6.10.2：隐函数求导

📚 Example 6.10.2: Implicit Differentiation | 例题6.10.2：隐函数求导

In this article, we will work through a classic AQA A-Level Mathematics example that demonstrates how to differentiate an implicit equation and find the equation of a tangent at a given point. This skill is essential for the Pure Mathematics component of the A-Level exam.

在本文中，我们将详解一道经典的AQA A-Level数学例题，该题演示如何对隐式方程求导，并求出曲线在指定点处的切线方程。这一技能是A-Level纯数学部分的考点核心，也是后续深入学习的基础。

1. Problem Statement | 题目概述

Consider the circle defined by the equation \(x^2 + y^2 = 25\). We are asked to find the equation of the tangent to this circle at the point \((3, 4)\).

考虑由方程 \(x^2 + y^2 = 25\) 所定义的圆。我们要求此圆在点 \((3, 4)\) 处的切线方程。

2. Understanding Implicit Differentiation | 理解隐函数求导

In many cases, a curve is given by an equation where \(y\) is not explicitly written as a function of \(x\). For example, \(x^2 + y^2 = 25\) relates \(x\) and \(y\) implicitly. To find the gradient \(\frac{dy}{dx}\), we differentiate both sides of the equation with respect to \(x\), using the chain rule for the \(y\)-terms.

在许多情况下，曲线的方程并未将 \(y\) 显式写成 \(x\) 的函数。例如，\(x^2 + y^2 = 25\) 以一种隐式的方式关联 \(x\) 与 \(y\)。为了求梯度 \(\frac{dy}{dx}\)，我们需要对等式两边同时关于 \(x\) 求导，并对含 \(y\) 的项使用链式法则。

3. Step 1: Differentiate Both Sides | 第一步：两边同时求导

The first step is to differentiate every term in the equation with respect to \(x\). For the left-hand side, we have the derivative of \(x^2\) with respect to \(x\) is \(2x\). The derivative of \(y^2\) with respect to \(x\) is not simply \(2y\); because \(y\) is a function of \(x\), we must apply the chain rule: \(\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}\). The right-hand side is a constant \(25\), whose derivative is \(0\).

第一步是对方程中的每一项关于 \(x\) 求导。对于左侧，\(x^2\) 关于 \(x\) 的导数为 \(2x\)。但 \(y^2\) 关于 \(x\) 的导数并不是简单的 \(2y\)，因为 \(y\) 是 \(x\) 的函数，我们需要应用链式法则：\(\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}\)。右侧是常数 \(25\)，其导数为 \(0\)。

4. Step 2: Apply the Chain Rule | 第二步：应用链式法则

After differentiating both sides, we obtain the equation:

2x + 2y · \(\frac{dy}{dx}\) = 0

This expression shows how the gradient \(\frac{dy}{dx}\) is buried inside the derivative of \(y^2\). The chain rule is what allows us to bring \(\frac{dy}{dx}\) into the equation.

两边同时求导后，我们得到方程：

2x + 2y · \(\frac{dy}{dx}\) = 0

该表达式显示梯度 \(\frac{dy}{dx}\) 是如何隐藏在 \(y^2\) 的导数之中的。正是链式法则让我们能够将 \(\frac{dy}{dx}\) 引入到方程中。

5. Step 3: Collect dy/dx Terms | 第三步：整理dy/dx项

We now isolate the term containing \(\frac{dy}{dx}\). Subtract \(2x\) from both sides:

2y · \(\frac{dy}{dx}\) = −2x

All \(\frac{dy}{dx}\) terms are already on the left; there are no extra terms to combine. This is a simple case because the equation only contains two terms.

现在我们将包含 \(\frac{dy}{dx}\) 的项分离出来。两边同时减去 \(2x\)：

2y · \(\frac{dy}{dx}\) = −2x

所有含 \(\frac{dy}{dx}\) 的项已经在左侧；没有其他项需要合并。这是一个简单的情形，因为原方程仅包含两个项。

6. Step 4: Solve for dy/dx | 第四步：解出dy/dx

Divide both sides by \(2y\) (assuming \(y \neq 0\)):

\(\frac{dy}{dx}\) = −\(\frac{x}{y}\)

This is the formula for the gradient of the tangent to the circle at any point \((x, y)\) on the curve, as long as \(y \neq 0\).

两边同时除以 \(2y\)（假设 \(y \neq 0\)）：

\(\frac{dy}{dx}\) = −\(\frac{x}{y}\)

这就是该圆上任意点 \((x, y)\) 处切线斜率的公式，但要求 \(y \neq 0\)。

7. Step 5: Evaluate at the Point | 第五步：在给定点求值

We are given the point \((3, 4)\). Substitute \(x = 3\) and \(y = 4\) into the gradient formula:

\(\frac{dy}{dx}\Big|_{(3,4)}\) = −\(\frac{3}{4}\)

Thus, the slope of the tangent at the point \((3, 4)\) is \(−\frac{3}{4}\).

我们已知点 \((3, 4)\)。将 \(x = 3\) 和 \(y = 4\) 代入斜率公式：

\(\frac{dy}{dx}\Big|_{(3,4)}\) = −\(\frac{3}{4}\)

因此，在点 \((3, 4)\) 处切线的斜率为 \(−\frac{3}{4}\)。

8. Gradient of the Tangent | 切线斜率

The gradient we have just found, \(m = −\frac{3}{4}\), is the same as the radius slope at that point multiplied by \(−1\) because the radius and the tangent are perpendicular. Notice that the radius from the origin \((0,0)\) to \((3,4)\) has slope \(\frac{4}{3}\), and the tangent slope is the negative reciprocal: \(−\frac{3}{4}\). This is a nice geometric check.

我们刚刚求得的斜率 \(m = −\frac{3}{4}\) 等于该点处半径斜率的相反数乘以 \(−1\)，因为半径与切线互相垂直。注意，从原点 \((0,0)\) 到 \((3,4)\) 的半径斜率为 \(\frac{4}{3}\)，而切线斜率是它的负倒数：\(−\frac{3}{4}\)。这是一个很好的几何验证。

9. Equation of the Tangent | 切线方程

Using the point-slope form of a line, \(y – y_1 = m(x – x_1)\), with \((x_1, y_1) = (3, 4)\) and \(m = −\frac{3}{4}\), we get:

y − 4 = −\(\frac{3}{4}\) (x − 3)

Multiply through by 4:

4y − 16 = −3x + 9

Rearrange to the standard form:

3x + 4y = 25

Thus, the equation of the tangent is \(3x + 4y = 25\).

利用直线的点斜式方程 \(y – y_1 = m(x – x_1)\)，其中 \((x_1, y_1) = (3, 4)\)，\(m = −\frac{3}{4}\)，可得：

y − 4 = −\(\frac{3}{4}\) (x − 3)

两边乘以4：

4y − 16 = −3x + 9

整理为标准形式：

3x + 4y = 25

所以，切线方程为 \(3x + 4y = 25\)。

10. Verification | 验证

We can verify that the point \((3, 4)\) lies on the tangent line: \(3(3) + 4(4) = 9 + 16 = 25\), which is correct. Also, the tangent line is indeed perpendicular to the radius, whose equation is \(y = \frac{4}{3}x\). The slopes multiply to \(−1\), confirming correctness.

我们可以验证点 \((3, 4)\) 是否在切线上：\(3(3) + 4(4) = 9 + 16 = 25\)，正确无误。同时，切线与半径方程 \(y = \frac{4}{3}x\) 垂直，两者斜率之积为 \(−1\)，进一步确认了解答的正确性。

11. Common Mistakes | 常见错误

There are several pitfalls that students often encounter when working with implicit differentiation. First, forgetting to apply the chain rule to \(y^2\) and writing only \(2y\) instead of \(2y \frac{dy}{dx}\). Second, making sign errors when moving terms across the equation. Third, incorrectly simplifying the fraction \(-x/y\) or forgetting to substitute both coordinates. Fourth, assuming the tangent line equation should have final \(y\) isolated; sometimes leaving it in standard form is acceptable and often expected.

在处理隐函数求导时，学生常常会陷入几个陷阱。第一，忘记对 \(y^2\) 使用链式法则，而只写成 \(2y\)，应写成 \(2y \frac{dy}{dx}\)。第二，在移项时出现符号错误。第三，错误地化简分数 \(-x/y\)，或者忘记代入两坐标值。第四，认为切线方程必须将 \(y\) 单独解出；实际上，保留标准形式也完全可接受，且常常是所期望的形式。

12. Practice Questions | 练习

To solidify your understanding, try solving the following similar problems ourselves without looking at the solutions.

为了巩固理解，请尝试解决以下类似问题，先不要看解答。

Find the gradient of the tangent to the curve \(x^2 + y^2 = 25\) at the point \((−3, 4)\).
找到曲线 \(x^2 + y^2 = 25\) 在点 \((−3, 4)\) 处的切线斜率。
The equation of a curve is given by \(x^2 + xy + y^2 = 3\). Find \(\frac{dy}{dx}\) and determine the tangent at \((1, 1)\).
曲线方程为 \(x^2 + xy + y^2 = 3\)。求 \(\frac{dy}{dx}\) 并在点 \((1, 1)\) 处确定切线方程。

These problems require the same technique shown above. Practice will help you gain speed and accuracy for the exam.

这些问题需要运用与上文相同的技巧。练习将帮助您在考试中提升速度和准确性。

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Example 6.10.2: Implicit Differentiation | 例题6.10.2：隐函数求导