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。

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 \(f'(f^{-1}(x)) \neq 0\):

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 \(f'(f^{-1}(x)) \neq 0\) 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: \(f'(f^{-1}(x)) \cdot (f^{-1})'(x) = 1\).

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：\(f'(f^{-1}(x)) \cdot (f^{-1})'(x) = 1\)。

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 \((f^{-1})'(5) = 1/3\)。

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。

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

Let \(f(x) = x^3 + 2x – 1\). Find \((f^{-1})'(2)\).

设 \(f(x) = x^3 + 2x – 1\)，求 \((f^{-1})'(2)\)。

Solution / 解：

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

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

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

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

(f⁻¹)'(2) = 1 / f'(1) = 1/5

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

Let \(f(x) = \sin x\) for \(-\pi/2 \le x \le \pi/2\). Find \((f^{-1})'(1/2)\).

设 \(f(x) = \sin x\)，其中 \(-\pi/2 \le x \le \pi/2\)。求 \((f^{-1})'(1/2)\)。

Solution / 解：

Since \(f^{-1}(x) = \arcsin x\), we know \(y = \arcsin(1/2) = \pi/6\).

因为 \(f^{-1}(x) = \arcsin x\)，所以 \(y = \arcsin(1/2) = \pi/6\)。

Then \(f'(x) = \cos x\), so \(f'(\pi/6) = \cos(\pi/6) = \sqrt{3}/2\).

而 \(f'(x) = \cos x\)，所以 \(f'(\pi/6) = \cos(\pi/6) = \sqrt{3}/2\)。

(f⁻¹)'(1/2) = 1 / (\sqrt{3}/2) = 2/\sqrt{3} = 2\sqrt{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。

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 \((f^{-1})'(x) = e^x\)。

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。

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。

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. \(f(x) = x^5 + 3x\). Find \((f^{-1})'(4)\).	Solve \(y^5+3y=4\) gives \(y=1\). \(f'(1)=5+3=8\), so answer \(1/8\).
2. \(f(x) = e^{2x}\). Find \((f^{-1})'(e^2)\).	\(y = 1\), \(f'(1)=2e^2\), answer \(1/(2e^2)\).
3. \(f(x) = \tan x\) on \((-\pi/2, \pi/2)\). Find \((f^{-1})'(\sqrt{3})\).	\(y=\pi/3\), \(f’= \sec^2 x\), at \(\pi/3\) gives 4, answer \(1/4\).

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：

\(\frac{d}{dx} \arcsin x = \frac{1}{\sqrt{1-x^2}}\)
\(\frac{d}{dx} \arccos x = -\frac{1}{\sqrt{1-x^2}}\)
\(\frac{d}{dx} \arctan x = \frac{1}{1+x^2}\)

For example, for \(y = \arcsin x\), we have \(x = \sin y\), so \(\frac{dx}{dy} = \cos y\). Then \(\frac{dy}{dx} = 1/\cos y\). Since \(\cos y = \sqrt{1-\sin^2 y} = \sqrt{1-x^2}\) (positive because \(y \in [-\pi/2, \pi/2]\)), the result follows.

例如，对于 \(y = \arcsin x\)，有 \(x = \sin y\)，所以 \(\frac{dx}{dy} = \cos y\)。于是 \(\frac{dy}{dx} = 1/\cos y\)。因为 \(\cos y = \sqrt{1-\sin^2 y} = \sqrt{1-x^2}\)（\(y \in [-\pi/2, \pi/2]\) 时取正），即得结果。

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。

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。

Published by TutorHao | IB Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

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