Derivatives of Inverse Functions | 反函数导数的求解方法

📚 Derivatives of Inverse Functions | 反函数导数的求解方法

In IB Mathematics, finding the derivative of an inverse function is a core skill that bridges algebraic manipulation, function composition, and the chain rule. This article provides a systematic method for solving such derivatives, including common pitfalls and exam-style examples.

在 IB 数学中，求解反函数的导数是连接代数变形、复合函数和链式法则的核心技能。本文提供一套系统性的求解方法，包括常见陷阱和考场风格例题。

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1. The Core Formula | 核心公式

If f is a differentiable, one-to-one function and f^-1 is its inverse, then for any point x in the domain of f^-1 where f'(f^-1(x)) ≠ 0:

若 f 是可微的一一对应函数，f^-1 是其反函数，则在 f^-1 定义域内任意一点 x 处，当 f'(f^-1(x)) ≠ 0 时：

(f⁻¹)'(x) = 1 / f'(f⁻¹(x))

This formula comes from differentiating the identity f(f^-1(x)) = x using the chain rule: f'(f^-1(x)) · (f^-1)'(x) = 1.

该公式来源于对恒等式 f(f^-1(x)) = x 两边求导，并应用链式法则：f'(f^-1(x)) · (f^-1)'(x) = 1。

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2. Understanding the Geometry | 几何意义

The graph of an inverse function is the reflection of the graph of f over the line y = x. Consequently, the slope of the tangent to y = f^-1(x) at x = a is the reciprocal of the slope of the tangent to y = f(x) at the corresponding point f^-1(a).

反函数图像是原函数图像关于直线 y = x 的对称。因此，y = f^-1(x) 在 x = a 处的切线斜率，等于 y = f(x) 在对应点 f^-1(a) 处切线斜率的倒数。

Example: If f(2) = 5 and f'(2) = 3, then (f^-1)'(5) = 1/3.

例如：若 f(2) = 5 且 f'(2) = 3，则 (f^-1)'(5) = 1/3。

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3. Step-by-Step Method | 分步求解法

To compute (f^-1)'(a) for a given a, follow these steps:

要计算 (f^-1)'(a)，请按以下步骤操作：

• Step 1: Solve f(y) = a for y. The solution is y = f^-1(a).
  第一步：解方程 f(y) = a，得到 y = f^-1(a)。
• Step 2: Compute f'(y).
  第二步：计算 f'(y)。
• Step 3: Take the reciprocal: (f^-1)'(a) = 1 / f'(y).
  第三步：取倒数：(f^-1)'(a) = 1 / f'(y)。

This method avoids explicitly finding the inverse function, which may be difficult or impossible algebraically.

该方法避免了显式求解反函数，因为反函数有时在代数上很难甚至无法表达。

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4. Worked Example 1 (Polynomial) | 例题 1（多项式）

Let f(x) = x³ + 2x - 1. Find (f^-1)'(2).

设 f(x) = x³ + 2x - 1，求 (f^-1)'(2)。

Solution / 解：

First solve f(y) = 2: y³ + 2y - 1 = 2, so y³ + 2y - 3 = 0. By inspection, y = 1 works. Factoring gives (y - 1)(y² + y + 3) = 0, and the quadratic has no real roots. Hence y = 1.

首先解 f(y) = 2：y³ + 2y - 1 = 2，即 y³ + 2y - 3 = 0。观察得 y = 1 是根。因式分解得 (y - 1)(y² + y + 3) = 0，二次部分无实根。因此 y = 1。

Then f'(x) = 3x² + 2. At x = 1, f'(1) = 3(1)² + 2 = 5.

然后 f'(x) = 3x² + 2。在 x = 1 处，f'(1) = 3(1)² + 2 = 5。

(f⁻¹)'(2) = 1 / f'(1) = 1/5

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5. Worked Example 2 (Trigonometric) | 例题 2（三角函数）

Let f(x) = sin x for -π/2 ≤ x ≤ π/2. Find (f^-1)'(1/2).

设 f(x) = sin x，其中 -π/2 ≤ x ≤ π/2。求 (f^-1)'(1/2)。

Solution / 解：

Since f^-1(x) = arcsin x, we know y = arcsin(1/2) = π/6.

因为 f^-1(x) = arcsin x，所以 y = arcsin(1/2) = π/6。

Then f'(x) = cos x, so f'(π/6) = cos(π/6) = √3/2.

而 f'(x) = cos x，所以 f'(π/6) = cos(π/6) = √3/2。

(f⁻¹)'(1/2) = 1 / (√3/2) = 2/√3 = 2√3/3

This matches the known derivative of arcsin x evaluated at 1/2.

这与 arcsin x 的已知导数在 1/2 处的值一致。

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6. Using Inverse Function Explicitly | 显式利用反函数求导

When the inverse function is easy to express, you can differentiate it directly and compare with the formula. For example, if f(x) = e^x, then f^-1(x) = ln x, and (f^-1)'(x) = 1/x. The formula gives (f^-1)'(x) = 1 / e^{ln x} = 1/x.

当反函数容易显式表达时，可以直接对其求导并与公式结果对照。例如，若 f(x) = e^x，则 f^-1(x) = ln x，且 (f^-1)'(x) = 1/x。公式给出 (f^-1)'(x) = 1 / e^{ln x} = 1/x。

Similarly, for f(x) = ln x, f^-1(x) = e^x, and (f^-1)'(x) = e^x.

类似地，对于 f(x) = ln x，反函数为 f^-1(x) = e^x，且 (f^-1)'(x) = e^x。

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7. Relationship with Tangent Lines | 与切线的关系

If P(a, b) lies on the graph of y = f(x), then the point Q(b, a) lies on the graph of y = f^-1(x). The tangent line at P has slope f'(a), while the tangent at Q has slope 1/f'(a).

若点 P(a, b) 在 y = f(x) 的图像上，则点 Q(b, a) 在 y = f^-1(x) 的图像上。点 P 处的切线斜率为 f'(a)，而点 Q 处的切线斜率为 1/f'(a)。

This property is often tested in IB questions asking for the equation of a tangent to an inverse function.

这一性质在 IB 考试中常以“求反函数图像的切线方程”的形式出现。

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8. Common Pitfalls | 常见错误

IB students frequently make the following mistakes when dealing with derivatives of inverse functions:

IB 学生在处理反函数导数时常犯以下错误：

• Using x instead of f^-1(x): Forgetting to evaluate f' at the inverse value, not at x.
  把 x 当作 f^-1(x) 代入：忘记 f' 应在反函数值处取值，而不是在 x 处。
• Confusing reciprocal with negative exponent: (f^-1)'(x) is not the same as [f(x)]^-1.
  混淆倒数与负指数：(f^-1)'(x) 并不等于 [f(x)]^-1。
• Forgetting the chain rule when deriving the formula: The derivation requires the chain rule, not the quotient rule.
  推导公式时忘记链式法则：该公式由链式法则推出，与商法则无关。
• Not checking one-to-one: The formula only applies if f is one-to-one on the relevant interval.
  未检查一一对应性：仅当 f 在相应区间上是一一对应时，公式才适用。

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9. Practice Problems | 练习题目

Try these problems on your own, then check the solutions.

请先自己尝试以下题目，再核对解答。

Problem / 题目	Answer / 答案
1. f(x) = x5 + 3x. Find (f-1)'(4).	Solve y5+3y=4 gives y=1. f'(1)=5+3=8, so answer 1/8.
2. f(x) = e2x. Find (f-1)'(e2).	y = 1, f'(1)=2e2, answer 1/(2e2).
3. f(x) = tan x on (-π/2, π/2). Find (f-1)'(√3).	y=π/3, f'= sec2 x, at π/3 gives 4, answer 1/4.

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10. Inverse Trigonometric Derivatives | 反三角函数的导数

The derivative formula for inverse functions directly yields the derivatives of inverse trigonometric functions:

反函数导数公式可直接推出反三角函数的导数：

• ddx arcsin x = ¹⁄_√1-x²
• ddx arccos x = -¹⁄_√1-x²
• ddx arctan x = ¹⁄_1+x²

For example, for y = arcsin x, we have x = sin y, so dxdy = cos y. Then dydx = 1/cos y. Since cos y = √1-sin² y = √1-x² (positive because y ∈ [-π/2, π/2]), the result follows.

例如，对于 y = arcsin x，有 x = sin y，所以 dxdy = cos y。于是 dydx = 1/cos y。因为 cos y = √1-sin² y = √1-x²（y ∈ [-π/2, π/2] 时取正），即得结果。

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11. Exam Tips | 考试技巧

In IB exams, the problem often gives you a specific value and asks for the derivative of the inverse at that value. Follow these tips:

在 IB 考试中，题目通常会给出一个具体数值，要求反函数在该值处的导数。请遵循以下技巧：

• Always write down the formula (f^-1)'(a) = 1/f'(f^-1(a)) before substituting.
• 先写出公式 (f^-1)'(a) = 1/f'(f^-1(a))，然后再代入。
• When solving f(y)=a, use trial and error for simple roots, or the factor theorem.
• 解 f(y)=a 时，可用试值法找简单根，或使用因式定理。
• Show your substitution clearly: state y=f^-1(a), then f'(y), then the reciprocal.
• 清晰地展示代入过程：写出 y=f^-1(a)，然后 f'(y)，再取倒数。
• If the inverse is a standard function (e.g., ln x, arcsin x), verify your result using the known derivative.
• 如果反函数是标准函数（如 ln x、arcsin x），可用已知导数检验结果。

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12. Summary | 总结

The derivative of an inverse function is found by the simple reciprocal rule: (f^-1)'(x) = 1/f'(f^-1(x)). The main challenge is finding f^-1(x) or the corresponding y-value. With practice, this becomes a fast and reliable tool for differentiation.

反函数的导数可由简洁的倒数规则求得：(f^-1)'(x) = 1/f'(f^-1(x))。主要难点在于求 f^-1(x) 或对应的 y 值。通过练习，这一工具会变得快速而可靠。

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