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。

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1. Problem Statement | 题目概述

Consider the circle defined by the equation x² + y² = 25. We are asked to find the equation of the tangent to this circle at the point (3, 4).

考虑由方程 x² + y² = 25 所定义的圆。我们要求此圆在点 (3, 4) 处的切线方程。

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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, x² + y² = 25 relates x and y implicitly. To find the gradient dydx, 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，x² + y² = 25 以一种隐式的方式关联 x 与 y。为了求梯度 dydx，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。

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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 x² with respect to x is 2x. The derivative of y² with respect to x is not simply 2y; because y is a function of x, we must apply the chain rule: ddx(y²) = 2y · dydx. 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，x² 关于 x 的导数为 2x。但 y² 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：ddx(y²) = 2y · dydx。右侧是常数 25，其导数为 0。

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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 · dydx = 0

This expression shows how the gradient dydx is buried inside the derivative of y². The chain rule is what allows us to bring dydx into the equation.

两边同时求导后，我们得到方程：

2x + 2y · dydx = 0

该表达式显示梯度 dydx 是如何隐藏在 y² 的导数之中的。正是链式法则让我们能够将 dydx 引入到方程中。

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

We now isolate the term containing dydx. Subtract 2x from both sides:

2y · dydx = −2x

All dydx 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 dydx 的项分离出来。两边同时减去 2x：

2y · dydx = −2x

所有含 dydx 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。

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

Divide both sides by 2y (assuming y ≠ 0):

dydx = −xy

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 ≠ 0.

两边同时除以 2y（假设 y ≠ 0）：

dydx = −xy

这就是该圆上任意点 (x, y) 处切线斜率的公式，但要求 y ≠ 0。

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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:

dydx|_(3,4) = −34

Thus, the slope of the tangent at the point (3, 4) is −34.

我们已知点 (3, 4)。将 x = 3 和 y = 4 代入斜率公式：

dydx|_(3,4) = −34

因此，在点 (3, 4) 处切线的斜率为 −34。

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8. Gradient of the Tangent | 切线斜率

The gradient we have just found, m = −34, 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 43, and the tangent slope is the negative reciprocal: −34. This is a nice geometric check.

我们刚刚求得的斜率 m = −34 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 43，而切线斜率是它的负倒数：−34。这是一个很好的几何验证。

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9. Equation of the Tangent | 切线方程

Using the point-slope form of a line, y - y₁ = m(x - x₁), with (x₁, y₁) = (3, 4) and m = −34, we get:

y − 4 = −34 (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 - y₁ = m(x - x₁)，其中 (x₁, y₁) = (3, 4)，m = −34，可得：

y − 4 = −34 (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。

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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 = 43x. 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 = 43x 垂直，两者斜率之积为 −1，进一步确认了解答的正确性。

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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 y² and writing only 2y instead of 2y dydx. 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 y² 使用链式法则，而只写成 2y，应写成 2y dydx。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。

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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 x² + y² = 25 at the point (−3, 4).
• 找到曲线 x² + y² = 25 在点 (−3, 4) 处的切线斜率。
• The equation of a curve is given by x² + xy + y² = 3. Find dydx and determine the tangent at (1, 1).
• 曲线方程为 x² + xy + y² = 3。求 dydx 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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